Worksheet on Area of the Path | Area of the Path Worksheet with Solutions

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1. A rectangular field is of dimensions 30 m × 15 m. Two paths run parallel to the sides of the rectangle through the centre of the field. The width of the longer path is 3 m and that of the shorter path is 2 m.

Find that:

(a) area of the paths

(b) area of the remaining portion of the field.

(c) cost of constructing the roads at the rate of $10 per m².

Solution:

Given that,

The dimensions of the rectangular field is 30 x 15

The width of the longer path is 3 m, the shorter path is 2 m.

The area of the rectangle = length x breadth

= 30 x 15 = 450

Area of the shorter path = 15 x 2 = 30 m

Area of the longer path = 30 x 3 = 90 m

Area of the middle path = 2 x 3 = 6

Area of paths = [Area of the shorter path] + [Area of the londer path] – [Area of the middle common path]

= 30 + 90 – 6

= 120 – 6 = 114

Area of the remaining portion of the field = The area of the rectangle – Area of path

= 450 – 114 = 336

Cost of constructing the roads at the rate of $10 per m²

The total cost of path = 114 x 10 = $1140

∴ The area of path is 114 m², The area of the remaining portion of the field is 336 m², Cost of constructing the roads at the rate of $10 per m² is $1140.


2. Find the area and perimeter of the following path.

Solution:

The perimeter of the path is the sum of all faces.

So, path perimeter = 4(20) + 2(4) + 2(8)

= 80 + 8 + 16

= 104

Area of ABLK = 20 x 4 = 80

Area of EFHG = 20 x 4 = 80

Area of CDIJ = 4 x 8 = 32

Area of the path = Area of ABLK + Area of EFGH + Area of CDIJ

= 80 + 80 + 32

= 192

∴ The area of path is 192 cm², the perimeter of the path is 104 cm.


3. A garden is 90 m long and 75 m board. A path 5 m wide is to be built outside all around it along its border. Find the area of the path?

Solution:

Given that,

The garden length is 90 m, width is 75 m.

Let us take ABCD as the garden.

Then, the area of path = Area of EFGH – Area of ABCD

Area of EFCGH = (90 + 5 + 5) x (75 + 5 + 5)

= 100 x 85 = 8500

Area of ABCD = 90 x 75

= 6750

Area of path = 8500 – 6750 = 1750

∴ The area of the path is 1750 m².


4. A grassy plot is 100 m x 60 m. Two cross paths each 5 m wide are constructed at right angles through the center of the field, such that each path is parallel to one of the sides of the rectangle. Find the total area used as a path.

Solution:

Area of the longer path (EFGH) = 100 x 5 = 500

Area of the shorter path (PQRS) = 5 x 60 = 300

Area of the centre field (IJKL) = 5 x 5 = 25

Area of the path = (Area of the longer path + Area of the shorter path – Area of the center field)

= 500 + 300 – 25

= 800 – 25 = 775 m².

∴ The total area used as path is 775 m².


5. A floor is 15 m long and 9 m wide. A square carpet of side 5 m is laid on the floor. Find the area of the floor not carpeted?

Solution:

Given that,

The length of the floor = 15 m

The breadth of the floor = 9 m

Side of the square = 5 m

Area of the square = side²

= 5² = 5 x 5

= 25 m²

Area of the floor = length x breadth

= 15 x 9 = 135

The area of floor not carpeted = Area of the floor – Area of carpet

= 135 – 25 = 110 m²

∴ The area of the floor not carpeted is 110 m².


6. Two crossroads, each of width 10 cm, run at right angles through the center of a rectangular garden of length 600 m and breadth 300 m and parallel to its sides.

Find:

(i) The area of the garden excluding crossroads.

(ii) The area of the crossroads

(iii) Convert these areas into hectares and ares.

Solution:

Given that,

(i) Length of the garden excluding crossroads = 600 – 10 = 590 m

Breadth of the garden excluding crossroads = 300 – 10 = 290

The area of the garden excluding crossroads = (length x breadth)

= 590 x 290 = 171100 m²

(ii) Length of the rectangular garden = 600 m

Breadth of the rectangular garden = 300 m

The total area of the rectangular garden = (length x breadth)

= (600 x 300) = 180000 m²

Area of the crossroads = The total area of the rectangular garden – The area of the garden excluding crossroads

= 180000 – 171100

= 8900 m²

(iii) We know that,

1 square metre = 0.0001

The area of the garden excluding crossroads = 171100 x 0.0001 = 17.11 hectares

Area of the crossroads = 8900 x 0.0001 = 0.89 hectares

1 square metre = 0.01 area

The area of the garden excluding crossroads = 171100 x 0.01 = 1711 ares

Area of the crossroads = 8900 x 0.01 = 89 ares


7. A square lawn is surrounded by a path 3 m wide. If the area of the path is 360 m². Find the area of the lawn.

Solution:

Given that,

Area of the path = 360 m²

Let the side of the lawn be x m.

Then, the area of the lawn = x²

Length of outer side = (x + 6) m

Area of outer square = (x + 6)²

= x² + 12x + 36

Area of the path = Area of the outer square – Area of the lower square

360 = x² + 12x + 36 – x²

360 = 12x + 36

12x = 360 – 36

12x = 324

x = 324/12

x = 27

Side of the lawn = 27 m

Area of the lawn = 27 x 27

= 729 m².

∴ Area of the lawn = 729 m²


8. A strip of width 3 m is cut out all round from a sheet of paper with dimensions 50 m × 30 m. Find the area of the strip cut out and the area of the remaining sheet.

Solution:

Given that,

Length of the sheet = 50 m

Breadth of the sheet = 30 m

Area of the sheet = 50 x 30 = 1500 m²

A strip of width 3 cm is cut out all round from a sheet of paper

Let that form a rectangle PQRS

Length of the rectangle PQRS = 50 – 3 – 3 = 50 – 6 = 44 m

Breadth of the rectangle = 30 – 3 – 3 = 30 – 6 = 24 m

Area of the rectangle PQRS = 44 x 24 = 1056 m²

Area of the strip cut out = Area of ABCD – Area of PQRS

= 1500 – 1056

= 444 m²

∴ The area of the strip cut out and the area of the remaining sheet is 444 m².


9. A path 5 m wide runs along inside a rectangular field. The length of the rectangular field is three times the breadth of the field. If the area of the path is 500 m², then find the length and breadth of the field.

Solution:

Let l be the length of the field, breadth of the field is b.

Breadth of the path = 5 m

The area of the path is 500 m².

2 x 5(l – 5) + 2 x 5(b – 5) = 500

10(l – 5) + 10(b – 5) = 500

10(l – 5 + b – 5) = 500l + b – 10 = 50

l + b = 50 + 10

l + b = 60 —- (i)

The length of the rectangular field is three times the breadth of the field

l = 3b —- (ii)

Substitute, l = 3b in equtaion (i)

3b + b = 60

4b = 60

b = 60/4

b = 15 m

Putting b = 15 in equation (i)

l + 15 = 60

l = 60 – 15

l = 45 m

∴ length and breadth of the field is 45 m, 15 m.


10. Four square flowerbeds each of the sides 4 m are dug on a piece of land 15 m long and 8 m wide. Find the area of the remaining portion of the land. Find the cost of leveling this land at a rate of $5 per 100 cm².

Solution:

Length of the land = 15 m

Breadth of the land = 8 m

Area of the land = 15 x 8

= 120 m²

Side of the square flowerbed = 4 m

Area of 1 square bed = side²

= 4² = 4 x 4 = 16

Area of 4 square bed = 16² = 16 x 16

= 256 m²

Area of the remaining portion = 256 – 16

= 240 m²

Cost of levelling 100 cm² = $5

Cost of levelling 0.0001 m² = $5

Cost of levelling of 240 m² = 240 x 50000 = 12000000

Hence, the area of the remaining portion of the land is 240 m² and the cost of leveling is $12000000.


Worksheet on Area and Perimeter of Squares | Questions on Area and Perimeter of a Square

Worksheet on Area and Perimeter of Squares contains different questions related to the square that helps the students to prepare for the exam. We have covered each and every model related to the square area, perimeter, and diagonal on this page. You can also get the step by step process to solve the Area and Perimeter Word Problems. Practice all the questions without fail and verify your answers from here.

Practicing questions from the Area and Perimeter of Squares Worksheet makes you feel comfortable at the exam. So that you can attempt all questions with 100% confidence and score better grades.

1. Find the perimeter and area of the following squares whose dimensions are:

(a) 8 m

(b) 6.8 cm

(c) 12 m

Solution:

(a)

Given that,

Square side length s = 8 m

Square perimeter p = 4s

p = 4 x 8 = 32 m

Square area A = s²

A = 8² = 8 x 8

A = 64 m²

∴ Square area is 64 m², perimeter is 32 m.

(b)

Given that,

Square side length s = 6.8 cm

Square perimeter p = 4s

p = 4 x 6.8

p = 27.2 cm

Square area A = s²

A = 6.8² = 6.8 x 6.8

A = 46.24 cm²

∴ The Square area is 46.24 cm², perimeter is 27.2 cm.

(c)

Given that,

Square side length s = 12 m

Square perimeter p = 4s

p = 4 x 12 = 48 m

Square area A = s²

A = 12²

A = 12 x 12 = 144 m²

∴ The Square area is 144 m², perimeter is 48 m.


2. Each side of a square is 2.6 cm. Find its perimeter, area?

Solution:

Given that,

Square side = 2.6 cm

Square perimeter = 4 x side length

P = 4 x 2.6

P = 10.4 cm

Square area = side²

= 2.6² = 2.6 x 2.6

= 6.76 cm²

∴ Square perimeter is 10.4 cm, area is 6.76 cm².


3. Find the perimeter of a square whose area is 120 m².

Solution:

Given that,

Square area A = 120 m²

Area A = Side²

120 = Side²

Side = √120

Side = 10.95

Square Perimeter P = 4side

P = 4 x 10.95

P = 43.817 m

∴ The perimeter of the square is 43.817 m.


4. Find the area of the square field whose perimeter is 240 m.

Solution:

The perimeter of the square p = 240 m

p = 4 x side

side = p / 4

side = 240 / 4

side = 60

Area of the square = side²

A = 60²

A = 60 x 60 = 3600

∴ Area of the square is 3600 m².


5. A rope of length of 104 m is used to fence a square garden. What is the length of the side of the garden?

Solution:

Given that,

The perimeter of the garden P = 104 m

We know that perimeter of a square = 4 × length of a side

So, 4 × length of a side = 104

The length of a side = 104/4

Side length = 26 m

∴ The length of the side of the garden is 26 m.


6. Lila has 16 square stamps of side 4 cm each. She glues them onto an envelope to form a bigger square. What area of the envelope does the bigger square cover?

Solution:

16 square-shaped stamps can be arranged as 4 in each row. So it forms 4 rows and 4 columns.

Side of the formed square s = 4 + 4 + 4 + 4

s = 16 cm

Area of the formed square A = 16 x 16

= 256

∴ The area of the bigger square is 256 cm².


7. If the diagonal length of a square is 7 cm, find the square area, perimeter?

Solution:

Given that,

The diagonal length of a square d = 7 cm

We know that, when you draw a diagonal in the square, it forms a right-angled triangle.

By using the Pythagorean theorem,

side² + side² = diagonal²

2side² = 7²

2side² = 49

side² = 49/2

side² = 24.5

side = √24.5

= 4.94 cm

The square perimeter p = 4 x side

p = 4 x 4.94

= 19.79 cm

The square area A = side²

A = 4.94² = 4.94 x 4.94

A = 24.4034 cm²

∴ The square area is 24.4034 cm², side length is 4.94 cm, and perimeter is 19.79 cm.


8. The diagonals of two squares are in the ratio 2:5. Find the ratio of their areas.

Solution:

Let us take the diagonals of two squares as 2x, 5x

Area of the square formula when diagonal is given,

A = (1/2) x d²

Area of the first square = (1/2) x (2x)²

= (1/2)(4x²)

= 2x²

Area of the second square = (1/2) x (5x)²

= (1/2) x (25x²)

= 12.5x²

The ratio of their areas = 2x² : 12.5x²

= 4 : 25

So, the ratio of the two squares is 4 : 25.


9. The areas of a square and rectangle are equal. If the side of the square is 15 cm and the breadth of the rectangle 10 cm, find the length of the rectangle and its perimeter.

Solution:

Given that,

Side of the square s = 15 cm

The breadth of the rectangle b = 10 cm

The areas of a square and rectangle are equal

Area of square = Area of Rectangle

s² = l x b

15² = l x 10

225 = l x 10

l = 225/10

l = 22.5

The perimeter of the rectangle p = 2(l + b)

= 2(22.5 + 10)

= 2(32.5) = 65

The perimeter of a square = 4 x side

= 4 x 15 = 60 cm

∴ The square, rectangle area is 225 cm², square perimeter is 60 cm, rectangle perimeter is 65 cm.


10. A wire is in the shape of a rectangle whose width is 12 cm is bent to form a square of side 17 cm. Find the rectangle length and also find which shape encloses more area.

Solution:

Given that,

Rectangle width w = 12 cm

Square side = 17 cm

The perimeter of a rectangle = Perimeter of a square

2(l + w) = 4side

2(l + 12) = 4 x 17

2l + 24 = 68

2l = 68 – 24

2l = 44

l = 44/2

l = 22

Area of square = side²

= 17² = 17 x 17

= 289

Area of the rectangle = l x b

= 22 x 12 = 264

∴ Rectangle length is 22 cm, and square has more area.


11. The area of a square field is 49 hectares. Find the cost of fencing the field with a wire at the rate of $5 per m.

Solution:

Given that,

Area of the square field = 49 hectares

1 hectare = 10000 sq.m.

So, 49 hectares = 49 x 10,000 = 4,90,000

So, the area of the square field = 4,90,000

Square area = side²

4,90,000 = side²

side = √(4,90,000)

side = 700

The perimeter of the square = 4 x side

P = 4 x 700 = 2800

Cost of fencing 1 m = $5

Cost of fencing 2800 m = 2800 x 5 = 14000

Hence The cost of fencing the square field is $14000.


12. How many square tiles of side 6 cm will be needed to fit in a square floor of a bathroom of side 600 cm. Find the cost of tiling at the rate of $65 per tile.

Solution:

Given that,

Side length of a tile = 6 cm

Side length of bathroom = 600 cm

Area of the square = 600 x 600 = 360000

Area of the tile = 6 x 6 = 36

Number of tiles = Area of the square / Area of the tile

= 360000 / 36 = 10000

Cost of 1 tile = $65

Cost of tiling = 65 x 10000 = 650000

∴ Number of square tiles required is 10000, cost of tiles is $650000.


13. If it costs $420 to fence a square field at the rate of $4 per m, find the length of the side and the area of the field.

Solution:

Given that,

The total cost of fencing = $420

The cost of fencing per m = $4

So, the Perimeter of the square field = 420/4

P = 105

Length of square = 105 / 4

Side = 26.25

Area of the square A = side²

A = (26.25)²

A = 26.25 x 26.25 = 689.0625

∴ The area of the field is 689.0625 m², side length is 26.25 m.


Worksheet on Area and Perimeter of Rectangles | Rectangle Area and Perimeter Worksheets with Answers

Refer to Worksheet on Area and Perimeter of Rectangles while preparing for the exam. You can perform well by practicing various questions from Rectangle Area and Perimeter Worksheets. Multiple models of questions along with the detailed solutions are given here. This Area and Perimeter of Rectangles Worksheet are designed as per the latest syllabus. So, students can learn the rectangles topic and score well in the examinations. Check out all the questions and rectangle area, rectangle perimeter formulas in the following sections of this page.

1. Find the perimeter and area of the following rectangles whose dimensions are:

(i) length = 11 cm, breadth = 5 cm

(ii) length = 6.7 cm, breadth = 5.1 cm

(iii) length = 14 m, breadth = 6 m

(iv) length = 8 feet, breadth = 3 feet

Solution:

(i)

Given that,

length = 11 cm, breadth = 5 cm

Rectangle Area = length x breadth

= 11 x 5 = 55 cm²

Rectangle Perimeter = 2(length + breadth)

= 2(11 + 5) = 16 x 2 cm

= 32 cm

∴ The Rectangle area is 55 cm², perimeter is 32 cm.

(ii)

Given that,

length = 6.7 cm, breadth = 5.1 cm

Rectangle Area = length x breadth

= 6.7 x 5.1 = 34.17 cm²

Rectangle Perimeter = 2(length + breadth)

= 2(6.7 + 5.1) = 2 x 11.8 cm

= 23.6

∴ The Rectangle area is 34.17 cm², perimeter is 23.6 cm.

(iii)

Given that,

length = 14 m, breadth = 6 m

Rectangle Area = length x breadth

= 14 x 6 = 84 m²

Rectangle Perimeter = 2(length + breadth)

= 2(14 + 6) = 20 x 2 = 40 m

∴ The Rectangle area is 84 m², perimeter is 40 m.

(iv)

Given that,

length = 8 feet, breadth = 3 feet

Rectangle Area = length x breadth

= 8 x 3 = 24 sq feet

Rectangle Perimeter = 2(length + breadth)

= 2(8 + 3) = 11 x 2 = 22 feet

∴ The Rectangle area is 24 sq feet, perimeter is 22 feet.


2. The area of a rectangle is 92 m², its length is 8 m. Find the rectangle breadth and perimeter?

Solution:

Given that,

Rectangle area = 92 m²

Rectangle length = 8 m

The rectangle area formula is

Area = length x breadth

So, breadth = area / length

Breadth = 92 / 8

= 11.5 m

Rectangle Perimeter = 2(Length + breadth)

= 2(8 + 11.5) = 2(19.5) = 39 m

∴ The Rectangle breadth is 11.5 m, perimeter is 39 m.


3. If the rectangle perimeter is 28 cm, its width is 18 cm. Find the rectangle length and area?

Solution:

Given that,

Rectangle Perimeter p = 28 cm

Width w = 18 cm

Rectangle perimeter p = 2(l + w)

28 = 2l + 36

2l = 28 – 36

2l = 8

l = 8/2

l = 4 cm

Rectangle area A = (l x w)

= 18 x 4 = 72 cm²

∴ The Rectangle length is 4 cm, area is 72 cm².


4. Find the cost of tiling a rectangular plot of land 250 m long and 500 m wide at the rate of $8 per hundred square m?

Solution:

Given that,

The rectangular plot length = 250 m

Rectangular plot width = 500 m

Area of rectangular polt = length x width

= 250 x 500 = 125000 sq. m

Cost of tiling = $8 per 100 sq. m = $8/100 per 1 sq. m

Cost of tiling of rectangular polt of 125000 sq. m = (8/100) x 125000 = 10,000

∴ The cost of tiling of the rectangular plot is Rs. 10,000/-.


5. A room is 4 feet long and 6 feet wide. How many square feet of carpet is needed to cover the floor of the room?

Solution:

Given that,

Rectangle length l = 4 feet

Rectangle width w = 6 feet

Area of the rectangle A = l x w

A = 4 x 6

A = 24 sq feet

So, 24 sq feet of carpet is needed to cover the room floor.


6. A table-top measures 5 m by 3 m 50 cm. What is the area and perimeter of the table?

Solution:

Given that,

Table length l = 5 m

width w = 3 m 50 cm

= 3 + 50 x (1/100)

= 3 + 1/2

= 7/2 = 3.5 m

Table Perimeter p = 2(l + w)

= 2(5 + 3.5) = 2(8.5) m

= 17 m

Table area A = l x w

A = 5 x 3.5

= 17.5 m²

∴ The table area is 17.5 m², the perimeter is 8.5 m.


7. A floor is 25 m long and 14 m wide. A square carpet of sides 8 m is laid on the floor. Find the area of the floor that is not carpeted and floor perimeter?

Solution:

Given that,

Rectangular floor-length l = 25 m

Rectangular floor breadth b = 14 m

Square carpet side s = 8 m

Rectangular floor area A = l x b

A = 25 x 14

A = 350 m²

Area of the square carpet a = s x s

a = 8 x 8

a = 64 m²

Area of the floor that is not carpeted = Rectangular floor area – Area of the square carpet

= A – a = 350 – 64

= 286 m²

Rectangular flooe perimeter P = 2(l + b)

P = 2(25 + 14)

P = 2(39) = 78 m

∴ Area of the floor that is not carpeted is 286 m², rectangular floor perimeter is 39 m.


8. How many tiles whose length and breadth are 15 cm and 6 cm respectively are needed to cover a rectangular region whose length and breadth are 510 cm and 135 cm?

Solution:

Given that,

Length of the tile l = 15 cm

Breadth of the tile b = 6 cm

Rectangular region length = 510

Rectangular region breadth = 135

Area of the tiles = l x b

= 15 x 6 = 90 cm²

Area of the plot = 510 x 135

= 68,850 cm²

Number of tiles required = Area of plot / Area of tiles

= 68850 / 90

= 765

Therefore, the required number of tiles are 765.


9. How many rectangles can be drawn with 22 cm as a perimeter? Also, find the dimensions of the rectangle whose area will be maximum?

Solution:

Given that,

The perimeter of the rectangle = 22 cm

2(l + w) = 22

(l + w) = 22/2

= 11

Possible dimensions of the rectangle are (1, 10), (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (10, 1), (9, 2), (8, 3), (7, 4).

∴ The dimensions of the rectangle whose area is maximum is (6, 5) or (5, 6) and 11 rectangles can be drawn.


10. The perimeter of a rectangular pool is 140 m, and its length is 60 m. Find the pool width and area?

Solution:

The rectangular pool perimeter p = 140 m

Rectangular pool length l = 60 m

Rectangular pool width w = ?

Rectangular pool perimeter p = 2(l + w)

140 = 2(60 + w)

140 = 120 + 2w

2w = 140 – 120

2w = 20

w = 20/2

w = 10

Rectangular pool area A = l x w

A = 60 x 10

A = 600

∴ The rectangular pool width is 10 m, area is 600 m².


11. The length of a rectangular wooden board is four times its width. If the width of the board is 155 cm, find the cost of framing it at the rate of $5 for 20 cm?

Solution:

Given that,

Width of a rectangular wooden board w = 155 cm

The cost of framing = $5 for 20 cm

The length of a rectangular wooden board l = 4w

l = 4 x 155

l = 620

Wooden board perimeter P = 2(l + w)

P = 2(620 + 155)

= 2(775) = 1550

The cost of framing = $5/20 for 1 cm

The cost of wooden board framing = 1550 x (5/20)

= 7750/20

= 387.5

∴ The cost of framing is $387.5.


Worksheet on Factoring Quadratic Trinomials | Factoring Trinomials Worksheet with Answers

The best resource to learn Factorization of Quadratic Trinomials Problems is Worksheet on Factoring Quadratic Trinomials. We have given different problems according to the updated syllabus on Factoring and Solving Quadratic Equations Worksheets. Solve all problems to get a complete grip on the Factorization of Quadratic Trinomials problems. Also, to learn complete factorization problems, check Factorization Worksheets, and improve your preparation level.

1. Factorize the quadratic trinomials

(i) a2 + 5a + 6
(ii) a2 + 10a + 24
(iii) a2 + 12a + 27
(iv) a2 + 15a + 56
(v) a2 + 19a + 60
(vi) a2 + 13a + 40
(vii) a2 – 10a + 24
(viii) a2 – 23a + 42
(ix) a2 – 17a + 16
(x) a2 – 21a + 90

Solution:

(i) The Given expression is a2 + 5a + 6.
By comparing the given expression a2 + 5a + 6 with the basic expression x^2 + ax + b.
Here, a = 1, b = 5, and c = 6.
The sum of two numbers is m + n = b = 5 = 3 + 2.
The product of two number is m * n = a * c = 1 * (6) = 6 = 3 * 2
From the above two instructions, we can write the values of two numbers m and n as 3 and 2.
Then, a2 + 5a + 6 = a2 + 3a + 2a + 6.
= a (a+ 3) + 2(a + 3).
Factor out the common terms.
(a + 3) (a + 2)

Then, a2 + 5a + 6  = (a + 3) (a + 2).

(ii) The Given expression is a2 + 10a + 24.
By comparing the given expression a2 + 5a + 6 with the basic expression x^2 + ax + b.
Here, a = 1, b = 10, and c = 24.
The sum of two numbers is m + n = b = 10 = 6 + 4.
The product of two number is m * n = a * c = 1 * (24) = 24 = 6 * 4
From the above two instructions, we can write the values of two numbers m and n as 6 and 4.
Then, a2 + 10a + 24 = a2 + 6a + 4a + 24.
= a (a+ 6) + 4(a + 6).
Factor out the common terms.
(a + 6) (a + 4)

Then, a2 + 10a + 24  = (a + 6) (a + 4).

(iii) The Given expression is a2 + 12a + 27.
By comparing the given expression a2 + 12a + 27 with the basic expression x^2 + ax + b.
Here, a = 1, b = 12, and c = 27.
The sum of two numbers is m + n = b = 12 = 9 + 3.
The product of two number is m * n = a * c = 1 * (27) = 27 = 9 * 3
From the above two instructions, we can write the values of two numbers m and n as 9 and 3.
Then, a2 + 12a + 27 = a2 + 9a + 3a + 27.
= a (a+ 9) + 3(a + 9).
Factor out the common terms.
(a + 9) (a + 3)

Then, a2 + 12a + 27  = (a + 9) (a + 3).

(iv) The Given expression is a2 + 15a + 56.
By comparing the given expression a2 + 15a + 56 with the basic expression x^2 + ax + b.
Here, a = 1, b = 15, and c = 56.
The sum of two numbers is m + n = b = 15 = 8 + 7.
The product of two number is m * n = a * c = 1 * (56) = 56 = 8 * 7
From the above two instructions, we can write the values of two numbers m and n as 8 and 7.
Then, a2 + 15a + 56 = a2 + 8a + 7a + 56.
= a (a+ 8) + 7(a + 8).
Factor out the common terms.
(a + 8) (a + 7)

Then, a2 + 15a + 56  = (a + 8) (a + 7).

(v) The Given expression is a2 + 19a + 60.
By comparing the given expression a2 + 19a + 60 with the basic expression x^2 + ax + b.
Here, a = 1, b = 19, and c = 60.
The sum of two numbers is m + n = b = 19 = 15 + 4.
The product of two number is m * n = a * c = 1 * (60) = 60 = 15 * 4
From the above two instructions, we can write the values of two numbers m and n as 15 and 4.
Then, a2 + 19a + 60 = a2 + 15a + 4a + 60.
= a (a+ 15) + 4(a + 15).
Factor out the common terms.
(a + 15) (a + 4)

Then, a2 + 19a + 60 = (a + 15) (a + 4).

(vi) The Given expression is a2 + 13a + 40.
By comparing the given expression a2 + 13a + 40 with the basic expression x^2 + ax + b.
Here, a = 1, b = 13, and c = 40.
The sum of two numbers is m + n = b = 13 = 8 + 5.
The product of two number is m * n = a * c = 1 * (40) = 40 = 8 * 5
From the above two instructions, we can write the values of two numbers m and n as 8 and 5.
Then, a2 + 13a + 40 = a2 + 8a + 5a + 40.
= a (a + 8) + 5(a + 8).
Factor out the common terms.
(a + 8) (a + 5)

Then, a2 + 13a + 40 = (a + 8) (a + 5).

(vii) The Given expression is a2 – 10a + 24.
By comparing the given expression a2 – 10a + 24 with the basic expression x^2 + ax + b.
Here, a = 1, b = -10, and c = 24.
The sum of two numbers is m + n = b = -10 = -6 – 4.
The product of two number is m * n = a * c = 1 * (24) = 24 = -6 * -4
From the above two instructions, we can write the values of two numbers m and n as -6 and -4.
Then, a2 – 10a + 24 = a2 – 6a – 4a + 24.
= a (a – 6) – 4(a – 6).
Factor out the common terms.
(a – 6) (a – 4)

Then, a2 – 10a + 24 = (a – 6) (a – 4).

(viii) The Given expression is a2 – 23a + 42.
By comparing the given expression a2 – 23a + 42 with the basic expression x^2 + ax + b.
Here, a = 1, b = -23, and c = 42.
The sum of two numbers is m + n = b = -23 = -21 – 2.
The product of two number is m * n = a * c = 1 * (42) = 42 = -21 * -2
From the above two instructions, we can write the values of two numbers m and n as -21 and -2.
Then, a2 – 23a + 42 = a2 – 21a – 2a + 42.
= a (a – 21) – 2(a – 21).
Factor out the common terms.
(a – 21) (a – 2)

Then, a2 – 23a + 42 = (a – 21) (a – 2).

(ix) The Given expression is a2 – 17a + 16.
By comparing the given expression a2 – 17a + 16 with the basic expression x^2 + ax + b.
Here, a = 1, b = -17, and c = 16.
The sum of two numbers is m + n = b = -17 = -16 – 1.
The product of two number is m * n = a * c = 1 * (16) = 16 = -16 * -1
From the above two instructions, we can write the values of two numbers m and n as -16 and -1.
Then, a2 – 17a + 16 = a2 – 16a -a + 16.
= a (a – 16) – 1(a – 16).
Factor out the common terms.
(a – 16) (a – 1)

Then, a2 – 17a + 16 = (a – 16) (a – 1).

(x) The Given expression is a2 – 21a + 90.
By comparing the given expression a2 – 21a + 90 with the basic expression x^2 + ax + b.
Here, a = 1, b = -21, and c = 90.
The sum of two numbers is m + n = b = -21 = -15 – 6.
The product of two number is m * n = a * c = 1 * (90) = 90 = -15 * -6
From the above two instructions, we can write the values of two numbers m and n as -15 and -6.
Then, a2 – 21a + 90 = a2 – 15a – 6a + 90.
= a (a – 15) – 6(a – 15).
Factor out the common terms.
(a – 15) (a – 6)

Then, a2 – 21a + 90 = (a – 15) (a – 6).


2. Factorize the expressions completely

(i) a2 – 22a + 117
(ii) a2 – 9a + 20
(iii) a2 + a – 132
(iv) a2 + 5a – 104
(v) b2 + 7b – 144
(vi) c2 + 19c – 150
(vii) b2 + b – 72
(viii) a2 + 6a – 91
(ix) a2 – 4a -77
(x) a2 – 6a – 135

Solution:

(i) The Given expression is a2 – 22a + 117.
By comparing the given expression a2 – 22a + 117 with the basic expression x^2 + ax + b.
Here, a = 1, b = -22, and c = 117.
The sum of two numbers is m + n = b = -22 = -13 – 9.
The product of two number is m * n = a * c = 1 * (117) = 117 = -13 * -9
From the above two instructions, we can write the values of two numbers m and n as -13 and -9.
Then, a2 – 22a + 117 = a2 – 13a – 9a + 117.
= a (a – 13) – 9(a – 13).
Factor out the common terms.
(a – 13) (a – 9)

Then, a2 – 22a + 117 = (a – 13) (a – 9).

(ii) The Given expression is a2 – 9a + 20.
By comparing the given expression a2 – 9a + 20 with the basic expression x^2 + ax + b.
Here, a = 1, b = -9, and c = 20.
The sum of two numbers is m + n = b = -9 = -5 – 4.
The product of two number is m * n = a * c = 1 * (20) = 20 = -5 * -4
From the above two instructions, we can write the values of two numbers m and n as -5 and -4.
Then, a2 – 9a + 20 = a2 – 5a -4a + 20.
= a (a – 5) – 4(a – 5).
Factor out the common terms.
(a – 5) (a – 4)

Then, a2 – 9a + 20 = (a – 5) (a – 4).

(iii) The Given expression is a2 + a – 132.
By comparing the given expression a2 + a – 132 with the basic expression x^2 + ax + b.
Here, a = 1, b = 1, and c = -132.
The sum of two numbers is m + n = b = 1 = 12 – 11.
The product of two number is m * n = a * c = 1 * (-132) = -132 = 12 * -11
From the above two instructions, we can write the values of two numbers m and n as 12 and -11.
Then, a2 + a – 132 = a2 + 12a – 11a – 132.
= a (a + 12) – 11(a + 12).
Factor out the common terms.
(a + 12) (a – 11)

Then, a2 + a – 132 = (a + 12) (a – 11).

(iv) The Given expression is a2 + 5a – 104.
By comparing the given expression a2 + 5a – 104 with the basic expression x^2 + ax + b.
Here, a = 1, b = 5, and c = -104.
The sum of two numbers is m + n = b = 5 = 13 – 8.
The product of two number is m * n = a * c = 1 * (-104) = -104 = 13 * -8
From the above two instructions, we can write the values of two numbers m and n as 13 and -8.
Then, a2 + 5a – 104 = a2 + 13a – 8a – 104.
= a (a + 13) – 8(a + 13).
Factor out the common terms.
(a + 13) (a – 8)

Then, a2 + 5a – 104 = (a + 13) (a – 8).

(v) The Given expression is b2 + 7b – 144.
By comparing the given expression b2 + 7b – 144 with the basic expression x^2 + ax + b.
Here, a = 1, b = 7, and c = -144.
The sum of two numbers is m + n = b = 7 = 16 – 9.
The product of two number is m * n = a * c = 1 * (-144) = -144 = 16 * -9
From the above two instructions, we can write the values of two numbers m and n as 16 and -9.
Then, b2 + 7b – 144 = b2 + 16b – 9b – 144.
= b (b + 16) – 9(b + 16).
Factor out the common terms.
(b + 16) (b – 9)

Then, b2 + 7b – 144 = (b + 16) (b – 9).

(vi) The Given expression is c2 + 19c – 150.
By comparing the given expression c2 + 19c – 150 with the basic expression x^2 + ax + b.
Here, a = 1, b = 19, and c = -150.
The sum of two numbers is m + n = b = 19 = 25 – 6.
The product of two number is m * n = a * c = 1 * (-150) = -150 = 25 * -6
From the above two instructions, we can write the values of two numbers m and n as 25 and -6.
Then, c2 + 19c – 150 = c2 + 25c – 6c – 150.
= c (c + 25) – 6(c + 25).
Factor out the common terms.
(c + 25) (c – 6)

Then, c2 + 19c – 150 = (c + 25) (c – 6).

(vii) The Given expression is b2 + b – 72.
By comparing the given expression b2 + b – 72 with the basic expression x^2 + ax + b.
Here, a = 1, b = 1, and c = -72.
The sum of two numbers is m + n = b = 1 =  9 – 8.
The product of two number is m * n = a * c = 1 * (-72) = -72 = 9 * -8
From the above two instructions, we can write the values of two numbers m and n as 9 and -8.
Then, b2 + b – 72 = b2 + 9b – 8b – 72.
= b (b + 9) – 8(b + 9).
Factor out the common terms.
(b + 9) (b – 8)

Then, b2 + b – 72 = (b + 9) (b – 8).

(viii) The Given expression is a2 + 6a – 91.
By comparing the given expression a2 + 6a – 91 with the basic expression x^2 + ax + b.
Here, a = 1, b = 6, and c = -91.
The sum of two numbers is m + n = b = 6 =  13 – 7.
The product of two number is m * n = a * c = 1 * (-91) = -91 = 13 * -7
From the above two instructions, we can write the values of two numbers m and n as 13 and -7.
Then, a2 + 6a – 91 = a2 + 13a – 7a – 91.
= a (a + 13) – 7(a + 13).
Factor out the common terms.
(a + 13) (a – 7)

Then, a2 + 6a – 91 = (a + 13) (a – 7).

(ix) The Given expression is a2 – 4a -77.
By comparing the given expression a2 – 4a -77 with the basic expression x^2 + ax + b.
Here, a = 1, b = -4, and c = -77.
The sum of two numbers is m + n = b = -4 =  -11 + 7.
The product of two number is m * n = a * c = 1 * (-77) = -77 = -11 * 7
From the above two instructions, we can write the values of two numbers m and n as -11 and 7.
Then, a2 – 4a -77 = a2 – 11a + 7a -77.
= a (a – 11) + 7(a – 11).
Factor out the common terms.
(a – 11) (a + 7)

Then, a2 – 4a -77 = (a – 11) (a + 7).

(x) The Given expression is a2 – 6a – 135.
By comparing the given expression a2 – 6a – 135 with the basic expression x^2 + ax + b.
Here, a = 1, b = -6, and c = -135.
The sum of two numbers is m + n = b = -6 =  -15 + 9.
The product of two number is m * n = a * c = 1 * (-135) = -135 = -15 * 9
From the above two instructions, we can write the values of two numbers m and n as -15 and 9.
Then, a2 – 6a – 135 = a2 – 15a + 9a – 135.
= a (a – 15) + 9(a – 15).
Factor out the common terms.
(a – 15) (a + 9)

Then, a2 – 6a – 135 = (a – 15) (a + 9).


3. Factor by splitting the middle term

(i) a2 – 11a – 42
(ii) a2 – 12a – 45
(iii) a2 – 7a – 30
(iv) a2 – 5a – 24
(v) 3a2 + 10a + 8
(vi) 3a2 + 14a + 8
(vii) 2a2 + a – 45
(viii) 6a2 + 11a – 10
(ix) 3a2 – 10a + 8
(x) 2a2 – 17a – 30

Solution:

(i) The Given expression is a2 – 11a – 42.
By comparing the given expression a2 – 11a – 42 with the basic expression x^2 + ax + b.
Here, a = 1, b = -11, and c = -42.
The sum of two numbers is m + n = b = -11 =  -14 + 3.
The product of two number is m * n = a * c = 1 * (-42) = -42 = -14 * 3
From the above two instructions, we can write the values of two numbers m and n as -14 and 3.
Then, a2 – 11a – 42 = a2 – 14a + 3a- 42.
= a (a – 14) + 3(a – 14).
Factor out the common terms.
(a – 14) (a + 3)

Then, a2 – 11a – 42 = (a – 14) (a + 3).

(ii) The Given expression is a2 – 12a – 45.
By comparing the given expression a2 – 12a – 45 with the basic expression x^2 + ax + b.
Here, a = 1, b = -12, and c = -45.
The sum of two numbers is m + n = b = -12 =  -15 + 3.
The product of two number is m * n = a * c = 1 * (-45) = -45 = -15 * 3
From the above two instructions, we can write the values of two numbers m and n as -15 and 3.
Then, a2 – 12a – 45 = a2 – 15a + 3a- 45.
= a (a – 15) + 3(a – 15).
Factor out the common terms.
(a – 15) (a + 3)

Then, a2 – 12a – 45 = (a – 15) (a + 3).

(iii) The Given expression is a2 – 7a – 30.
By comparing the given expression a2 – 7a – 30 with the basic expression x^2 + ax + b.
Here, a = 1, b = -7, and c = -30.
The sum of two numbers is m + n = b = -7 =  -10 + 3.
The product of two number is m * n = a * c = 1 * (-30) = -30 = -10 * 3
From the above two instructions, we can write the values of two numbers m and n as -10 and 3.
Then, a2 – 7a – 30 = a2 – 10a + 3a – 30.
= a (a – 10) + 3(a – 10).
Factor out the common terms.
(a – 10) (a + 3)

Then, a2 – 7a – 30 = (a – 10) (a + 3).

(iv) The Given expression is a2 – 5a – 24.
By comparing the given expression a2 – 5a – 24 with the basic expression x^2 + ax + b.
Here, a = 1, b = -5, and c = -24.
The sum of two numbers is m + n = b = -5 =  -8 + 3.
The product of two number is m * n = a * c = 1 * (-24) = -24 = -8 * 3
From the above two instructions, we can write the values of two numbers m and n as -8 and 3.
Then, a2 – 5a – 24 = a2 – 8a + 3a – 24.
= a (a – 8) + 3(a – 8).
Factor out the common terms.
(a – 8) (a + 3)

Then, a2 – 5a – 24 = (a – 8) (a + 3).

(v) The Given expression is 3a2 + 10a + 8.
By comparing the given expression 3a2 + 10a + 8 with the basic expression ax^2 + bx + c.
Here, a = 3, b = 10, and c = 8.
The sum of two numbers is m + n = b = 10 =  6 + 4.
The product of two number is m * n = a * c = 3 * (8) = 24 = 6 * 4
From the above two instructions, we can write the values of two numbers m and n as 6 and 4.
Then, 3a2 + 10a + 8 = 3a2 + 6a + 4a + 8.
= 3a (a + 2) + 4(a + 2).
Factor out the common terms.
(a + 2) (3a + 4)

Then, 3a2 + 10a + 8 = (a + 2) (3a + 4).

(vi) The Given expression is 3a2 + 14a + 8.
By comparing the given expression 3a2 + 14a + 8 with the basic expression ax^2 + bx + b.
Here, a = 3, b = 14, and c = 8.
The sum of two numbers is m + n = b = 14 = 12 + 2.
The product of two number is m * n = a * c = 3 * (8) = 24 = 8 * 3
From the above two instructions, we can write the values of two numbers m and n as 8 and 3.
Then, 3a2 + 14a + 8 = 3a2 + 8a + 3a + 8.
= a (3a + 8) + 1 (3a + 8).
Factor out the common terms.
(3a + 8) (a + 1)

Then, 3a2 + 14a + 8 = (3a + 8) (a + 1).

(vii) The Given expression is 2a2 + a – 45.
By comparing the given expression 2a2 + a – 45 with the basic expression ax^2 + bx + b.
Here, a = 2, b = 1, and c = -45.
The sum of two numbers is m + n = b = 1 = 10 – 9.
The product of two number is m * n = a * c = 2 * (-45) = -90 = 10 * -9
From the above two instructions, we can write the values of two numbers m and n as 10 and -9.
Then, 2a2 + a – 45 = 2a2 + 10a – 9a – 45.
= 2a (a + 5) – 9 (a + 5).
Factor out the common terms.
(a + 5) (2a – 9)

Then, 2a2 + a – 45 = (a + 5) (2a – 9).

(viii) The Given expression is 6a2 + 11a – 10.
By comparing the given expression 6a2 + 11a – 10 with the basic expression ax^2 + bx + b.
Here, a = 6, b = 11, and c = -10.
The sum of two numbers is m + n = b = 11 = 15 – 4.
The product of two number is m * n = a * c = 6 * (-10) = -60 = 15 * -4
From the above two instructions, we can write the values of two numbers m and n as 15 and -4.
Then, 6a2 + 11a – 10 = 6a2 + 15a – 4a – 10.
= 3a (2a + 5) – 2 (2a + 5).
Factor out the common terms.
(2a + 5) (3a – 2)

Then, 6a2 + 11a – 10 = (2a + 5) (3a – 2).

(ix) The Given expression is 3a2 – 10a + 8.
By comparing the given expression 3a2 – 10a + 8 with the basic expression ax^2 + bx + b.
Here, a = 3, b = -10, and c = 8.
The sum of two numbers is m + n = b = -10 = -6 – 4.
The product of two number is m * n = a * c = 3 * (8) = 24 = -6 * -4
From the above two instructions, we can write the values of two numbers m and n as -6 and -4.
Then, 3a2 – 10a + 8 = 3a2 – 6a – 4a + 8.
= 3a (a – 2) – 4 (a – 2).
Factor out the common terms.
(a – 2) (3a – 4)

Then, 3a2 – 10a + 8 = (a – 2) (3a – 4).

(x) The Given expression is 2a2 – 17a – 30.
By comparing the given expression 2a2 – 17a – 30 with the basic expression ax^2 + bx + b.
Here, a = 2, b = -17, and c = -30.
The sum of two numbers is m + n = b = -17 = -20 + 3.
The product of two number is m * n = a * c = 2 * (-30) = -60 = -20 * 3
From the above two instructions, we can write the values of two numbers m and n as -20 and 3.
Then, 2a2 – 17a – 30 = 2a2 – 20a + 3a – 30.
= 2a (a – 10) + 3 (a – 10).
Factor out the common terms.
(a – 10) (2a + 3)

Then, 2a2 – 17a – 30 = (a – 10) (2a + 3).


Worksheet on Problems on Simultaneous Linear Equations | Simultaneous Equations Problems with Solutions

Students who want to get complete knowledge on Simultaneous Linear Equations Word problems can check this Worksheet on Problems on Simultaneous Linear Equations. Our System of Linear Equations Word Problems Worksheet is helpful to improve your preparation levels. It contains a number of examples of Simultaneous Linear Equations. So, practice all the questions from the Simultaneous Linear Equations Worksheet and develop your skills. Also, have a look at Worksheet on Simultaneous Linear Equations and prepare well for the exam.

1. The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number.

Solution:

Let the two-digit number be xy.

According to the data provided in the question,

The sum of the digits of a two-digit number is 7.

x + y = 7 —– (i)

If the numbers are reversed, then the number is increased by 27.

The sm of two digits of the number as can be written as 10x + y

If the numbers are reversed, then the number can be written as 10y + x

10y + x = 10x + y + 27

10x + y + 27 – x – 10y = 0

9x – 9y = -27

9(x – y) = -27

x – y = -3 —– (ii)

Adding equations (i) & (ii)

x + y + x – y = 7 + (-3)

2x = 7 – 3

2x = 4

x = 4/2

x = 2

Substituting x = 2 in equation (i)

2 + y = 7

y = 7 – 2

y = 5

Therefore, the required 2 digit number is 25.


2. Mahesh bought 13 bushes and 4 trees from the nursery for the first time and its total cost was $487. For the second time, he bought 6 bushes and 2 trees and totaled $232. The bills do not list the per-item price. What were the costs of one bush and of one tree?

Solution:

Let the cost of each bush be x, one tree is y.

As per the question,

The first condition is he bought 13 bushes and 4 trees from the nursery for the first time and its total cost was $487

13x + 4y = 487 —- (i)

The second condition is he bought 6 bushes and 2 trees and totaled $232

6x + 2y = 232 —– (ii)

Multiply the second equation by 2.

2(6x + 2y = 232)

12x + 4y = 464 —– (iii)

Subtract equation (iii) from equation (i)

12x + 4y – (13x + 4y) = 464 – 487

12x + 4y – 13x – 4y = -23

-x = -23

x = 23

Putting x = 23 in equation (i)

13(23) + 4y = 487

299 + 4y = 487

4y = 487 – 299

4y = 188

y = 188/4

y = 47

So, costs of one bush is $23, and of one tree is $47.


3. The first number is six times the second number. The difference between the numbers is 60. Find those two numbers?

Solution:

Let the first number be x, the second number be y.

As per the given data,

The first condition is the first number is six times the second number.

x = 6y —- (i)

The second condition is the difference between the numbers is 60.

x – y = 60 —– (ii)

Put equation (i) in equation (ii)

6y – y = 60

5y = 60

y = 60/6

y = 10

Substituting y = 10 in equation (i)

x = 6 x 10

x = 60

Therefore, the two numbers are 60, 10.


4. One number is three times the other number. The sum of two numbers is 24. Find the two numbers?

Solution:

Assume that two numbers are x, y.

Given that,

One number is three times the other number.

x = 3y —– (i)

The sum of two numbers is 24.

x + y = 24 —– (ii)

Substituting x = 3y in equation (ii)

3y + y = 24

4y = 24

y = 24/4

y = 6

Putting y = 6 in equation (ii)

x + 6 = 24

x = 24 – 6

x = 18

so, the two numbers are 18, 6.


5. It takes 3 hours for a boat to travel 27 miles upstream. The same boat can travel 30 miles downstream in 2 hours. Find the speeds of the boat and the current?

Solution:

Let x be the speed of the boat (without current), let y be the speed of the current.

According to the given problem,

The boat can travel 27 miles upstream in 3 hours

Upstream speed = (Upstream distance)/(Upstream time)

Upstream speed = 27/3 = 9 miles/hour

The boat can travel 30 miles downstream in 2 hours.

Downstream speed = (Downstream distance) / (Downstream time)

Downstream speed = 30/2 = 15 miles/hour

The linear equation for upstream speed is x – y = 9 —— (i)

downstream speed is x + y = 15 ——- (ii)

Adding equations (i) and (ii)

x – y + x + y = 9 + 15

2x = 24

x = 24/2
x = 12

Substituting x = 12 in equation (ii)

12 + y = 15

y = 15 – 12

y = 3

Therefore, the speed of boat is 12 miles/hour, current is 3 miles/hour.


6. The sum of two numbers is 28 and their difference is 4. Find the two numbers?

solution

Let us take the two numbers as x, y.

As per the data given in the question,

The sum of two numbers is 28.

x + y = 28 —— (i)

The difference between the two numbers is 4.

x – y = 4 ——- (ii)

Adding equations (i) and (ii)

x + y + x – y = 28 + 4

2x = 32

x = 32/2

x = 16

Substitute x = 16 in equation (i)

16 + y = 28

y = 28 – 16

y = 12

So, the two numbers are 16, 12.


7. The sum of two numbers is 14. The difference in their squares is 28. Find the numbers?

Solution:

Assume that the two numbers are x and y.

According to the first condition in the question,

The sum of two numbers is 14.

x + y = 14 —— (i)

y = 14 – x

According to the second condition in the question,

x² – y² = 28 —- (ii)

Substitute y = 14 – x in equation (ii)

x² – (14 – x)² = 28

x² – (14² + x² – 28x) = 28

x² – 196 – x² + 28x = 28

28x – 196 = 28

28x = 28 + 196

28x = 224

x = 224/28

x = 4.6

Substitute x = 4.6 in equation (i)

4.6 + y = 14

y = 14 – 4.6

y = 9.4

Therefore, the two numbers are 4.6, 9.4.


8. The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended?

Solution:

Let us take the number of children who attended the fair as x, the number of adults who attended the fair as y.

The total number of people who attended the fair is 2200.

x + y = 2200 —- (i)

x = 2200 – y

The admission fee at a small fair is $1.50 for children and $4.00 for adults and the total amount collected is $5050.

1.5x + 4y = 5050 —— (ii)

Putting x = 2200 – y in the second equation.

1.5(2200 – y) + 4y = 5050

3300 – 1.5y + 4y = 5050

3300 + 2.5y = 5050

2.5y = 5050 – 3300

2.5y = 1750

y = 1750/2.5

y = 700

Substitute y = 700 in equation (i)

x + 700 = 2200

x = 2200 – 700

x = 1500

So, the number of children attended the small fair is 1500, the number of adults attended the fair is 700.


9. Emma has 23 notes of £20 and £5 in her handbag. The amount of money she has in the bag is £340.00. Find the number of notes of each type.

Solution:

Let us take the number of £20 notes as x, £5 notes as y.

The total number of notes in the handbag is 23.

x + y = 23 —- (i)

As per the second condition in the question,

20x + 5y = 340 —– (ii)

Multiply the equation (i) by 5.

5(x + y) = 23 x 5

5x + 5y = 115 —— (iii)

Subtracting equation (i) from equation (ii)

5x + 5y – (20x + 5y) = 115 – 340

5x + 5y – 20x – 5y = -225

-15x = -225

x = 225/15

x = 15

Putting x = 15 in equation (i)

15 + y = 23

y = 23 – 15

y = 8

So, Emma has 15 notes of £20, 8 notes of £5 in her handbag.


10. A family goes to the cinema. 4 adults and 2 child tickets cost $47, 1 adult and 3 child tickets cost $25.50. Calculate the costs of each adult and child ticket.

Solution:

Let us take the cost of the ticket for adults as x, cost of the ticket for the child as y.

According to the question,

4 adults and 2 child tickets cost $47, 1 adult and 3 child tickets cost $25.50.

4x + 2y = 47 —— (i)

x + 3y = 25.5 —– (ii)

x = 25.5 – 3y

Putting x = 25.5 – 3y in equation (i)

4(25.5 – 3y) + 2y = 47

102 – 12y + 2y = 47

102 – 10y = 47

102 – 47 = 10y

10y = 55

y = 55/10

y = 5.5

Substitute y = 5.5 in equation (ii)

x + 3(5.5) = 25.5

x + 16.5 = 25.5

x = 25.5 – 16.5

x = 9

Therefore, the ticket cost for adults is $9, the ticket cost for child is $5.5.


11. The area of a rectangle gets reduced by 10 square units if its length is reduced by 4 units and breadth is increase by 2 units. If we increased the length by 3 units and breadth by 4 units, the area is increased by 96 square units. Find the length and breadth of the rectangle.

Solution:

Let the length and breadth of the rectangle is l and b respectively.

The area of a rectangle is lb.

The area of a rectangle gets reduced by 10 square units if its length is reduced by 4 units and breadth is increased by 2 units.

(l – 4)(b + 2) = (lb – 10) sq units

lb + 2l – 4b – 8 = lb – 10

lb – lb – 2l + 4b = 10 – 8

4b – 2l = 2 —- (i)

If we increased the length by 3 units and breadth by 4 units, the area is increased by 96 square units.

(l + 3)(b + 4) = (lb + 96)

lb + 3b + 4l + 12 = lb + 96

lb + 3b + 4l – lb = 96 – 12

3b + 4l = 84 —— (ii)

Multiplying equation (i) by 2

4b – 2l = 2 x 2

8b – 4l = 4 —- (iii)

Adding equation (iii) in equation (ii)

8b – 4l + 3b + 4l = 84 + 4

11b = 88

b = 88/11

b = 8

Substituting b = 11 in equation (i)

4(11) – 2l = 2

44 – 2 = 2l

42 = 2l

l = 42/2

l = 21

So, the length and breadth of the rectangle is 21 units, 8 units.


12. If 5 is added to the numerator and 9 is subtracted from the denominator it becomes 21/12 and if 3, 8 is subtracted from the numerator and denominator it becomes 4/5. Find fractions.

Solution:

Let the numerator of the fraction be x, the denominator of the fraction be y.

As stated in the question,

If 5 is added to the numerator and 9 is subtracted from the denominator it becomes 21/12.

(x + 5)/(y – 9) = 21/12

Cross multiply the fractions.

12(x + 5) = 21(y – 9)

12x + 60 = 21y – 189

12x – 21y + 189 + 60 = 0

12x – 21y + 249 = 0 —- (ii)

If 3, 8 is subtracted from the numerator and denominator it becomes 4/5.

(x – 3)/(y – 8) = 4/5

Cross multiply the fractions.

5(x – 3) = 4(y – 8)

5x – 15 = 4y – 32

5x – 4y = -32 + 15

5x – 4y = -17

5x – 4y + 17 = 0

5x = 4y – 17

x = (4y – 17)/5 — (ii)

substitute equation (ii) value in equation (i)

12[(4y – 17)/5] – 21y + 249 = 0

[(48y – 204)/5] – 21y + 249 = 0

48y – 204 – 105y + 1245 = 0

-57y + 1041 = 0

57y = 1041

y = 1041/57

y = 18.26

Putting y = 18.26 in equation (ii)

x = (4(18.26) – 17)/5

x = (73.052 – 17)/5

x = 56.0526/5

x = 11.210

Therefore, the required fraction is 1121/1826.


13. Chole and Tino have a combined age of 48. Three years ago Chole was double the age Tino is now. Find the present ages of Chole and Tino.

Solution:

Let us take the age of chole and Tino as x, y.

As per the data given in the question,

The combined age of Chole and Tino is 48.

x + y = 48 —- (i)

Three years ago Chole was double the age Tino is now

x – 2y = 3

x = 3 + 2y —– (ii)

Putting x value in equation (i)

3 + 2y + y = 48

3 + 3y = 48

3y = 48 – 3

3y = 45

y = 45/3

y = 15

Substituting y = 15 in equation (ii)

x = 3 + 2(15)

x = 3 + 30

x = 33

So, the present age of Chole is 33 years, Tino is 15 years.


Worksheet on Simultaneous Linear Equations | Word Problems on System of Linear Equations

Worksheet on Simultaneous Linear Equations is available here. Students can get various methods to solve the system of linear equations from this page. One can learn how to solve two simultaneous equations in two variables using comparison, substitution, elimination, and cross-multiplication methods. You can even find different types of problems on linear equations in two variables. So, practice as many questions as possible for a better understanding of the concept.

This System of Linear Equations in Two Variables Worksheet page includes the questions with detailed step by step solution. Therefore, have a look at these problems and practice them to score good marks in the examination easily.

1. Use the comparison method to solve the following simultaneous linear equations:

(i) x + y = 1, x – 2y = 5

(ii) 6x + 7y = 5, 2x – 3y – 8 = 0

(iii) 9x – 6y = 12, 4x + 6y = 14

(iv) x – y = -1, 2y + 3x = 12

Solution:

(i) x + y = 1, x – 2y = 5

Given pair of the system of linear equations are

x + y = 1 ——– (1)

x – 2y = 5 ——- (2)

Express x in terms of y

From equation (1) x + y = 1, we get

x = 1 – y

From equation (2), we get

x = 5 + 2y

Equate the values of x obtained from both equations.

1 – y = 5 + 2y

2y + y = 1 – 5

3y = -4

y = -4/5

Substitute y = -4/5 in equation (1)

x – 4/5 = 1

x = 1 + 4/5

x = (5 + 4)/5 = 9/5

Therefore, the required solution is x = 9/5, y = -4/5.

(ii) 6x + 7y = 5, 2x – 3y – 8 = 0

Given simultaneous linear equations are

6x + 7y = 5 ——- (1)

2x – 3y – 8 = 0 ——– (2)

Express x in terms of y

From equation (1) 6x + 7y = 5, we get

6x = 5 – 7y

x = (5 – 7y)/6

From equation (2) 2x – 3y – 8 = 0, we get

2x = 3y + 8

x = (3y + 8)/2

Equate the values of x obtained from both equations.

(5 – 7y)/6 = (3y + 8)/2

2(5 – 7y) = 6(3y + 8)

10 – 14y = 18y + 48

10 – 48 = 18y + 14y

32y = -38

y = -38/32 = -19/16

Putting y = -19/16 in equation (2)

2x – 3(-19/16) – 8 = 0

2x + 57/16 – 8 = 0

2x + (57 – 128)/16 = 0

2x -71/16 = 0

2x = 71/16

x = 71/32

Therefore, the required solution set is x = 71/32, y = -19/16.

(iii) 9x – 6y = 12, 4x + 6y = 14

Given the system of linear equations are

9x – 6y = 12 —— (i)

4x + 6y = 14 —— (ii)

Express y in terms of x

From equation (i), we get

9x – 12 = 6y

y = (9x – 12)/6 —— (iii)

From equation (ii), we get

6y = (14 – 4x)

y = (14 – 4x)/6 ——- (iv)

Equation equation (iii) and equation (iv)

(9x – 12)/6 = (14 – 4x)/6

Cross multiply the fractions

6(9x – 12) = 6(14 – 4x)

9x – 12 = 14 – 4x

9x + 4x = 14 + 12

13x = 26

x = 26/13

x = 2

Putting x = 2 in equation (ii)

4(2) + 6y = 14

8 + 6y = 14

6y = 14 – 8

6y = 6

y = 6/6

y = 1

Therefore, the required solution set is x = 2, y = 1.

(iv) x – y = -1, 2y + 3x = 12

Given pair of linear equations are

x – y = -1 —— (i)

2y + 3x = 12 —— (ii)

Express y in terms of x

From equation (i), we get

x + 1 = y

From equation (ii), we get

2y = (12 – 3x)

y = (12 – 3x)/2

Equate the values of x obtained from both equations.

x + 1 = (12 – 3x)/2

2(x + 1) = (12 – 3x)

2x + 2 = 12 – 3x

2x + 3x = 12 – 2

5x = 10

x = 10/5

x = 2

Substituting x = 2 in equation (ii)

2y + 3(2) = 12

2y + 6 = 12

2y = 12 – 6

2y = 6

y = 6/2

y = 3

Therefore, the required solution is x = 2, y = 3.


2. Solve the following simultaneous equations by using the Elimination Method:

(i) 2x – y = 5, x + 3y – 9 = 0

(ii) 2x – 3y = 1, 3x – 4y = 1

(iii) (2x/3) + (y/2) = -1, (-x/3) + y = 3

Solution:

(i) 2x – y = 5, x + 3y – 9 = 0

Given the system of linear equations are

2x – y = 5 —– (i)

x + 3y = 9 —— (ii)

Multiply the equation (i) by 3.

3(2x – y) = 3(5)

6x – 3y = 15 —– (iii)

Add equation (ii) and equation (iii)

x + 3y + 6x – 3y = 9 + 15

7x = 24

x = 24/7

Substitute x = 24/7 in equation (ii)

24/7 + 3y = 9

3y = 9 – 24/7

3y = (63 – 24)/7

y = 39/21

Therefore, the required solution is x = 24/7, y = 39/21.

(ii) 2x – 3y = 1, 3x – 4y = 1

Given the system of linear equations are

2x – 3y = 1 —- (i)

3x – 4y = 1 —- (ii)

Multiply equation (i) by 3

3(2x – 3y) = 3(1)

6x – 9y = 3 —— (iii)

Multiply equation (ii) by 2

2(3x – 4y) = 2(1)

6x – 8y = 2 —– (iv)

Subtract equation (iii) from equation (iv)

(6x – 9y) – (6x – 8y) = 3 – 2

6x – 9y – 6x + 8y = 1

-y = 1

y = -1

Putting y = -1 in equation (iv)

6x – 8(-1) = 2

6x + 8 = 2

6x = 2 – 8

6x = -6

x = -1

Therefore, the required solution set is x = -1, y = -1.

(iii) (2x/3) + (y/2) = -1, (-x/3) + y = 3

Given pair of linear equations are

(2x/3) + (y/2) = -1

(4x + 3y)/6 = -1

(4x + 3y) = -6 —– (i)

(-x/3) + y = 3

(-x + 3y)/3 = 3

-x + 3y = 9 —– (ii)

Subtract equation (i) from equation (ii)

-x + 3y – (4x + 3y) = 9 – (-6)

-x + 3y – 4x – 3y = 9 + 6

-5x = 15

x = -15/5

x = -3

Put x = -3 in equation (ii)

-(-3) + 3y = 9

3 + 3y = 9

3y = 9 – 3

3y = 6

y = 6/3

y = 2

Therefore, the required solution is x = -3, y = 2.


3. Solve each other pair of the equation given below using the Substitution Method:

(i) x + y = 12, x – y = 2

(ii) 2x – 5y = 9, 4x – y = 9

(iii) x – y = 5, 2x – y = 11

Solution:

(i) x + y = 12, x – y = 2

Given pair of simultaneous linear equations are

x + y = 12 —— (i)

x – y = 2 —— (ii)

From equation (ii)

x = y + 2

Substitute the obtained x value in equation (i)

(y + 2) + y = 12

2y  + 2 = 12

2y = 12 – 2

2y = 10

y = 10/2

y = 5

Putting y = 5 in equation (ii)

x – 5 = 2

x = 2 + 5

x = 7

Therefore, the required solution is x = 7, y = 5.

(ii) 2x – 5y = 9, 4x – y = 9

Given pair of linear equations are

2x – 5y = 9 —– (i)

4x – y = 9 —– (ii)

From equation (ii), we can write

4x – 9 = y

Substituting the obtained y value in equation (i)

2x – 5(4x – 9) = 9

2x – 20x + 45 = 9

-18x = 9 – 45

-18x = -36

x = 36/18

x = 2

Putting x = 2 in equation (ii)

4(2) – y = 9

8 – 9 = y

y = -1

Therefore, the required solution is x = 2, y = -1.

(iii) x – y = 5, 2x – y = 11

Given simultaneous linear equations are

x – y = 5 —— (i)

2x – y = 11 —— (ii)

From equation (i), we can write as

x = 5 + y

Substituting the new x value in equation (ii)

2(5 + y) – y = 11

10 + 2y – y = 11

10 + y = 11

y = 11 – 10

y = 1

Substitute y = 1 in equation (i)

x – 1 = 5

x = 5 + 1

x = 6

Therefore, the solution is x = 6, y = 1.


4. Solve the below-mentioned simultaneous linear equations by the method of cross-multiplication:

(i) 2x + y = 4, x + y = 6

(ii) 4x – 3y = 7, 3x – y = 4

(iii) 3x + y = 8, 4x + 3y = 14

Solution:

(i) 2x + y = 4, x + y = 6

The transposition of the given simultaneous linear equations are

2x + y – 4 = 0 ——– (i)

x + y – 6 = 0 ———- (ii)

Multiply equation (i) by 1 and equation (ii) by 1, we get

1(2x + y – 4) = 1 x 0

2x + y – 4 = 0 ——- (iii)

1(x + y – 6) = 1x 0

x + y – 6 = 0 ——- (iv)

Subtract equation (iv) from equation (iii)

[2x + y – 4 = 0] – [x + y – 6 = 0]

2x + y – 4 – x – y + 6 = 0

x + 2 = 0

x = -2

Substitute x = -2 in equation (ii)

-2 + y – 6 = 0

y – 8 = 0

y = 8

Therefore, the required solution set is x = -2, y = 8.

(ii) 4x – 3y = 7, 3x – y = 4

The transposition of the given system of linear equations are

4x – 3y – 7 = 0 —— (i)

3x – y – 4 = 0 ——– (ii)

Multiply the equation (i) by -1 and equation (ii) by -3, we get

-1(4x – 3y – 7) = -1 x 0

-4x + 3y + 7 = 0 ——— (iii)

-3(3x – y – 4) = -3 x 0

-9x + 3y + 12 = 0 ——- (iv)

Subtract equation (iv) from equation (iii)

[-4x + 3y + 7] – [-9x + 3y + 12] = 0

-4x + 3y + 7 + 9x – 3y – 12 = 0

5x – 5 = 0

5x = 5

x = 5/5

x = 1

Putting x = 1 in equation (i)

4(1) – 3y – 7 = 0

4 – 3y – 7 = 0

-3y – 3 = 0

-3y = 3

y = -3/3

y = -1

Therefore, the required solution set is x = 1, y = -1.

(iii) 3x + y = 8, 4x + 3y = 14

The transposition of the given linear equations are

3x + y – 8 = 0 —– (i)

4x + 3y – 14 = 0 —— (ii)

Multiplying the equation (i) by 3, we get

3(3x + y – 8) = 0 x 3

9x + 3y – 24 = 0 —– (iii)

Multiplying the equation (ii) by 1, we get

1(4x + 3y – 14) = 1 x 0

4x + 3y – 14 = 0 —— (iv)

Subtracting equation (iv) from equation (iii)

[9x + 3y – 24 = 0] – [4x + 3y – 14 = 0]

(9x + 3y – 24) – (4x + 3y – 14) = 0

9x + 3y – 24 – 4x – 3y + 14 = 0

5x – 10 = 0

5x = 10

x = 10/5

x = 2

Putting x = 2 in equation (ii)

4(2) + 3y – 14 = 0

8 + 3y – 14 = 0

3y – 6 = 0

3y = 6

y = 6/3

y = 2

Therefore, the required solution set is x = 2, y = 2.


5. Solve the following simultaneous equations:

(i) 4/(x – 3) + 6/(y – 4) = 5, 5/(x – 3) – 3/(y – 4) = 1

(ii) (y/6) – (x/15) = 4, (y/3) – (x/12) = 19/4

Solution:

(i) 4/(x – 3) + 6/(y – 4) = 5, 5/(x – 3) – 3/(y – 4) = 1

Given simultaneous linear equations are

4/(x – 3) + 6/(y – 4) = 5 —– (i)

5/(x – 3) – 3/(y – 4) = 1

5/(x – 3) – 1 = 3/(y – 4) —— (ii)

Putting equation (ii) in equation (i)

4/(x – 3) + 2(5/(x – 3) – 1) = 5

4/(x – 3) + 10/(x – 3) – 2 = 5

14/(x – 3) = 5 + 2

14/(x – 3) = 7

2/(x – 3) = 1

2 = (x – 3)

x = 2 + 3

x = 5

Substituting x = 5 in equation (i)

4/(5 – 3) + 6/(y – 4) = 5

4/2 + 6/(y – 4) = 5

2 + 6/(y – 4) = 5

6/(y – 4) = 5 – 2

6/(y – 4) = 3

2/(y – 4) = 1

2 = y – 4

y = 2 + 4

y = 6

Therefore, the required solution set is x = 5, y = 6

(ii) (y/6) – (x/15) = 4, (y/3) – (x/12) = 19/4

Given linear equations are

(y/6) – (x/15) = 4

(15y – 6x) /90 = 4

15y – 6x = 90 x 4

15y – 6x = 360 ——- (i)

(y/3) – (x/12) = 19/4

(12y – 3x)/36 = 19/4

4(12y – 3x) = 19 x 36

(12y – 3x) = 171 —- (ii)

Express y in terms of x

12y = 171 + 3x

y = (171 + 3x) / 12

Substitute the obtained y value in equation (i)

15[(171 + 3x) / 12] – 6x = 360

3(5[(171 + 3x) / 12] – 2x) = 360

5[(171 + 3x) / 12] – 2x = 120

(855 + 15x)/12 – 2x = 120

855 + 15x – 24x = 120 x 12

855 – 9x = 1440

9x = 855 – 1440

9x = -585

x = -585/9

x = -65

Putting the value of x in equation (ii)

(12y – 3(-65)) = 171

12y + 195 = 171

12y = 171 – 195

12y = -24

y = -24/12

y = -2

Therefore, the required solution is x = -65, y = -2.


Worksheet on Factoring Trinomials by Substitution | Solving Trinomials Worksheets

For a clear understanding of the factorization trinomials by substitution, take the help of our worksheets. Follow the questions on Worksheet on Factoring Trinomials by Substitution for reference and grade up your skills level on the concept. Clear all your doubts on the factorization concept by following the below-solved examples of factoring trinomials by substitution.

We know the factorization process for x^2 + ax + b or ax^2 + bx + c. But, for the trinomial expression, we need to substitute the terms in the place of common factors. So that, we will get the expression in the form of x^2 + ax + b or ax^2 + bx + c. Check Factorization Worksheets to understand the complete factorization concept.

Solved Examples of Factoring Trinomials by Substitution

1. Factor the following trinomials using the substitution method

(i) 4(x – y)^2 – 14(x – y) – 8.
(ii) 3(3a + 2)^2 + 5(3a + 2) – 2.
(iii) 2(x + 2y)^2 + (x + 2y) – 1.
(iv) (x + y)^2 – (x + y) – 6.
(v) (a2 – 3a)^2 – 38(a^2 -3a) – 80.
(vi) 6(a – b)^2 – a + b – 15.
(vii) (x^2 – 3y^2)^2 – 16(x^2 – 3y^2) + 63.
(viii) (x^2 + 2x)^2 – 22(x^2 + 2x) + 72.
(ix) (a^2 – 8a)^2 – 29(a^2 – 8a) + 180.
(x) (p + q)^2 – 8p – 8q + 7.

Solution:

(i) The given expression is 4(x – y)^2 – 14(x – y) – 8.
Replace the (x – y) with p. That is 4p^2 – 14p – 8.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 4, b = – 14, c = – 8.
a*c = 4 * ( – 8) = – 32 and b = – 14.
So, – 32 = – 16 * 2 and – 14 = – 16 + 2.
Then, 4p^2 – 16 p + 2p – 8 = 4p(p – 4) + 2(p – 4).
Factor out the common terms from the above expression. That is,
(p – 4) (4p + 2).
Now, replace the term p with the (x – y). That is,
(x – y – 4) (4(x – y) + 2) = (x – y – 4) 2(2x – 2y + 1).

Then, 4(x – y)^2 – 14(x – y) – 8 is equal to 2(x – y – 4) (2x – 2y + 1).

(ii) The given expression is 3(3a + 2)^2 + 5(3a + 2) – 2.
Replace the (3a + 2) with p. That is,
3p^2 + 5p – 2.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 3, b = 5, c = – 2.
a*c = 3 * ( – 2) = – 6 and b = 5.
So, – 6 = 6 * (- 1) and 5= 6 – 1.
Then, 3p^2 + 5p – 2 = 3p^2 + 6p – p – 2.
= 3p(p + 2) – (p + 2).
Factor out the common terms from the above expression. That is,
(3p – 1)(p + 2).
Now, replace the term p with the (3a + 2). That is,
(3(3a + 2) – 1) (3a + 2 + 2) = (9a + 6 – 1) (3a + 4)
= (9a + 5) (3a + 4)

Then, 3(3a + 2)^2 + 5(3a + 2) – 2 is equal to (9a + 5) (3a + 4).

(iii) The given expression is 2(x + 2y)^2 + (x + 2y) – 1.
Replace the (x + 2y) with p. That is,
2p^2 + p – 1.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 2, b = 1, c = – 1.
a*c = 2 * ( – 1) = – 2 and b = 1.
So, – 2 = 2 * ( – 1) and 1 = 2 – 1.
Then, 2p^2 + p – 1 = 2p^2 + 2p – p – 1.
= 2p(p + 1) – (p + 1).
Factor out the common terms from the above expression. That is,
(2p – 1) (p + 1).
Now, replace the term p with the (x + 2y). That is,
(2p – 1) (p + 1) = (2(x + 2y) – 1) (x + 2y + 1).
= (2x + 4y – 1) (x + 2y + 1).

Then, 2(x + 2y)^2 + (x + 2y) – 1 is equal to (2x + 4y – 1) (x + 2y + 1).

(iv) The given expression is (x + y)^2 – (x + y) – 6.
Replace the (x + y) with p. That is,
p^2 – p – 6.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 1, c = – 6.
a*c = 1 * ( – 6) = – 6 and b = – 1.
So, – 6 = – 3 * 2 and – 1 = – 3 + 2.
Then, p^2 – p – 6 = p^2 – 3p + 2p – 6.
= p(p – 3) + 2(p – 3).
Factor out the common terms from the above expression. That is,
(p – 3) (p + 2).
Now, replace the term p with the (x + y). That is,
(p – 3) (p + 2) = (x + y – 3) (x + y + 2).

Then, (x + y)^2 – (x + y) – 6 is equal to (x + y – 3) (x + y + 2).

(v) The given expression is (a^2 – 3a)^2 – 38(a^2 -3a) – 80.
Replace the (a^2 – 3a) with p. That is,
p^2 – 38p – 80.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 38, c = – 80.
a*c = 1 * ( – 80) = – 80 and b = – 38.
So, – 80 = – 40 * 2 and – 38 = – 40 + 2.
Then, p^2 – 38p – 80 = p^2 – 40p + 2p – 80.
= p(p – 40) + 2(p – 40).
Factor out the common terms from the above expression. That is,
(p – 40) (p + 2).
Now, replace the term p with the (a^2 – 3a). That is,
(p – 40) (p + 2) = (a^2 – 3a – 40) ( a^2 – 3a + 2).

Then, (a2 – 3a)^2 – 38(a^2 -3a) – 80 is equal to (a^2 – 3a – 40) ( a^2 – 3a + 2).

(vi) The given expression is 6(a – b)^2 – a + b – 15.
We can write it as 6(a – b)^2 – (a – b) – 15.
Replace the (a – b) with p. That is,
6p^2 – p – 15.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 6, b = – 1, c = – 15.
a*c = 6 * ( – 15) = – 90 and b = – 1.
So, – 90 = – 10 * 9 and – 1 = – 10 + 9.
Then, 6p^2 – p – 15 = 6p^2 – 10p + 9p – 15.
= 2p(3p – 5) + 3(3p – 5).
Factor out the common terms from the above expression. That is,
(3p – 5) (2p + 3).
Now, replace the term p with the (a – b). That is,
(3p – 5) (2p + 3) = (3(a – b) – 5) (2(a – b) + 3).

Then, 6(a – b)^2 – a + b – 15 is equal to (3(a – b) – 5) (2(a – b) + 3).

(vii) The given expression is (x^2 – 3y^2)^2 – 16(x^2 – 3y^2) + 63.
Replace the (x^2 – 3y^2) with p. That is,
p^2 – 16p + 63.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 16, c = 63.
a*c = 1 * 63 = 63 and b = – 16.
So, 63 = – 7 * ( – 9) and – 16 = – 7 – 9.
Then, p^2 – 16p + 63 = p^2 – 9p – 7p + 63.
= p(p – 9) – 7(p – 9).
Factor out the common terms from the above expression. That is,
(p – 9) (p – 7).
Now, replace the term p with the (x^2 – 3y^2). That is,
(p – 9) (p – 7) = (x^2 – 3y^2 – 9) (x^2 – 3y^2 – 7).

Then, (x^2 – 3y^2)^2 – 16(x^2 – 3y^2) + 63 is equal to (x^2 – 3y^2 – 9) (x^2 – 3y^2 – 7).

(viii) The given expression is (x^2 + 2x)^2 – 22(x^2 + 2x) + 72.
Replace the (x^2 + 2x) with p. That is,
p^2 – 22p + 72.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 22, c = 72.
a*c = 1 * 72 = 72 and b = – 22.
So, 72 = – 18 * ( – 4) and – 22 = – 18 + ( – 4).
Then, p^2 – 22p + 72 = p^2 – 18p – 4p + 72.
= p(p – 18) – 4(p – 18).
Factor out the common terms from the above expression. That is,
(p – 18) (p – 4).
Now, replace the term p with the (x^2 + 2x). That is,
(p – 18) (p – 4) = (x^2 + 2x – 18) (x^2 + 2x – 4).

Then, (x^2 + 2x)^2 – 22(x^2 + 2x) + 72 is equal to (x^2 + 2x – 18) (x^2 + 2x – 4).

(ix) The given expression is (a^2 – 8a)^2 – 29(a^2 – 8a) + 180.
Replace the (a^2 – 8a) with p. That is,
p^2 – 29p + 180.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 29, c = 180.
a*c = 1 * 180 = 180 and b = – 29.
So, 180 = – 20 * (- 9) and – 29 = – 20 – 9.
Then, p^2 – 29p + 180 = p^2 – 9p – 20p + 180.
= p(p – 9) – 20(p – 9).
Factor out the common terms from the above expression. That is,
(p – 9) (p – 20).
Now, replace the term p with the (a^2 – 8a). That is,
(p – 9) (p – 20) = (a^2 – 8a – 9) (a^2 – 8a – 20).

Then, (a^2 – 8a)^2 – 29(a^2 – 8a) + 180 is equal to (a^2 – 8a – 9) (a^2 – 8a – 20).

(x) (p + q)2 – 8p – 8q + 7.
Solution: The given expression is
(p + q)2 – 8p – 8q + 7.
We can write it as (p + q)^2 – 8(p + q) + 7.
Replace the (p + q) with x. That is,
x^2 – 8x + 7.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 8, c = 7.
a*c = – 1 * (-7) = 7 and b = – 8.
So, 7 = – 7 * – 1and – 8 = – 7 – 1.
Then, x^2 – 8x + 7 = x^2 – 7x – x + 7.
= x(x – 7) – (x – 7).
Factor out the common terms from the above expression. That is,
(x – 7) (x – 1).
Now, replace the term x with the (p + q). That is,
(x – 7) (x – 1) = (p + q – 7) (p + q – 1).
Then, (p + q)2 – 8p – 8q + 7is equal to (p + q – 7) (p + q – 1).


2. Factor trinomials using substitution

(i) (a – 2b)^2 + 7(a – 2b) – 18
(ii) 3(x – y)^2 – (x – y) – 44
(iii) (5x – 3y)^2 + 8(5x – 3y) + 16
(iv) (a – 4b)^2 – 10(a – 4b) + 25
(v) (3x – 4)^2 – 4(3x – 4) – 12
(vi) (7x – 1)^2 + 12(7x – 1) – 45

Solution:

(i) The given expression is (a – 2b)^2 + 7(a – 2b) – 18.
Replace the (a – 2b) with x. That is,
x^2 + 7x – 18.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = 7, c = – 18.
a*c = 1 * (- 18) = – 18 and b = 7.
So, – 18 = 9 * ( -2) and 7 = 9 + ( – 2).
Then, x^2 + 7x – 18 = x^2 – 2x + 9x – 18.
= x(x – 2) + 9(x – 2).
Factor out the common terms from the above expression. That is,
(x – 2) (x + 9).
Now, replace the term x with the (a – 2b). That is,
(x – 2) (x + 9) = (a – 2b – 2) (a – 2b+ 9).

Then, (a – 2b)^2 + 7(a – 2b) – 18 is equal to (a – 2b – 2) (a – 2b + 9).

(ii) The given expression is 3(x – y)^2 – (x – y) – 44.
Replace the (x – y) with p. That is,
3p^2 – p – 44.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 3, b = – 1, c = – 44.
a*c = 3 * (- 44) = – 132 and b = – 1.
So, – 132 = – 12 * 11and – 1 = – 12 + 11.
Then, 3p^2 – p – 44 = 3p^2 – 12p + 11p – 44.
= 3p(p – 4) + 11(p – 4)
Factor out the common terms from the above expression. That is,
(p – 4) (3p + 11).
Now, replace the term p with the (x – y). That is,
(p – 4) (3p + 11) = (x – y – 4) (3(x – y) + 11).

Then, 3(x – y)^2 – (x – y) – 44 is equal to (x – y – 4) (3(x – y) + 11).

(iii) The given expression is (5x – 3y)^2 + 8(5x – 3y) + 16.
Replace the (5x – 3y) with p. That is,
p^2 + 8p + 16.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = 8, c = 16.
a*c = 1 * 16 = 16 and b = 8.
So, 16 = 4 * 4and 8 = 4 + 4.
Then, p^2 + 8p + 16 = p^2 + 4p + 4p + 16.
= p(p + 4) + 4(p + 4).
Factor out the common terms from the above expression. That is,
(p + 4) (p + 4).
Now, replace the term p with the (5x – 3y). That is,
(p +4) (p + 4) = (5x – 3y + 4) (5x – 3y + 4).

Then, (5x – 3y)^2 + 8(5x – 3y) + 16 is equal to (5x – 3y + 4) (5x – 3y + 4).

(iv) The given expression is (a – 4b)^2 – 10(a – 4b) + 25.
Replace the (a – 4b) with p. That is,
p^2 – 10p + 25.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 10, c = 25.
a*c = 1 * 25 = 25 and b = – 10.
So, 25 = – 5 * ( – 5) and – 10 = – 5 + ( – 5).
Then, p^2 – 10p + 25 = p^2 – 5p – 5p + 25.
= p(p – 5) – 5(p – 5).
Factor out the common terms from the above expression. That is,
(p – 5) (p – 5).
Now, replace the term p with the (a – 4b). That is,
(p – 5) (p – 5) = (a – 4b – 5) (a – 4b – 5).

Then, (a – 4b)^2 – 10(a – 4b) + 25 is equal to (a – 4b – 5) (a – 4b – 5).

(v) The given expression is (3x – 4)^2 – 4(3x – 4) – 12.
Replace the (3x – 4) with p. That is,
p^2 – 4p – 12.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 4, c = – 12.
a*c = 1 * ( – 12) = – 12 and b = – 4.
So, – 12 = – 6 * 2 and – 4 = – 6 + 2.
Then, p^2 – 4p – 12 = p^2 – 6p + 2p – 12.
= p(p – 6) + 2(p – 6).
Factor out the common terms from the above expression. That is,
(p – 6) (p + 2).
Now, replace the term p with the (3x – 4). That is,
(p – 6) (p + 2) = (3x – 4 – 6) (3x – 4 + 2).
= (3x – 10) (3x- 2).

Then, (3x – 4)^2 – 4(3x – 4) – 12 is equal to (3x – 10) (3x- 2).

(vi) The given expression is (p + q)^2 – 8p – 8q + 7.
Replace the (7x – 1) with p. That is,
p^2 + 12p – 45.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = 12, c = – 45.
a*c = 1 * ( – 45) = – 45 and b = 12.
So, – 45 = 15 * ( – 3) and 12 = 15 + ( – 3).
Then, p^2 + 12p – 45 = p^2 + 15p – 3p – 45.
= p(p + 15) – 3(p + 15).
Factor out the common terms from the above expression. That is,
(p + 15) (p – 3).
Now, replace the term p with the (7x – 1). That is,
(p + 15) (p – 3) = (7x – 1 + 15) (7x – 1 – 3).
= (7x + 14) (7x – 4).

Then, (p + q)^2 – 8p – 8q + 7 is equal to (7x + 14) (7x – 4).


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1. Factorize each of the following expression

(i) – 5a^2 + 5ab – 52a
(ii) x^2yz + xy^2z + xyz^2
(iii) – 4x^5 – 16x^3y – 20x^2y^2
(iv) x^3yz + 4xy^3 + 14x^3
(v) – a^2 + 3a + a – p – 2
(vi) x^2p + y^2p + z^2p – x^2q – y^2q – z^2q

Solution:

(i) The given expression is – 5a^2 + 5ab – 52a.
Factor out the common term from the above expression.That is,
a(- 5a + 5b – 52).

– 5a^2 + 5ab – 52a is equal to a(- 5a + 5b – 52).

(ii) The given expression is x^2yz + xy^2z + xyz^2
Factor out the common term from the above expression.That is,
xyz(x + y + z).

x^2yz + xy^2z + xyz^2 is equal to xyz(x + y + z).

(iii) The given expression is – 4x^5 – 16x^3y – 20x^2y^2.
Factor out the common term from the above expression. That is,
– 4x^2(x^3 + 4xy + 5y^2).

– 4x^5 – 16x^3y – 20x^2y^2 is equal to 4x^2(x^3 + 4xy + 5y^2).

(iv) The given expression is x^3yz + 4xy^3 + 14x^3
Factor out the common term from the above expression.That is,
x(x^2yz + 4y^2 + 14x^2).

x^3yz + 4xy^3 + 14x^3is equal to x(x^2yz + 4y^2 + 14x^2).

(v) The given expression is – a^2 + 3a + a – p – 2.
Factor out the common term from the above expression.That is,
a(-a + 3 + 1) – (p + 2).
(a – 1) (p – a + 2).

– a^2 + 3a + a – p – 2 is equal to (a – 1) (p – a + 2).

(vi) The given expression is x^2p + y^2p + z^2p – x^2q – y^2q – z^2q.
Factor out the common term from the above expression.That is,
x^2(p – q) + y^2(p – q) + z^2(p – q) = (p – q) (x^2 + y^2 + z^2).

x^2p + y^2p + z^2p – x^2q – y^2q – z^2q is equal to (p – q) (x^2 + y^2 + z^2).


2. Factorize using the formula of the difference of two squares

(i) 25a^2 – 36a^2b^2
(ii) 49x^2 – y^4
(iii) 81x^4 – y^2
(iv) a^3 – 25a.
(v) 32x^2y – 200y^3
(vi) 4x^2 + 12xy + 9y^2 – 9.

Solution:

(i) The given expression is 25a^2 – 36a^2b^2.
We can write it as (5a)^2 – (6ab)^2.
Factor out the common term from the above expression. That is,
a^2[(5)^2 – (6b)^2].
By comparing the above expression, it matches with the basic expression (a)^2 – (b)^2 = (a + b) (a – b).
So, a^2[(5)^2 – (6b)^2] = a^2(5 + 6b) (5 – 6b).

25a^2 – 36a^2b^2 is equal to a^2(5 + 6b) (5 – 6b).

(ii) The given expression is 49x^2 – y^4.
We can write it as (7x)^2 – (y^2)^2.
By comparing the above expression, it matches with the basic expression (a)^2 – (b)^2 = (a + b) (a – b).
So, (7x)^2 – (y^2)^2= (7x + y^2) (7x – y^2).

49x^2 – y^4 is equal to (7x + y^2) (7x – y^2).

(iii) The given expression is 81x^4 – y^2.
We can write it as (9x^2)^2 – (y)^2.
By comparing the above expression, it matches with the basic expression (a)^2 – (b)^2 = (a + b) (a – b).
So, (9x^2)^2 – (y)^2 = (9x^2 + y) (9x^2 – y).

81x^4 – y^2 is equal to (9x^2 + y) (9x^2 – y).

(iv) The given expression is a^3 – 25a
factor out the common terms from the above expression. That is,
a (a^2 – 25).
We can write it as a (a^2 – 5^2).
By comparing the above expression, it matches with the expression a^2 – b^2 = (a + b) (a – b).
So, a (a^2 – 5^2) = a (a + 5) (a – 5).

a^3 – 25a is equal to a(a + 5) (a – 5).

(v) The given expression is 32x^2y – 200y^3.
Factor out the common terms from the above expression. That is,
8y (4x^2 – 25y^2).
We can write it as 8y ((2x)^2 – (5y)^2).
By comparing the above expression, it matches the expression a^2 – b^2 = (a + b) (a – b).
So,8y ((2x)^2 – (5y)^2) = 8y (2x + 5y) (2x – 5y).

32x^2y – 200y^3 is equal to 8y(2x + 5y) (2x – 5y).

(vi) The given expression is 4x^2 + 12xy + 9y^2 – 9.
We can write it as (2x)^2 + 2(2x) (3y) + (3y)^2 – 9.
By comparing the above expression. It matches with the expression a^2 + 2ab + b^2 = (a + b)^2.
Then, (2x)^2 + 2(2x) (3y) + (3y)^2 – 9 = (2x + 3y)^2 – 9.
(2x + 3y)^2 – 9 = (2x + 3y)^2 – (3)^2
By comparing the above expression, it matches the expression a^2 – b^2 = (a + b) (a – b).
So, (2x + 3y)^2 – (3)^2 = (2x + 3y + 3) (2x + 3y – 3).

4x^2 + 12xy + 9y^2 – 9 is equal to (2x + 3y + 3) (2x + 3y – 3).


3. Find the values of

(i) (8)^2 – (6)^2
(ii) (27^2/3)^2 – (8 1/3)^2
(iii) (42)^2 – (14)^2
(iv) (10003)^2 – (9997)^2
(v) (9.2)^2 – (0.8)^2

Solution:

(i) The given expression is (8)^2 – (6)^2.
By comparing the above equation, it matches with the basic expression
(a)^2 – (b)^2 = (a + b) (a – b).
(8)^2 – (6)^2 = (8 + 6) (8 – 6).
= (14) (2) = 28.

(8)^2 – (6)^2 = 28.

(ii) The given expression is (27^2 / 3)^2 – (8 ^1/3)^2.
We can write it as
(3^3 * 2 / 3)^2 – (2^3 * 1 / 3)^2 = (3^2)^2 – (2)^2.
By comparing the above equation, it matches with the basic expression
(a)^2 – (b)^2 = (a + b) (a – b).
(9)^2 – (2)^2 = (9 + 2) (9 – 2) = (11) (7)= 77.

(27^2 / 3)^2 – (8 ^1/3)^2 is equal to 77

(iii) The given expression is (42)^2 – (14)^2.
By comparing the above expression with the basic expressions, it matches with the expression a^2 – b^2 = (a + b) (a – b).
(42)^2 – (14)^2 = (42 + 14) (42 – 14) = (56) (28) = 1568.

(42)^2 – (14)^2 is equal to 1568.

(iv) The given expression is (10003)^2 – (9997)^2.
By comparing the above expression with the basic expressions, it matches with the expression a^2 – b^2 = (a + b) (a – b).
(10003)^2 – (9997)^2 = (10003 + 9997) (10003 – 9997).
= (20,000) (6) = 1,20,000.

(10003)^2 – (9997)^2 is equal to 1,20,000.

(v) The given expression is (9.2)^2 – (0.8)^2.
By comparing the above expression with the basic expressions, it matches the expression a^2 – b^2 = (a + b) (a – b).
(9.2)^2 – (0.8)^2 = (9.2 + 0.8) (9.2 – 0.8) = (10) (8.4) = 84.

(9.2)^2 – (0.8)^2 = 84.


4. Factorization of trinomial

(i) 10x^2y^2 – 11xy + 3
(ii) 12a^2 + 11a – 5
(iii) 15x^2y^2 – 21xy + 6
(iv) a^4 – 10a^2 + 9

Solution:

(i) The given expression is 10x^2y^2 – 11xy + 3.
If xy = a, then 10a^2 – 11a + 3.
By comparing the above expression, it matches with the expression ax^2 + bx + c.
Here, a = 10, b = – 11, c = 3
a * c = 10 * 3 = 30, b = – 11.
30 = – 6 * ( – 5) and – 11 = – 6 + ( – 5).
So, 10a^2 – 11a + 3 = 10a^2 – 5a – 6a + 3.
10a^2 – 5a – 6a + 3 = 5a(2a – 1) – 3(2a – 1).
Factor out the common terms from the above expression. That is,
(2a – 1) (5a – 3).
Here, replace a with the xy, then
(2xy – 1) (5xy – 3).

10x^2y^2 – 11xy + 3 is equal to (2xy – 1) (5xy – 3).

(ii) The given expression is 12a^2 + 11a – 5.
By comparing the above expression, it matches with the expression ax^2 + bx + c.
Here, a = 12, b = 11, c = – 5.
a * c = 12 * (- 5) = – 60, b = 11.
-60 = 15 * ( – 4) and 11 = 15 + ( – 4).
So, 12a^2 + 11a – 5= 12a^2 – 4a + 15a – 5.
= 4a(3a – 1) + 5(3a – 1).
Factor out the common terms from the above expression. That is,
(3a – 1) (4a + 5).

12a^2 + 11a – 5 is equal to (3a – 1) (4a + 5).

(iii) The given expression is 15x^2y^2 – 21xy + 6.
If xy = a, then 15a^2 – 21a + 6.
By comparing the above expression, it matches with the expression ax^2 + bx + c.
Here, a = 15, b = – 21, c = 6.
a * c = 15 * 6 = 90 and b = – 21.
90 = – 15 * ( – 6) and – 21 = – 15 + (- 6).
15a^2 – 21a + 6 = 15a^2 – 15a – 6a + 6.
= 15a(a – 1) – 6(a – 1).
Factor out the common terms from the above expression. That is,
(a – 1) (15a – 6).
Replace a with the xy. That is,
(xy – 1) (15xy – 6).

15x^2y^2 – 21xy + 6 is equal to (xy – 1) (15xy – 6).

(iv) The given expression is a^4 – 10a^2 + 9.
We can write it as (a^2)^2 – 10a^2 + 9.
If a^2 = x, then x^2 – 10x + 9.
By comparing the above expression with the basic expressions, it matches with the ax^2 + bx + c.
Here, a = 1, b = – 10, c = 9.
a * c = 1 * 9 = 9 and b = – 10.
9 = – 9 * ( – 1) and – 10 = – 9 – 1.
x^2 – 9x – x + 9 = x(x – 9) – (x – 9).
Factoring out common terms from the above expression. That is,
(x – 9) (x – 1).
Replace the x with a^2. That is,
(a^2 – 9) (a^2 – 1).

a^4 – 10a^2 + 9 is equal to (a^2 – 9) (a^2 – 1).


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1. Factorize the following expression:

(i) a2 + 9a + 20
(ii) m2 + 15m + 54
(iii) y2 + 3y – 4
(iv) n2 + 2n – 24
(v) x2 – 5x + 4
(vi) a2 – 15a + 14

Solution:

(i) The Given expression is a2 + 9a + 20.
By comparing the given expression a2 + 9a + 20 with the basic expression x^2 + ax + b.
Here, a = 1, b = 9, and c = 20.
The sum of two numbers is m + n = b = 9 = 5 + 4.
The product of two number is m * n = a * c = 1 * (20) = 20 = 5 * 4
From the above two instructions, we can write the values of two numbers m and n as 5 and 4.
Then, a2 + 9a + 20 = a2 +5a + 4a + 20.
= a (a+ 5) + 4(a + 5).
Factor out the common terms.
(a + 5) (a + 4)

Then, a2 + 9a + 20 = (a + 5) (a + 4).

(ii) The Given expression is m2 + 15m + 54.
By comparing the given expression m2 + 15m + 54 with the basic expression x^2 + ax + b.
Here, a = 1, b = 15, and c = 54.
The sum of two numbers is m + n = b = 15 = 6 + 9.
The product of two number is m * n = a * c = 1 * (54) = 54 = 6 * 9
From the above two instructions, we can write the values of two numbers m and n as 6 and 9.
Then, m2 + 15m + 54 = m2 + 6m + 9m + 54.
= m (m + 6) + 9(m + 6).
Factor out the common terms.
(m + 6) (m + 9)

Then, m2 + 15m + 54 = (m + 6) (m + 9).

(iii) The Given expression is y2 + 3y – 4.
By comparing the given expression y2 + 3y – 4 with the basic expression x^2 + ax + b.
Here, a = 1, b = 3, and c = -4.
The sum of two numbers is m + n = b = 3 = 4 – 1.
The product of two number is m * n = a * c = 1 * (-4) = -4 = 1 * -4
From the above two instructions, we can write the values of two numbers m and n as 1 and -4.
Then, y2 + 3y – 4 = y2 + 4y -y – 4.
= y (y + 4) – 1(y + 4).
Factor out the common terms.
(y + 4) (y – 1)

Then, y2 + 3y – 4 = (y + 4) (y – 1).

(iv) The Given expression is n2 + 2n – 24.
By comparing the given expression n2 + 2n – 24 with the basic expression x^2 + ax + b.
Here, a = 1, b = 2, and c = -24.
The sum of two numbers is m + n = b = 2 = 6 – 4.
The product of two number is m * n = a * c = 1 * (-24) = -24 = 6 * -4
From the above two instructions, we can write the values of two numbers m and n as 6 and -4.
Then, n2 + 2n – 24 = n2 + 6n – 4n – 24.
= n (n + 6) – 4(n + 6).
Factor out the common terms.
(n + 6) (n – 4)

Then, n2 + 2n – 24 = (n + 6) (n – 4).

(v) The Given expression is x2 – 5x + 4.
By comparing the given expression x2 – 5x + 4 with the basic expression x^2 + ax + b.
Here, a = 1, b = -5, and c = 4.
The sum of two numbers is m + n = b = -5 = -1 – 4.
The product of two number is m * n = a * c = 1 * (4) = 4 = -1 * -4
From the above two instructions, we can write the values of two numbers m and n as -1 and -4.
Then, x2 – 5x + 4 = x2 – x – 4x + 4.
= x (x – 1) – 4(x – 1).
Factor out the common terms.
(x – 1) (x – 4)

Then, x2 – 5x + 4 = (x – 1) (x – 4).

(vi) The Given expression is a2 – 15a + 14.
By comparing the given expression a2 – 15a + 14 with the basic expression x^2 + ax + b.
Here, a = 1, b = -15, and c = 14.
The sum of two numbers is m + n = b = -15 = -1 – 14.
The product of two number is m * n = a * c = 1 * (14) = 14 = -1 * -14
From the above two instructions, we can write the values of two numbers m and n as -1 and -14.
Then, a2 – 15a + 14 = a2 – a – 14a + 14.
= a (a – 1) – 14(a – 1).
Factor out the common terms.
(a – 1) (a – 14)

Then, a2 – 15a + 14 = (a – 1) (a – 14).


2. Resolve into factors

(i) a2 + 3a – 10
(ii) m2 – 18m – 63
(iii) a2 + 6a + 8
(iv) a2 + 12a + 32
(v) x2 – 8x + 15
(vi) m2 – 12m + 35

Solution:

(i) The Given expression is a2 + 3a – 10.
By comparing the given expression a2 + 3a – 10 with the basic expression x^2 + ax + b.
Here, a = 1, b = 3, and c = -10.
The sum of two numbers is m + n = b = 3 = 5 – 2.
The product of two number is m * n = a * c = 1 * (-10) = -10 = 5 * -2
From the above two instructions, we can write the values of two numbers m and n as 5 and -2.
Then, a2 + 3a – 10 = a2 + 5a – 2a – 10.
= a (a + 5) – 2(a + 5).
Factor out the common terms.
(a + 5) (a – 2)

Then, a2 + 3a – 10 = (a + 5) (a – 2).

(ii) The Given expression is m2 – 18m – 63.
By comparing the given expression m2 – 18m – 63 with the basic expression x^2 + ax + b.
Here, a = 1, b = -18, and c = -63.
The sum of two numbers is m + n = b = -18 = 3 – 21.
The product of two number is m * n = a * c = 1 * (-63) = -63 = 3 * -21
From the above two instructions, we can write the values of two numbers m and n as 3 and -21.
Then, m2 – 18m – 63 = m2 + 3m – 21m – 63.
= m (m + 3) – 21(m + 3).
Factor out the common terms.
(m + 3) (m – 21)

Then, m2 – 18m – 63 = (m + 3) (m – 21).

(iii) The Given expression is a2 + 6a + 8.
By comparing the given expression a2 + 6a + 8 with the basic expression x^2 + ax + b.
Here, a = 1, b = 6, and c = 8.
The sum of two numbers is m + n = b = 6 = 2 + 4.
The product of two number is m * n = a * c = 1 * (8) = 8 = 2 * 4
From the above two instructions, we can write the values of two numbers m and n as 2 and 4.
Then, a2 + 6a + 8 = a2 + 2a + 4a + 8.
= a (a + 2) + 4(a + 2).
Factor out the common terms.
(a + 2) (a + 4)

Then, a2 + 6a + 8 = (a + 2) (a + 4).

(iv) The Given expression is a2 + 12a + 32.
By comparing the given expression a2 + 12a + 32 with the basic expression x^2 + ax + b.
Here, a = 1, b = 12, and c = 32.
The sum of two numbers is m + n = b = 12 = 8 + 4.
The product of two number is m * n = a * c = 1 * (32) = 32 = 8 * 4
From the above two instructions, we can write the values of two numbers m and n as 8 and 4.
Then, a2 + 12a + 32 = a2 + 8a + 4a + 32.
= a (a + 8) + 4(a + 8).
Factor out the common terms.
(a + 8) (a + 4)

Then, a2 + 12a + 32 = (a + 8) (a + 4).

(v) The Given expression is x2 – 8x + 15.
By comparing the given expression x2 – 8x + 15 with the basic expression x^2 + ax + b.
Here, a = 1, b = -8, and c = 15.
The sum of two numbers is m + n = b = -8 = – 3 – 5.
The product of two number is m * n = a * c = 1 * (15) = 15 = -3 * -5
From the above two instructions, we can write the values of two numbers m and n as -3 and -5.
Then, x2 – 8x + 15 = x2 – 3x – 5x + 15.
= x (x – 3) – 5(x – 3).
Factor out the common terms.
(x – 3) (x – 5)

Then, x2 – 8x + 15 = (x – 3) (x – 5).

(vi) The Given expression is m2 – 12m + 35.
By comparing the given expression m2 – 12m + 35 with the basic expression x^2 + ax + b.
Here, a = 1, b = -12, and c = 35.
The sum of two numbers is m + n = b = -12 = – 5 – 7.
The product of two number is m * n = a * c = 1 * (35) = 35 = -5 * -7
From the above two instructions, we can write the values of two numbers m and n as -5 and -7.
Then, m2 – 12m + 35 = m2 – 5m – 7m + 35.
= m (m – 5) – 7(m – 5).
Factor out the common terms.
(m – 7) (m – 5)

Then, m2 – 12m + 35 = (m – 7) (m – 5).


3. Factor the middle term

(i) m2 – 4m – 12
(ii) a2 – 4a – 45
(iii) x2 + 15x + 56
(iv) p2 – 13p + 36
(v) q2 + 5q – 24
(vi) r2 + 17r – 84
(vii) a2 – 15a + 44
(viii) m2 – 5m – 24
(ix) x2 – 4x – 77
(x) a2 – 12a + 20

Solution:

(i) The Given expression is m2 – 4m – 12.
By comparing the given expression m2 – 4m – 12 with the basic expression x^2 + ax + b.
Here, a = 1, b = -4, and c = -12.
The sum of two numbers is m + n = b = -4 = 2 – 6.
The product of two number is m * n = a * c = 1 * (-12) = -12 = 2 * -6
From the above two instructions, we can write the values of two numbers m and n as 2 and -6.
Then, m2 – 4m – 12 = m2 – 2m + 6m – 12.
= m (m – 2) + 6(m – 2).
Factor out the common terms.
(m – 2) (m + 6)

Then, m2 – 4m – 12 = (m – 2) (m + 6).

(ii) The Given expression is a2 – 4a – 45.
By comparing the given expression a2 – 4a – 45 with the basic expression x^2 + ax + b.
Here, a = 1, b = -4, and c = -45.
The sum of two numbers is m + n = b = -4 = 5 – 9.
The product of two number is m * n = a * c = 1 * (-45) = -45 = 5 * -9
From the above two instructions, we can write the values of two numbers m and n as 5 and -9.
Then, a2 – 4a – 45 = a2 + 5a – 9a – 45.
= a (a + 5) – 9(a + 5).
Factor out the common terms.
(a + 5) (a – 9)

Then, a2 – 4a – 45 = (a + 5) (a – 9).

(iii) The Given expression is x2 + 15x + 56.
By comparing the given expression x2 + 15x + 56 with the basic expression x^2 + ax + b.
Here, a = 1, b = 15, and c = 56.
The sum of two numbers is m + n = b = 15 = 8 + 7.
The product of two number is m * n = a * c = 1 * (56) = 56 = 8 * 7
From the above two instructions, we can write the values of two numbers m and n as 8 and 7.
Then, x2 + 15x + 56 = x2 + 8x + 7x + 56.
= x (x + 8) + 7(x + 8).
Factor out the common terms.
(x + 8) (x + 7)

Then, x2 + 15x + 56 = (x + 8) (x + 7).

(iv) The Given expression is p2 – 13p + 36.
By comparing the given expression p2 – 13p + 36 with the basic expression x^2 + ax + b.
Here, a = 1, b = -13, and c = 36.
The sum of two numbers is m + n = b = -13 = -9 – 4.
The product of two number is m * n = a * c = 1 * (36) = 36 = -9 * -4
From the above two instructions, we can write the values of two numbers m and n as -9 and -4.
Then, p2 – 13p + 36 = p2 – 9p – 4p + 36.
= p (p – 9) – 4(p – 9).
Factor out the common terms.
(p – 9) (p – 4)

Then, p2 – 13p + 36 = (p – 9) (p – 4).

(v) The Given expression is q2 + 5q – 24.
By comparing the given expression q2 + 5q – 24 with the basic expression x^2 + ax + b.
Here, a = 1, b = 5, and c = -24.
The sum of two numbers is m + n = b = 5 = -3 + 8.
The product of two number is m * n = a * c = 1 * (-24) = -24 = -3 * 8
From the above two instructions, we can write the values of two numbers m and n as -3 and 8.
Then, q2 + 5q – 24 = q2 – 3q + 8q – 24.
= q (q – 3) + 8(q – 3).
Factor out the common terms.
(q – 3) (q + 8)

Then, q2 + 5q – 24 = (q – 3) (q + 8).

(vi) The Given expression is r2 + 17r – 84.
By comparing the given expression r2 + 17r – 84 with the basic expression x^2 + ax + b.
Here, a = 1, b = 17, and c = -84.
The sum of two numbers is m + n = b = 17 = 21 – 4.
The product of two number is m * n = a * c = 1 * (-84) = -84 = 21 * -4
From the above two instructions, we can write the values of two numbers m and n as 21 and -4.
Then, r2 + 17r – 84 = r2 + 21r -4r – 84.
= r (r + 21) – 4(r + 21).
Factor out the common terms.
(r + 21) (r – 4)

Then, r2 + 17r – 84 = (r + 21) (r – 4).

(vii) The Given expression is a2 – 15a + 44.
By comparing the given expression a2 – 15a + 44 with the basic expression x^2 + ax + b.
Here, a = 1, b = -15, and c = 44.
The sum of two numbers is m + n = b = -15 = -11 – 4.
The product of two number is m * n = a * c = 1 * (44) = 44 = -11 * -4
From the above two instructions, we can write the values of two numbers m and n as -11 and -4.
Then, a2 – 15a + 44 = a2 – 11a – 4a + 44.
= a (a – 11) – 4(a – 11).
Factor out the common terms.
(a – 11) (a – 4)

Then, a2 – 15a + 44 = (a – 11) (a – 4).

(viii) The Given expression is m2 – 5m – 24.
By comparing the given expression m2 – 5m – 24 with the basic expression x^2 + ax + b.
Here, a = 1, b = -5, and c = -24.
The sum of two numbers is m + n = b = -5 = 3 – 8.
The product of two number is m * n = a * c = 1 * (-24) = -24 = 3 * -8
From the above two instructions, we can write the values of two numbers m and n as 3 and -8.
Then, m2 – 5m – 24 = m2 + 3m – 8m – 24.
= m (m + 3) – 8(m – 3).
Factor out the common terms.
(m – 3) (m – 8)

Then, m2 – 5m – 24 = (m – 3) (m – 8).

(ix) The Given expression is x2 – 4x – 77.
By comparing the given expression x2 – 4x – 77 with the basic expression x^2 + ax + b.
Here, a = 1, b = -4, and c = -77.
The sum of two numbers is m + n = b = -4 = -11 + 7.
The product of two number is m * n = a * c = 1 * (-77) = -77 = -11 * 7
From the above two instructions, we can write the values of two numbers m and n as -11 and 7.
Then, x2 – 4x – 77 = x2 – 11x +7x – 77.
= x (x – 11) + 7(x – 11).
Factor out the common terms.
(x – 11) (x + 7)

Then, x2 – 4x – 77 = (x – 11) (x + 7).

(x) The Given expression is a2 – 12a + 20.
By comparing the given expression a2 – 12a + 20 with the basic expression x^2 + ax + b.
Here, a = 1, b = -12, and c = 20.
The sum of two numbers is m + n = b = -12 = -10 – 2.
The product of two number is m * n = a * c = 1 * (20) = 20 = -10 * -2
From the above two instructions, we can write the values of two numbers m and n as -10 and -2.
Then, a2 – 12a + 20 = a2 – 10a – 2a + 20.
= a (a – 10) – 2(a – 10).
Factor out the common terms.
(a – 10) (a – 2)

Then, a2 – 12a + 20 = (a – 10) (a – 2).


Worksheet on Factorization using Formula | Factorization Worksheet with Solutions

Assess your preparation levels using the Worksheet on Factorization using Formula. To help you we have included different problems with a clear explanation here. Practice them on a regular basis and get the step by step solution listed here. We have covered all the topics related to the concept in Factorization using Formula Worksheet according to the new syllabus. You can always look up to our Factorization Worksheets to clear all your queries.

I. Worksheet on factorization using formula when a binomial is the difference of two squares

1. a2 – 36
2. 4x2 – 9
3. 81 – 49a2
4. 4a2 – 9b2
5. 16m2 – 225n2
6. 9a2b2 – 25
7. 16a2 – 1/144
8. (2x + 3y)2 – 16z2
9. 1 – (m – n)2
10. 9(a + b)2 – a2
11. 25(x + y)2 – 16(x – y)2
12. 20x2 – 45y2
13. a3 – 64a
14. 12a2 – 27
15. 3a5 – 48a3
16. 63x2y2 – 7
17. m2 – 2mn + n2 – r2
18. a2 – b2 – 2ab – 1
19. 9a2 – b2 + 4b – 4

Solution:

1. Given expression is a2 – 36
Rewrite the given expression in the form of a2 – b2.
(a)2 – (6)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a and b = 6
[a + 6] [a – 6]

The final answer is [a + 6] [a – 6]

2. Given expression is 4x2 – 9
Rewrite the given expression in the form of a2 – b2.
(2x)2 – (3)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 2x and b = 3
[2x + 3] [2x – 3]

The final answer is [2x + 3] [2x – 3]

3. Given expression is 81 – 49a2
Rewrite the given expression in the form of a2 – b2.
(9)2 – (7a)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 9 and b = 7a
[9 + 7a] [9 – 7a]

The final answer is [9 + 7a] [9 – 7a]

4. Given expression is 4a2 – 9b2
Rewrite the given expression in the form of a2 – b2.
(2a)2 – (3b)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 2a and b = 3b
[2a + 3b] [2a – 3b]

The final answer is [2a + 3b] [2a – 3b]

5. Given expression is 16m2 – 225n2
Rewrite the given expression in the form of a2 – b2.
(4m)2 – (15n)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 4m and b = 15n
[4m + 15n] [4m – 15n]

The final answer is [4m + 15n] [4m – 15n]

6. Given expression is 9a2b2 – 25
Rewrite the given expression in the form of a2 – b2.
(3ab)2 – (5)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 3ab and b = 5
[3ab + 5] [3ab – 5]

The final answer is [3ab + 5] [3ab – 5]

7. Given expression is 16a2 – 1/144
Rewrite the given expression in the form of a2 – b2.
(4a)2 – (1/12)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 4a and b = 1/12
[4a + 1/12] [4a – 1/12]

The final answer is [4a + 1/12] [4a – 1/12]

8. Given expression is (2x + 3y)2 – 16z2
Rewrite the given expression in the form of a2 – b2.
(2x + 3y)2 – (4z)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 2x + 3y and b = 4z
[2x + 3y + 4z] [2x + 3y – 4z]

The final answer is [2x + 3y + 4z] [2x + 3y – 4z]

9. Given expression is 1 – (m – n)2
Rewrite the given expression in the form of a2 – b2.
(1)2 – (m – n)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 1 and b = m – n
[1 + m – n] [1 – (m – n)]
[1 + m – n] [1 – m + n]

The final answer is [1 + m – n] [1 – m + n]

10. Given expression is 9(a + b)2 – a2
Rewrite the given expression in the form of a2 – b2.
(3(a + b))2 – (a)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 3(a + b) and b = a
[3a + 3b + a] [3a + 3b – a]
[4a + 3b] [2a + 3b]

The final answer is [4a + 3b] [2a + 3b]

11. Given expression is 25(x + y)2 – 16(x – y)2
Rewrite the given expression in the form of a2 – b2.
(5(x + y))2 – (4(x – y))2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 5(x + y) and b = 4(x – y)
[5x + 5y + 4x – 4y] [5x + 5y – 4x + 4y]
[9x + y] [x + 9y]

The final answer is [9x + y] [x + 9y]

12. Given expression is 20x2 – 45y2
Rewrite the given expression in the form of a2 – b2.
5{(2x)2 – (3y)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 2x and b = 3y
5{[2x + 3y] [2x – 3y]}

The final answer is 5{[2x + 3y] [2x – 3y]}

13. Given expression is a3 – 64a
Rewrite the given expression in the form of a2 – b2.
a{(a)2 – (8)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a and b = 8
a{[a + 8] [a – 8]}

The final answer is a{[a + 8] [a – 8]}

14. Given expression is 12a2 – 27
Rewrite the given expression in the form of a2 – b2.
3{(2a)2 – (3)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 2a and b = 3
3{[2a + 3] [2a – 3]}

The final answer is 3{[2a + 3] [2a – 3]}

15. Given expression is 3a5 – 48a3
Rewrite the given expression in the form of a2 – b2.
3a3{(a)2 – (4)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a and b = 4
3a3{[a + 4] [a – 4]}

The final answer is 3a3{[a + 4] [a – 4]}

16. Given expression is 63x2y2 – 7
Rewrite the given expression in the form of a2 – b2.
7{(3xy)2 – (1)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 3xy and b = 1
7{[3xy + 1] [3xy – 1]}

The final answer is 7{[3xy + 1] [3xy – 1]}

17. Given expression is m2 – 2mn + n2 – r2
Rewrite the given expression in the form of a2 – b2.
m2 – 2mn + n2 is in the form of a2 – 2ab + b2
{(m – n)2 – (r)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = m – n and b = r
{[m – n + r] [m – n – r]}

The final answer is 7{[3xy + 1] [3xy – 1]}

18. Given expression is a2 – b2 – 2ab – 1
Rewrite the given expression in the form of a2 – b2.
a2 – b2 – 2ab is in the form of a2 – 2ab + b2
{(a – b)2 – (1)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a – b and b = 1
{[a – b + 1] [a – b – 1]}

The final answer is {[a – b + 1] [a – b – 1]}

19. Given expression is 9a2 – b2 + 4b – 4
9a2 – b2 + 4b – 4 = 9a2 – (b2 – 4b + 22)
Rewrite the given expression in the form of a2 – b2.
b2 – 4b + 22 is in the form of a2 – 2ab + b2
{(3a)2 – (b – 2)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 3a and b = b – 2
{[3a + b – 2] [3a – b + 2]}

The final answer is {[3a + b – 2] [3a – b + 2]}


II. Solved Problems on Factorization Using Formula

1. a2 – 2ab + b2 – c2
2. 25 – x2 – y2 -2xy
3. 16b3 – 4b
4. 3a5 – 48a
5. (3a – 4b)2 – 25c2
6. (2m + 3n)2 – 1
7. 16z2 – (5x + y)2
8. 100 – (a – 5)2
9. Evaluate:
(i) (13)2 – (12)2
(ii) (6.3)2 – (4.2)2

Solution:

1. Given expression is a2 – 2ab + b2 – c2
Rewrite the given expression in the form of a2 – b2.
a2 – b2 – 2ab is in the form of a2 – 2ab + b2
{(a – b)2 – (c)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a – b and b = c
{[a – b + c] [a – b – c]}

The final answer is {[a – b + c] [a – b – c]}

2. Given expression is 25 – x2 – y2 -2xy
Rewrite the given expression in the form of a2 – b2.
25 – x2 – y2 -2xy = 25 – (x2 + y2 + 2xy)
x2 + y2 + 2xy is in the form of a2 + 2ab + b2
{(5)2 – (x + y)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 5 and b = x + y
{[5 + x + y] [5 – x – y]}

The final answer is {[5 + x + y] [5 – x – y]}

3. Given expression is 16b3 – 4b
Rewrite the given expression in the form of a2 – b2.
b{(4b)2 – (2)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 4b and b = 2
b{[4b + 2] [4b – 2]}

The final answer is b{[4b + 2] [4b – 2]}

4. Given expression is 3a5 – 48a
Rewrite the given expression in the form of a2 – b2.
3a{(a2)2 – (22)2}
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a2 and b = 22
3a{[a2 + 22] [a2 – 22]}
From the above equation, [a2 – 22] is in the form of a2 – b2.
[(a)2 – (2)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a and b = 2
[a + 2] [a – 2]
Now, 3a{[a2 + 22] [a2 – 22]}
3a{[a2 + 4] [a + 2] [a – 2]}

The final answer is 3a{[a2 + 4] [a + 2] [a – 2]}

5. Given expression is (3a – 4b)2 – 25c2
Rewrite the given expression in the form of a2 – b2.
(3a – 4b)2 – (5c)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 3a – 4b and b = 5c
{[3a – 4b + 5c] [3a – 4b – 5c]}

The final answer is {[3a – 4b + 5c] [3a – 4b – 5c]}

6. Given expression is (2m + 3n)2 – 1
Rewrite the given expression in the form of a2 – b2.
(2m + 3n)2 – (1)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 2m + 3n and b = 1
{[2m + 3n + 1] [2m + 3n – 1]}

The final answer is {[2m + 3n + 1] [2m + 3n – 1]}

7. Given expression is 16z2 – (5x + y)2
Rewrite the given expression in the form of a2 – b2.
(4z)2 – (5x + y)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 4z and b = 5x + y
{[4z + 5x + y] [4z – (5x + y)]}
{[4z + 5x + y] [4z – 5x – y]}

The final answer is {[4z + 5x + y] [4z – 5x – y]}

8. Given expression is 100 – (a – 5)2
Rewrite the given expression in the form of a2 – b2.
(10)2 – (a – 5)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 10 and b = a – 5
{[10 + a – 5] [10 – (a – 5)]}
{[10 + a – 5] [10 – a + 5]}

The final answer is {[10 + a – 5] [10 – a + 5]}

9. (i) Given expression is (13)2 – (12)2
Rewrite the given expression in the form of a2 – b2.
(13)2 – (12)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 13 and b = 12
{[13 + 12] [13 – 12)]}
{[25] [1]} = 25

The final answer is 25

9. (ii) Given expression is (6.3)2 – (4.2)2
Rewrite the given expression in the form of a2 – b2.
(6.3)2 – (4.2)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 6.3 and b = 4.2
{[6.3 + 4.2] [6.3 – 4.2)]}
{[10.5] [2.1]} = 22.05

The final answer is 22.05


III. Worksheet on factorization using formula when the given expression is a perfect square

1. a2 + 8a + 16
2. a2 + 14a + 49
3. 1 + 2a + a2
4. 9 + 6c + c2
5. m2 + 6am + 9a2
6. 4a2 + 20a +25
7. 36a2 + 36a + 9
8. 9a2 + 24a + 16
9. a2 + a + 1/4
10. a2 – 6a + 9

Solution:

1. Given expression is a2 + 8a + 16
The given expression a2 + 8a + 16 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = a, b = 4
Apply the formula and substitute the a and b values.
a2 + 8a + 16
(a)2 + 2 (a) (4) + (4)2
(a + 4)2
(a + 4) (a + 4)

Factors of the a2 + 8a + 16 are (a + 4) (a + 4)

2. Given expression is a2 + 14a + 49
The given expression a2 + 14a + 49 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = a, b = 7
Apply the formula and substitute the a and b values.
a2 + 14a + 49
(a)2 + 2 (a) (7) + (7)2
(a + 7)2
(a + 7) (a + 7)

Factors of the a2 + 14a + 49 are (a + 7) (a + 7)

3. Given expression is 1 + 2a + a2
The given expression a2 + 2a + 1 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = a, b = 1
Apply the formula and substitute the a and b values.
a2 + 2a + 1
(a)2 + 2 (a) (1) + (1)2
(a + 1)2
(a + 1) (a + 1)

Factors of the 1 + 2a + a2 are (a + 1) (a + 1)

4. Given expression is 9 + 6c + c2
The given expression c2 + 6c + 9 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = c, b = 3
Apply the formula and substitute the a and b values.
c2 + 6c + 9
(c)2 + 2 (a) (3) + (3)2
(c + 3)2
(c + 3) (c + 3)

Factors of the 9 + 6c + c2 are (c + 3) (c + 3)

5. Given expression is m2 + 6am + 9a2
The given expression m2 + 6am + 9a2 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = m, b = 3a
Apply the formula and substitute the a and b values.
m2 + 6am + 9a2
(m)2 + 2 (m) (3a) + (3a)2
(m + 3a)2
(m + 3a) (m + 3a)

Factors of the m2 + 6am + 9a2 are (m + 3a) (m + 3a)

6. Given expression is 4a2 + 20a +25
The given expression 4a2 + 20a +25 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = 2a, b = 5
Apply the formula and substitute the a and b values.
4a2 + 20a +25
(2a)2 + 2 (2a) (5) + (5)2
(2a + 5)2
(2a + 5) (2a + 5)

Factors of the 4a2 + 20a +25 are (2a + 5) (2a + 5)

7. Given expression is 36a2 + 36a + 9
The given expression 36a2 + 36a + 9 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = 6a, b = 3
Apply the formula and substitute the a and b values.
36a2 + 36a + 9
(6a)2 + 2 (6a) (3) + (3)2
(6a + 3)2
(6a + 3) (6a + 3)

Factors of the 36a2 + 36a + 9 are (6a + 3) (6a + 3)

8. Given expression is 9a2 + 24a + 16
The given expression 9a2 + 24a + 16 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = 3a, b = 4
Apply the formula and substitute the a and b values.
9a2 + 24a + 16
(3a)2 + 2 (3a) (4) + (4)2
(3a + 4)2
(3a + 4) (3a + 4)

Factors of the 9a2 + 24a + 16 are (3a + 4) (3a + 4)

9. Given expression is a2 + a + 1/4
The given expression a2 + a + 1/4 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = a, b = 1/2
Apply the formula and substitute the a and b values.
a2 + a + 1/4
(a)2 + 2 (a) (1/2) + (1/2)2
(a + 1/2)2
(a + 1/2) (a + 1/2)

Factors of the a2 + a + 1/4 are (a + 1/2) (a + 1/2)

10. The given expression a2 – 6a + 9 is in the form a2 – 2ab + b2.
So find the factors of given expression using a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) where a = a, b = 3
Apply the formula and substitute the a and b values.
a2 – 6a + 9
(a)2 – 2 (a) (3) + (3)2
(a – 3)2
(a – 3) (a – 3)

Factors of the a2 – 6a + 9 are (a – 3) (a – 3)


IV. Solved Problems on Factorization Using Formula When the Given Expression Is a Perfect Square

1. a2 – 10a + 25
2. 9a2 – 12a + 4
3. 16a2 – 24a + 9
4. 1 – 2a + a2
5. 1 – 6a + 9a2
6. x2y2 – 6xyz + 9z2
7. a2 – 4ab + 4b2

Solution:

1. The given expression a2 – 10a + 25 is in the form a2 – 2ab + b2.
So find the factors of given expression using a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) where a = a, b = 5
Apply the formula and substitute the a and b values.
a2 – 10a + 25
(a)2 – 2 (a) (5) + (5)2
(a – 5)2
(a – 5) (a – 5)

Factors of the a2 – 10a + 25 are (a – 5) (a – 5)

2. The given expression 9a2 – 12a + 4 is in the form a2 – 2ab + b2.
So find the factors of given expression using a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) where a = 3a, b = 2
Apply the formula and substitute the a and b values.
9a2 – 12a + 4
(3a)2 – 2 (3a) (2) + (2)2
(3a – 2)2
(3a – 2) (3a – 2)

Factors of the 9a2 – 12a + 4 are (3a – 2) (3a – 2)

3. The given expression 16a2 – 24a + 9 is in the form a2 – 2ab + b2.
So find the factors of given expression using a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) where a = 4a, b = 3
Apply the formula and substitute the a and b values.
16a2 – 24a + 9
(4a)2 – 2 (4a) (3) + (3)2
(4a – 3)2
(4a – 3) (4a – 3)

Factors of the 16a2 – 24a + 9 are (4a – 3) (4a – 3)

4. The given expression 1 – 2a + a2 is in the form a2 – 2ab + b2.
So find the factors of given expression using a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) where a = a, b = 1
Apply the formula and substitute the a and b values.
1 – 2a + a2
(a)2 – 2 (a) (1) + (1)2
(a – 1)2
(a – 1) (a – 1)

Factors of the 1 – 2a + a2 are (a – 1) (a – 1)

5. The given expression 1 – 6a + 9a2 is in the form a2 – 2ab + b2.
So find the factors of given expression using a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) where a = 3a, b = 1
Apply the formula and substitute the a and b values.
1 – 6a + 9a2
(3a)2 – 2 (3a) (1) + (1)2
(3a – 1)2
(3a – 1) (3a – 1)

Factors of the 1 – 6a + 9a2 are (3a – 1) (3a – 1)

6. The given expression x2y2 – 6xyz + 9z2 is in the form a2 – 2ab + b2.
So find the factors of given expression using a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) where a = xy, b = 3z
Apply the formula and substitute the a and b values.
x2y2 – 6xyz + 9z2
(xy)2 – 2 (xy) (3z) + (3z)2
(xy – 3z)2
(xy – 3z) (xy – 3z)

Factors of the x2y2 – 6xyz + 9z2 are (xy – 3z) (xy – 3z)

7. The given expression a2 – 4ab + 4b2 is in the form a2 – 2ab + b2.
So find the factors of given expression using a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) where a = a, b = 2b
Apply the formula and substitute the a and b values.
a2 – 4ab + 4b2
(a)2 – 2 (a) (2b) + (2b)2
(a – 2b)2
(a – 2b) (a – 2b)

Factors of the a2 – 4ab + 4b2 are (a – 2b) (a – 2b)