Worksheet on Factoring Simple Quadratics is for those who are seriously searching to learn factorization problems. We have given different problems along with solutions on our Factorization Worksheets. Therefore, anyone who wants to start their practice immediately to learn factorization problems can utilize our Factoring Quadratic Equations Worksheet With Answers and get good marks. Concentrate on the areas you are lagging and improve your performance in the exams. Practice with Factoring Simple Quadratics Worksheet those have equations like x^2 + ax + b.

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

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

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)

## Worksheet on Factoring Trinomials | Factoring Trinomials Worksheet with Answers

Take the help of Worksheet on Factoring Trinomials and practice well for the exam. It’s really easy for all the students to score good marks in the exam if they practice with Factoring Trinomials Worksheets. Allot your time to practice all problems available on the Factoring Trinomials Practice Worksheets to improve your subject knowledge. You can easily clear all of your doubts by taking the reference of our Factorization Worksheets. Immediately begin your practice and become an expert in solving Factorization Problems.

1. Factor trinomials of the form ax² + bx + c

(i) 2a2 + 11a + 12
(ii) 3a2 + 8a + 4
(iii) 3a2 – 13a + 14
(iv) 4a2 – 7a + 3
(v) 5a2 – 11a – 12
(vi) 7a2 – 15a – 18

Solution:

(i) The given expression 2a2 + 11a + 12
Rewrite the equation
2a2 + 8a + 3a + 12
Group the first two terms and last two terms.
The first two terms are 2a2 + 8a and the second two terms are 3a + 12.
Take 2a common from the first two terms.
2a (a + 4)
Take 3 common from the second two terms.
3 (a + 4)
2a (a + 4) + 3 (a + 4)
Then, take (a + 4) common from the above expression.
(a + 4) (2a + 3)

The final answer is (a + 4) (2a + 3).

(ii) The given expression 3a2 + 8a + 4
Rewrite the equation
3a2 + 6a + 2a + 4
Group the first two terms and last two terms.
The first two terms are 3a2 + 6a and the second two terms are 2a + 4.
Take 3a common from the first two terms.
3a (a + 2)
Take 2 common from the second two terms.
2 (a + 2)
3a (a + 2) + 2 (a + 2)
Then, take (a + 2) common from the above expression.
(a + 2) (3a + 2)

The final answer is (a + 2) (3a + 2).

(iii) The given expression 3a2 – 13a + 14
Rewrite the equation
3a2 – 6a – 7a + 14
Group the first two terms and last two terms.
The first two terms are 3a2 – 6a and the second two terms are – 7a + 14.
Take 3a common from the first two terms.
3a (a – 2)
Take -7 common from the second two terms.
-7 (a – 2)
3a (a – 2) – 7 (a – 2)
Then, take (a – 2) common from the above expression.
(a – 2) (3a – 7)

The final answer is (a – 2) (3a – 7).

(iv) The given expression 4a2 – 7a + 3
Rewrite the equation
4a2 – 4a – 3a + 3
Group the first two terms and last two terms.
The first two terms are 4a2 – 4a and the second two terms are – 3a + 3.
Take 4a common from the first two terms.
4a (a – 1)
Take -3 common from the second two terms.
-3 (a – 1)
4a (a – 1) – 3 (a – 1)
Then, take (a – 1) common from the above expression.
(a – 1) (4a – 3)

The final answer is (a – 1) (4a – 3).

(v) The given expression 5a2 – 11a – 12
Rewrite the equation
5a2 – 15a + 4a – 12
Group the first two terms and last two terms.
The first two terms are 5a2 – 15a and the second two terms are 4a – 12.
Take 4a common from the first two terms.
5a (a – 3)
Take 4 common from the second two terms.
4 (a – 3)
5a (a – 3) + 4 (a – 3)
Then, take (a – 3) common from the above expression.
(a – 3) (5a + 4)

The final answer is (a – 3) (5a + 4).

(vi) The given expression 7a2 – 15a – 18
Rewrite the equation
7a2 – 21a + 6a – 18
Group the first two terms and last two terms.
The first two terms are 7a2 – 21a  and the second two terms are 6a – 18.
Take 7a common from the first two terms.
7a (a – 3)
Take 6 common from the second two terms.
6 (a – 3)
7a (a – 3) + 6 (a – 3)
Then, take (a – 3) common from the above expression.
(a – 3) (7a + 6)

The final answer is (a – 3) (7a + 6).

2. Factor trinomials

(i) 7a2 + 6a – 1
(ii) 9a2 + 35a – 4
(iii) 2a2 – 5a + 3
(iv) 7a – 6 – 2a2
(v) 11a2 – 54a + 63
(vi) a2 + 2a – 3

Solution:

(i) The given expression 7a2 + 6a – 1
Rewrite the equation
7a2 + 7a – a – 1
Group the first two terms and last two terms.
The first two terms are 7a2 + 7a and the second two terms are – a – 1.
Take 7a common from the first two terms.
7a (a + 1)
Take -1 common from the second two terms.
-1 (a + 1)
7a (a + 1) – 1 (a + 1)
Then, take (a + 1) common from the above expression.
(a + 1) (7a – 1)

The final answer is (a + 1) (7a – 1).

(ii) The given expression 9a2 + 35a – 4
Rewrite the equation
9a2 + 36a – a – 4
Group the first two terms and last two terms.
The first two terms are 9a2 + 36a and the second two terms are – a – 4.
Take 9a common from the first two terms.
9a (a + 4)
Take -1 common from the second two terms.
-1 (a + 4)
9a (a + 4) – 1 (a + 4)
Then, take (a + 4) common from the above expression.
(a + 4) (9a – 1)

The final answer is (a + 4) (9a – 1).

(iii) The given expression 2a2 – 5a + 3
Rewrite the equation
2a2 – 2a – 3a + 3
Group the first two terms and last two terms.
The first two terms are 2a2 – 2a and the second two terms are – 3a + 3.
Take 2a common from the first two terms.
2a (a – 1)
Take -3 common from the second two terms.
-3 (a – 1)
2a (a – 1) – 3 (a – 1)
Then, take (a – 1) common from the above expression.
(a – 1) (2a – 3)

The final answer is (a – 1) (2a – 3).

(iv) The given expression 7a – 6 – 2a2
Rewrite the equation
– 2a2 + 4a + 3a – 6
Group the first two terms and last two terms.
The first two terms are – 2a2 + 4a and the second two terms are 3a – 6.
Take -2a common from the first two terms.
-2a (a – 2)
Take 3 common from the second two terms.
3 (a – 2)
-2a (a – 2) + 3 (a – 2)
Then, take (a – 2) common from the above expression.
(a – 2) (-2a + 3)

The final answer is (a – 2) (-2a + 3).

(v) The given expression 11a2 – 54a + 63
Rewrite the equation
11a2 – 33a – 21a + 63
Group the first two terms and last two terms.
The first two terms are 11a2 – 33a and the second two terms are – 21a + 63.
Take 11a common from the first two terms.
11a (a – 3)
Take -21 common from the second two terms.
-21 (a – 3)
11a (a – 3) – 21 (a – 3)
Then, take (a – 3) common from the above expression.
(a – 3) (11a – 21)

The final answer is (a – 3) (11a – 21).

(vi) The given expression a2 + 2a – 3
Rewrite the equation
a2 + 3a – a – 3
Group the first two terms and last two terms.
The first two terms are a2 + 3a and the second two terms are – a – 3.
Take a common from the first two terms.
a (a + 3)
Take -1 common from the second two terms.
-1 (a + 3)
a (a + 3) – 1 (a + 3)
Then, take (a + 3) common from the above expression.
(a + 3) (a – 1)

The final answer is (a + 3) (a – 1).

(i) 2x2 + 5x + 2
(ii) 3a2 + 14a + 8
(iii) 2x2 + 7x + 6
(iv) 6a2 – a – 15
(v) 9s2 – s – 8
(vi) 12 + a – 6a2
(vii) 6 + 5x – 6x2
(viii) a2 + 8a – 105

Solution:

(i) The given expression 2x2 + 5x + 2
Rewrite the equation
2x2 + 4x + x + 2
Group the first two terms and last two terms.
The first two terms are 2x2 + 4x and the second two terms are x + 2.
Take 2x common from the first two terms.
2x (x + 2)
Take 1 common from the second two terms.
1 (x + 2)
2x (x + 2) + 1 (x + 2)
Then, take (x + 2) common from the above expression.
(x + 2) (2x + 1)

The final answer is (x + 2) (2x + 1).

(ii) The given expression 3a2 + 14a + 8
Rewrite the equation
3a2 + 12a + 2a + 8
Group the first two terms and last two terms.
The first two terms are 3a2 + 12a and the second two terms are 2a + 8.
Take 3a common from the first two terms.
3a (a + 4)
Take 2 common from the second two terms.
2 (a + 4)
3a (a + 4) + 2 (a + 4)
Then, take (a + 4) common from the above expression.
(a + 4) (3a + 2)

The final answer is (a + 4) (3a + 2).

(iii) The given expression 2x2 + 7x + 6
Rewrite the equation
2x2 + 4x + 3x + 6
Group the first two terms and last two terms.
The first two terms are 2x2 + 4x and the second two terms are 3x + 6.
Take 2x common from the first two terms.
2x (x + 2)
Take 3 common from the second two terms.
3 (x + 2)
2x (x + 2) + 3 (x + 2)
Then, take (x + 2) common from the above expression.
(x + 2) (2x + 3)

The final answer is (x + 2) (2x + 3).

(iv) The given expression 6a2 – a – 15
Rewrite the equation
6a2 + 9a -10a  – 15
Group the first two terms and last two terms.
The first two terms are 6a2 + 9a and the second two terms are -10a  – 15.
Take 3a common from the first two terms.
3a (2a + 3)
Take -5 common from the second two terms.
-5 (2a + 3)
3a (2a + 3) -5 (2a + 3)
Then, take (2a + 3) common from the above expression.
(2a + 3) (3a – 5)

The final answer is (2a + 3) (3a – 5).

(v) The given expression 9s2 – s – 8
Rewrite the equation
9s2 – 9s + 8s – 8
Group the first two terms and last two terms.
The first two terms are 9s2 – 9s and the second two terms are 8s – 8.
Take 9s common from the first two terms.
9s (s – 1)
Take 8 common from the second two terms.
8 (s – 1)
9s (s – 1) + 8 (s – 1)
Then, take (s – 1) common from the above expression.
(s – 1) (9s + 8)

The final answer is (s – 1) (9s + 8).

(vi) The given expression 12 + a – 6a2
Rewrite the equation
– 6a2 + 9a -8a + 12
Group the first two terms and last two terms.
The first two terms are – 6a2 + 9a and the second two terms are -8a + 12 .
Take -3a common from the first two terms.
-3a (2a – 3)
Take -4 common from the second two terms.
-4 (2a – 3)
-3a (2a – 3) – 4 (2a – 3)
Then, take (2a – 3) common from the above expression.
(2a – 3) (-3a – 4)

The final answer is (2a – 3) (-3a – 4).

(vii) The given expression 6 + 5x – 6x2
Rewrite the equation
– 6x2 + 9x – 4x +6
Group the first two terms and last two terms.
The first two terms are – 6x2 + 9x and the second two terms are – 4x +6.
Take -3x common from the first two terms.
-3x (2x – 3)
Take -2 common from the second two terms.
-2 (2x – 3)
-3x (2x – 3) -2 (2x – 3)
Then, take (2x – 3) common from the above expression.
(2x – 3) (-3x – 2)

The final answer is (2x – 3) (-3x – 2).

(viii) The given expression a2 + 8a – 105
Rewrite the equation
a2 + 15a – 7a – 105
Group the first two terms and last two terms.
The first two terms are a2 + 15a and the second two terms are – 7a – 105.
Take a common from the first two terms.
a (a + 15)
Take -7 common from the second two terms.
-7 (a + 15)
a (a + 15) – 7 (a + 15)
Then, take (a + 15) common from the above expression.
(a + 15) (a – 7)

The final answer is (a + 15) (a – 7).

## Worksheet on Factoring Identities | Factoring Polynomials using Algebraic Identities Worksheets

For a better practice and to get full of knowledge on Factoring Polynomials Using Algebraic Identities, use Worksheet on Factoring Identities. To grade up your preparation level, we are providing a number of examples on factoring identities topics. So, practice the below worksheet on Factoring Polynomials using Algebraic Identities and develop your skills on the concept. Also, check different problems on Factorization Worksheets and prepare better for the exam.

The basic expressions for the factorization of the identities are mentioned below. They are

(i) (x + y)^2 = x^2 + y^2 + 2xy.
(ii) (x – y)^2 = x^2 + y^2 – 2xy.
(iii) x^2 – y^2 = (x + y) (x – y).

## Solved Examples for Factoring Identities

1. Factor the given expressions using identity

(i) x^2 + 8x + 16.
(ii) 4a^2 – 4a + 1.
(iii) x^4 + 9y^4 + 6x^2y^2.
(iv) (x^4 – 8x^2y^2 + 16y^4) – 18.
(v) 256 – x^2 – 2xy – y^2.

Solution:

(i) The given expression is x^2 + 8x + 16.
Now, expand the given expression. That is,
x^2 + 8x + 16 = x^2 + 2(4x) + (4)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x + y)^2 = x^2 + y^2 + 2xy.
Here, x = x and y = 4.
So, (x + y)^2 = (x + 4)^2.

Then, x^2 + 8x + 16 is equal to (x + 4)^2.

(ii) The given expression is 4a^2 – 4a + 1.
Now, expand the given expression. That is,
4a^2 – 4a + 1 = (2a)^2 – 2(2a)(1) + (1)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x – y)^2 = x^2 + y^2 – 2xy.
Here, x = 2a and y = 1.
So, (x – y)^2 = (2a – 1)^2.

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

(iii) The given expression is x^4 + 9y^4 + 6x^2y^2.
Now, expand the given expression. That is,
x^4 + 9y^4 + 6x^2y^2 = (x^2)^2 + 2(x^2)(3y^2) + (3y^2)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x + y)^2 = x^2 + y^2 + 2xy.
Here, x = x^2 and y = 3y^2.
So, (x + y)^2 = (x^2 + 3y^2)^2.

Then, x^4 + 9y^4 + 6x^2y^2 is equal to (x^2 + 3y^2)^2.

(iv) The given expression is (x^4 – 8x^2y^2 + 16y^4) – 18.
Now, expand the given expression. That is,
(x^4 – 8x^2y^2 + 16y^4) – 18 = [(x^2)^2 – 2(x^2) (4y^2) + (4y^2)^2] – 18.
By comparing the given expression with the basic expressions, it matches with the expression (x – y)^2 = x^2 + y^2 – 2xy.
Here, x = x^2 and y = 4y^2.
So, (x – y)^2 = (x^2 – 4y^2)^2.

Then, (x^4 – 8x^2y^2 + 16y^4) – 18 is equal to (x^2 – 4y^2)^2 – 18.

(v) The given expression is 256 – x^2 – 2xy – y^2.
Now, expand the given expression. That is,
256 – x^2 – 2xy – y^2 = (16)^2 – (x^2 + 2xy + y^2)
By comparing the given expression with the basic expressions, it matches with the expression (x + y)^2 = x^2 + y^2 + 2xy.
Here, x = x and y = y.
So, (x + y)^2 = (x + y)^2.

Then, 256 – x^2 – 2xy – y^2 is equal to (16)^2 – (x + y)^2.

2. Factorize the expressions

(i) 4a^2 – 12ab + 9b^2.
(ii) 36x^2 – 84xy + 49y^2.
(iii) 9x^2 + 42xy + 49y^2.
(iv) (3x – 5y)² + 2 (3x – 5y ) (2y – x) + (2y – x)²
(v) 36a² + 36a + 9
(vi) 4x4 – y4

Solution:

(i) The given expression is 4a^2 – 12ab + 9b^2.
Now, expand the given expression. That is,
4a^2 – 12ab + 9b^2 = (2a)^2 – 2(2a)(3b) + (3b)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x – y)^2 = x^2 + y^2 – 2xy.
Here, x = 2a and y = 3b.
So, (x – y)^2 = (2a – 3b)^2.

Then, 4a^2 – 12ab + 9b^2 is equal to (2a – 3b)^2.

(ii) The given expression is 36a^2 – 84ab + 49b^2.
Now, expand the given expression. That is,
36a^2 – 84ab + 49b^2 = (6a)^2 – 2(6a)(7b) + (7b)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x – y)^2 = x^2 + y^2 – 2xy.
Here, x = 6a and y = 7b.
So, (x – y)^2 = (6a – 7b)^2.

Then, 36a^2 – 84ab + 49b^2 is equal to (6a – 7b)^2.

(iii) The given expression is 9x^2 + 42xy + 49y^2.
Now, expand the given expression. That is,
9x^2 + 42xy + 49y^2 = (3x)^2 + 2(3x)(7y) + (7y)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x + y)^2 = x^2 + y^2 + 2xy.
Here, x = 3x and y = 7y.
So, (x + y)^2 = (3x + 7y)^2.

Then, 9x^2 + 42xy + 49y^2 is equal to (3x+ 7y)^2.

(iv) The given expression is (3x – 5y)^2 + 2 (3x – 5y) (2y – x) + (2y – x)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x + y)^2 = x^2 + y^2 + 2xy.
Here, x = (3x – 5y) and y = (2y – x).
So, (x + y)^2 = [(3x – 5y) + (2y – x)]^2.

Then, (3x – 5y)^2 + 2 (3x – 5y) (2y – x) + (2y – x)^2 is equal to [(3x – 5y) + (2y – x)]^2.

(v) The given expression is 36a² + 36a + 9.
Now, expand the given expression. That is,
36a² + 36a + 9 = (6a)^2 + 2(6a)(3) + (3)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x + y)^2 = x^2 + y^2 + 2xy.
Here, x = 6a and y = 3.
So, (x + y)^2 = (6a + 3)^2.

Then, 36a2 + 36a + 9 is equal to (6a + 3)^2.

(vi) The given expression is
4x^4 – y^4.
Now, expand the given expression. That is,
4x^4 – y^4 = (2x)^2 – (y^2)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x)^2 – (y)^2 = (x + y) (x – y).
Here, x = 2x and y = y^2.
So, (x)^2 – (y)^2= (2x + y^2) (2x – y^2).

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

3. Factor the identities

(i) 4a² + 12ab + 9b²
(ii) a² + 22a + 121.
(iii) 9a² – 24ab + 16b²
(iv) 36a² – 36a + 9.
(v) 16a4 – 72a2b2 + 81b4
(vi) (x² + z² + 2xz) – y²

Solution:

(i) The given expression is 4a² + 12ab + 9b²
Now, expand the given expression. That is,
4a² + 12ab + 9b²= (2a)^2 + 2(2a)(3b) + (3b)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x + y)^2 = x^2 + 2xy + y^2.
Here, x = 2a and y = 3b.
So, (x + y)^2= (2a + 3b)^2.

Then, 4a2 + 12ab + 9b2 is equal to (2a + 3b)^2.

(ii) The given expression is a^2 + 22a + 121.
Now, expand the given expression. That is,
a^2 + 22a + 121= (a)^2 + 2(a)(11) + (11)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x + y)^2 = x^2 + 2xy + y^2.
Here, x = a and y = 11.
So, (x + y)^2= (a + 11)^2.

Then, a² + 22a + 121 is equal to (a + 11)^2.

(iii) The given expression is 9a² – 24ab + 16b²
Now, expand the given expression. That is,
9a² – 24ab + 16b²= (3a)^2 – 2(3a)(4b) + (4b)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x – y)^2 = x^2 – 2xy + y^2.
Here, x = 3a and y = 4b.
So, (x – y)^2= (3a – 4b)^2.

Then, 9a² – 24ab + 16b² is equal to (3a – 4b)^2.

(iv) The given expression is 36a² – 36a + 9..
Now, expand the given expression. That is,
36a² – 36a + 9 = (6a)^2 – 2(6a)(3) + (3)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x – y)^2 = x^2 – 2xy + y^2.
Here, x = 6a and y = 3.
So, (x – y)^2= (6a – 3)^2.

Then, 36a² – 36a + 9 is equal to (6a – 3)^2.

(v) The given expression is 16a^4 – 72a^2b^2 + 81b^4.
Now, expand the given expression. That is,
16a^4 – 72a^2b^2 + 81b^4= (4a^2)^2 – 2(4a^2)(9b^2) + (9b^2)^2.
By comparing the given expression with the basic expressions, it matches with the expression (x – y)^2 = x^2 – 2xy + y^2.
Here, x = 4a^2 and y = 9b^2.
So, (x – y)^2= (4a^2 – 9b^2)^2.

Then, 16a^4 – 72a^2b^2 + 81b^4 is equal to (4a^2 – 9b^2)^2.

(vi) The given expression is (x^2 + z^2 + 2xz) – y^2.
By comparing the given expression with the basic expressions, it matches with the expression (x + y)^2 = x^2 + 2xy + y^2.
Here, x = x and y = z.
So, (x + y)^2= (x + z)^2.
Now, (x^2 + z^2 + 2xz) – y^2 = (x + z)^2 – y^2.
By comparing the above expression, it matches the expression x^2 – y^2 = (x + y) (x – y).
So, (x + z)^2 – y^2 = (x + z + y) (x + z – y).

Then, (x^2 + z^2 + 2xz) – y^2 is equal to (x + z + y) (x + z – y).

4. Factor completely using the formula

(i) 100 – [121a² – 88ab + 16b²]
(ii) 36 – x² – y² – 2xy.
(iii) 25x² + 49y² -70xy – 15x + 21y.
(iv) 4a² – 4a – 3.
(v) 64 – x² – y² – 2xy.
(vi) 25a² – (3b + 4c)²

Solution:

(i) The given expression is 100 – [121a² – 88ab + 16b²].
Now, expand the given expression. That is,
100 – [121a² – 88ab + 16b²] = (10)^2 – [(11a)^2 – 2(11a)(4b) + (4b)^2].
By comparing the given expression with the basic expressions, it matches with the expression (x – y)^2 = x^2 – 2xy + y^2.
Here, x = 11a and y = 4b.
So, (x – y)^2= (11a – 4b)^2.
Then, 100 – [121a² – 88ab + 16b²] is equal to (10)^2 – (11a – 4b)^2.
By comparing the above expression, it matches with the expression x^2 – y^2 = (x + y) (x – y).
Here, x = 10 and y = (11a – 4b).
x^2 – y^2 = (x + y) (x – y).
(10)^2 – (11a – 4b)^2 = (10 + (11a – 4b)) (10 – (11a – 4b)).

Then, 100 – [121a^2 – 88ab + 16b^2] is equal to (10 + (11a – 4b)) (10 – (11a – 4b)).

(ii) The given expression is 36 – x² – y² – 2xy.
Now, expand the given expression. That is,
36 – x² – y² – 2xy = (6)^2 – [x^2 + y^2 + 2xy].
By comparing the given expression with the basic expressions, it matches with the expression (x + y)^2 = x^2 + 2xy + y^2.
Here, x = x and y = y.
So, (x + y)^2= (x + y)^2.
Then, 36 – x² – y² – 2xy is equal to (6)^2 – (x + y)^2.
By comparing the above expression, it matches with the expression x^2 – y^2 = (x + y) (x – y).
Here, x = 6 and y = (x + y).
x^2 – y^2 = (x + y) (x – y).
((6)^2 – (x + y)^2) = (6 + (x + y)) (6 – (x + y)).
Then, 36 – x² – y² – 2xy is equal to (6 + (x + y)) (6 – (x + y)).

(iii) The given expression is 25x^2 + 49y^2 – 70xy – 15x + 21y.
Now, expand the given expression. That is,
25x^2 + 49y^2 – 70xy – 15x + 21y = [(5x)^2 + (7y)^2 – 2(5x)(7y)] – 15x + 21y.
By comparing the given expression with the basic expressions, it matches with the expression (x – y)^2 = x^2 – 2xy + y^2.
Here, x = 5x and y = 7y.
So, (x – y)^2= (5x – 7y)^2.
Then, [(5x)^2 + (7y)^2 – 2(5x)(7y)] – 15x + 21y is equal to (5x – 7y)^2 – 15x + 21y.
(5x – 7y)^2 – 3(5x – 7y).
Factor out the common term from the above expression. That is,
(5x – 7y) (5x – 7y – 3).

So, 25x^2 + 49y^2 – 70xy – 15x + 21y is equal to (5x – 7y) (5x – 7y – 3).

(iv) The given expression is 4a^2 – 4a – 3.
Now, expand the given expression. That is,
4a^2 – 4a – 3 = (2a)^2 – 2(2a) – 3.
Factor out the common terms from the above expression. That is,
(2a)^2 – 2(2a) – 3 = (2a)(2a – 2) – 3.
= (2a – 2) (2a – 3).

4a^2 – 4a – 3 is equal to (2a – 2) (2a – 3).

(v) The given expression is 64 – x^2 – y^2 – 2xy.
Now, expand the given expression. That is,
64 – x^2 – y^2 – 2xy = (8)^2 – [x^2 + y^2 + 2xy].
By comparing the given expression with the basic expressions, it matches with the expression (x + y)^2 = x^2 + 2xy + y^2.
Here, x = x and y = y.
So, (x + y)^2= (x + y)^2.
Then, (8)^2 – [x^2 + y^2 + 2xy] is equal to (8)^2 – (x + y)^2.
By comparing the above expression with the basic expressions, it matches with the expression (x)^2 – (y)^2 = (x + y) (x – y).
Here, x = 8 and y = x + y.
So, (8)^2 – (x + y)^2 = (8 + (x + y)) (8 – (x + y)).

Then, 64 – x^2 – y^2 – 2xy is equal to (8 + (x + y)) (8 – (x + y)).

(vi) The given expression is 25a^2 – (3b + 4c)^2.
Now, expand the expression. That is,
25a^2 – (3b + 4c)^2 = (5a)^2 – (3b + 4c)^2.
By comparing the given expression with the basic expressions, it matches with the expression x^2 – y^2 = (x + y) (x – y).
Here, x = 5a and y = 3b + 4c.
So, x^2 – y^2 = (x + y) (x – y).
(5a)^2 – (3b + 4c)^2 = [5a + (3b + 4c)] [ 5a – (3b + 4c)].

Then, 25a^2 – (3b + 4c)^2 is equal to [5a + (3b + 4c)] [ 5a – (3b + 4c)].

## Worksheet on Factoring out a Common Binomial Factor | Factoring Binomials Worksheet with Answers

Worksheet on Factoring out a Common Binomial Factor is available with a number of different problems and solutions. The Binomial Factorization Worksheet is available as per the new syllabus instructions. Learn and practice more problems on the factorization of a common binomial factor with the help of the given worksheet. We have explained every problem with the simplest method to help students clearly understand how to factor out a common binomial factor of an algebraic expression.

Also, check out all Factorization Worksheets and practice all problems for better results.

## Solved Problems on Factoring Out a Common Binomial Factor

1. Factorize by taking binomial as a common factor

(i) 3(a + 2) + 7(a + 2)
(ii) (a + 3)a + (a + 3)4.
(iii) 2(5a + 3b) + c(5a + 3b).
(iv) 3a(b – 4c) – 5d(b – 4c).
(v) x(p – q) + y (q – p).

Solution:

(i) The given expression is 3(a + 2) + 7(a + 2).
Factor out the common terms from the above expression. That is,
(a + 2) (3 + 7)
10 (a + 2).

The final answer is 10 (a + 2).

(ii) The given expression is (a + 3)a + (a + 3)4.
Factor out the common term from the above expression. That is,
(a + 3) (a + 4).

The final answer is (a + 3) (a + 4).

(iii) The given expression is 2(5a + 3b) + c(5a + 3b).
Factor out the common term from the above expression. That is,
(5a + 3b) (2 + c).

The final answer is (5a + 3b) (2 + c).

(iv) The given expression is 3a(b – 4c) – 5d(b – 4c).
Factor out the common term from the above expression. That is,
(b – 4c) (3a – 5d).

The final answer is (b – 4c) (3a – 5d).

(v) The given expression is x(p – q) + y(q – p).
x(p – q) – y(- q + p).
x(p – q) – y(p – q).
Factor out the common terms from the above expression. That is,
(p – q) (x – y).

The final answer is (p – q) (x – y).

2. Factorize a common binomial factor from each of the following expression

(i) p(q + r) – s (q + r).
(ii) 12(ab + 1) + 3x (ab + 1).
(iii) x^2 + y^2 + 9p (x^2 + y^2).
(iv) 3(x + y) – 5(x + y)^2.

Solution:

(i) The given expression is p (q + r) – s (q + r).
Factor out the common terms from the above expression. That is,
(q + r) (p – s).

The final answer is (q + r) (p – s).

(ii) The given expression is 12(ab + 1) + 3x (ab + 1).
Factor out the common terms from the above expression. That is,
(ab + 1) (12 + 3x).

The final answer is (ab + 1) (12 + 3x).

(iii) The given expression is x^2 + y^2 + 9p(x^2 + y^2).
Factor out the common terms from the above expression. That is,
(x^2 + y^2) (1 + 9p).

The final answer is (x^2 + y^2) (1 + 9p).

(iv) The given expression is 3(x + y) – 5(x + y)^2.
Factor out the common terms from the above expression. That is,
(x + y) (3 – 5(x + y)).
(x + y) (3 – 5x – 5y).

The final answer is (x + y) (3 – 5x – 5y).

3. Factorize common binomial factor from each of the following expressions
(i) x (3y – 7z) – z (3y – 7z).
(ii) (2x – 6) (3p – 2q) – (2x – 6) (2q – 3p).
(iii) a (a + b) + (5a + 5b).
(iv) (6ab + 3a) + (2b + 1).
(v) a (b – c)^2 – d (c – b)^3.
(vi) (a – 3) + (3xy – xya)

Solution:

(i) The given expression is x (3y – 7z) – z (3y – 7z).
Factor out the common terms from the above expression. That is,
(3y – 7z) (x – z).

The final answer is (3y – 7z) (x – z).

(ii) The given expression is (2x – 6) (3p – 2q) – (2x – 6) (2q – 3p).
(2x – 6) (3p – 2q) – (2x – 6) ( – ) (3p – 2q).
(2x – 6) (3p – 2q) + (2x – 6) (3p – 2q).
Factor out the common terms from the above expression. That is,
2(2x – 6) (3p – 2q).

The final answer is 2(2x – 6) (3p – 2q).

(iii) The given expression is a(a + b) + (5a + 5b).
a(a + b) + 5(a + b).
Factor out the common terms from the above expression. That is,
(a + b) (a + 5).

The final answer is (a + b) (a + 5).

(iv) The given expression is (6ab + 3a) + (2b + 1).
3a(2b + 1) + (2b + 1).
Factor out the common terms from the above expression. That is,
(2b + 1) (3a + 1).

The final answer is (2b + 1) (3a + 1).

(v) The given expression is a(b – c)^2 – d(c – b)^3.
a(b – c)^2 + d(b – c)^3.
Factor out the common terms from the above expression. That is,
(b – c)^2 [a + d(b – c)].

The final answer is (b – c)^2 [a + d(b – c)].

(vi) The given expression is (a – 3) + (3xy – xya).
(a – 3) + xy( 3 – a).
(a – 3) – xy (a – 3).
Factor out the common terms from the above expression. That is,
(a – 3) (1 – xy).

The final answer is (a – 3) (1 – xy).

Provide a free source of Factorization of Algebraic Expressions Worksheets for your students. Each concept of Factorization Worksheet quick links are given here. Find your required concept and practice all the problems available on that particular worksheet. You can prepare with the Factorization Worksheets with Answers and get good marks in the exams. Easily estimate your preparation level with the help of Worksheet on Factorization. Effective preparation comes to your hand when you choose the best material.

One of such best materials is the Factorization Practice Worksheets provided on our website. All the topics and subtopics of factorization questions, answers are given along with detailed explanations. Therefore, without any delay, begin your practice with the available worksheets and score better grades in your exams.

## Factorization Questions Worksheets

Get Printable Factorization Worksheets all in one place. Every Factorization concept worksheet is attached through direct links. Click on the required link and check all the problems within it. You can check a solution for every problem available on the worksheet.

All the worksheets provided on our website are prepared and tested by a team of math experts. Also, the worksheets have practice questions and multiple-choice questions. So, it is easy for everyone to test their preparation level. We have included answers and explanations for each problem to help students while preparing for the exam.

Contact us at any time through the comment section to clear your doubts. Bookmark our website to get instant updates available on our website.

## Worksheet on Factoring by Grouping | Factorization by Grouping Worksheets

Do you want to have a better understanding of factoring by grouping topics? Here, we are to help you in this and you have reached the correct place. By going through this article, you can clearly understand the factorization of the grouping method. We are offering the Worksheet on Factoring by Grouping, which includes different algebraic equations factorization with a simple solution. Practice using the factorization by grouping worksheets and enhance your conceptual knowledge.

We are providing all Factorization Worksheets for free of cost on our website. Check every worksheet and improve your preparation level.

## Solved Examples on Factoring by Grouping

1. Factor the following

(i) 12p + 15
(ii) 14p – 21
(iii) 9q – 12q²

Solution:

(i) The given expression is 12p + 15.
Expand the given expression. That is,
(3 * 4)p + (3 * 5).
Factor out the greatest common factor. That is,
3(4p + 5).

Then, 12p + 15 is equal to 3(4p + 5).

(ii) The given expression is 14p – 21.
Expand the given expression. That is,
(7 * 2)p – (7 * 3).
Factor out the greatest common factor. That is,
7(2p – 3).

Then, 14p – 21 is equal to 7(2p – 3).

(iii) The given expression is 9q – 12q^2.
Expand the given expression. That is,
(3 * 3)q – (3 * 4)q^2.
Factor out the greatest common factor. That is,
3q(3 – 4q).

Then, 9q – 12q^2 is equal to 3q(3 – 4q).

2. Factor by grouping the expressions

(i) 16x² – 24xy
(ii) 15xy² – 20x²y
(iii) 12a²b³ – 21a³b²

Solution:

(i) The given expression is 16x^2 – 24xy.
Expand the given expression. That is,
(4 * 4)x^2 – (4 * 6)xy.
Factor out the greatest common factor. That is,
4x(4x – 6y).

Then, 16x^2 – 24xy is equal to 4x(4x – 6y).

(ii) The given expression is 15xy^2 – 20x^2y.
Expand the given expression. That is,
(5 * 3)xy^2 – (5 * 4)x^2y.
Factor out the greatest common factor. That is,
5xy(3y – 4x).

Then, 15xy^2 – 20x^2y is equal to 5xy(3y – 4x).

(iii) The given expression is 12a^2b^3 – 21a^3b^2.
Expand the given expression. That is,
(3 * 4)a^2b^3 – (3 * 7)a^3b^2.
Factor out the greatest common factor. That is,
3a^2b^2(4b – 7a).

Then, 12a^2b^3 – 21a^3b^2 is equal to 3a^2b^2(4b – 7a).

3. Factorize the expressions

(i) 24a³ – 36a²b.
(ii) 10a³ – 15a²
(iii) 36a³b – 60a²b³c

Solution:

(i) The given expression is 24a^3 – 36a^2b.
Expand the given expression. That is,
(6 * 4)a^3 – (6 * 6)a^2b.
Factor out the greatest common factor. That is,
6a^2(4a – 6b).

Then, 24a^3 – 36a^2b is equal to 6a^2(4a – 6b).

(ii) The given expression is 10a^3 – 15a^2.
Expand the given expression. That is,
(5 * 2)a^3 – (5 * 3)a^2.
Factor out the greatest common factor. That is,
5a^2(2a – 3).

Then, 10a^3 – 15a^2 is equal to 5a^2(2a – 3).

(iii) The given expression is 36a^3b – 60a^2b^3c.
Expand the given expression. That is,
(6 * 6)a^3b – (6 * 10)a^2b^3c.
Factor out the greatest common factor. That is,
6a^2b(6a – 10b^2c).

Then, 36a^3b – 60a^2b^3c is equal to 6a^2b(6a – 10b^2c).

4. Factorize

(i) 9a³ – 6a² + 12a.
(ii) 8a² – 72ab + 12a.
(iii) 18x³y³ – 27x²y³ + 36x³y²

Solution:

(i) The given expression is 9a³ – 6a² + 12a.
Expand the given expression. That is,
(3 * 3)a^3 – (3 * 2)a^2 + (3 * 4)a.
Factor out the greatest common factor. That is,
3a(3a^2 – 2a + 4).

Then, 9a^3 – 6a^2 + 12a is equal to 3a(3a^2 – 2a + 4).

(ii) The given expression is 8a^2 – 72ab + 12a.
Expand the given expression. That is,
(4 * 2)a^2 – (4 * 18)ab + (4 * 3)a.
Factor out the greatest common factor. That is,
4a(2a – 18b + 3a).

Then, 8a^2 – 72ab + 12a is equal to 4a(2a – 18b + 3a).

(iii) The given expression is 18x³y³ – 27x²y³ + 36x³y²
Expand the given expression. That is,
(9 * 2)x^3y^3 – (9 * 3)x^2y^3 + (9 * 4)x^3y^2.
Factor out the greatest common factor. That is,
9x^2y^2(2xy -3y + 4x).

Then, 18x³y³ – 27x²y³ + 36x³y² is equal to 9x^2y^2 (2xy -3y + 4x).

5. How to factor by grouping?

(i) 14a³ + 21a^4b – 28a²b²
(ii) -5 – 10x + 20x²

Solution:

(i) The given expression is 14a^3 + 21a^4b – 28a^2b^2.
Expand the given expression. That is,
(7 * 2)a^3 + (7 * 3)a^4b – (7 * 4)a^2b^2.
Factor out the greatest common factor. That is,
7a^2(2a + 3a^2b – 4b^2).

Then, 14a^3 + 21a^4b – 28a^2b^2 is equal to 7a^2(2a + 3a^2b – 4b^2).

(ii) The given expression is -5 – 10x + 20x^2.
Expand the given expression. That is,
-5 – (5 * 2)x + (5 * 4)x^2.
Factor out the greatest common factor. That is,
5( -1 – 2x + 4x^2).

Then, -5 – 10x + 20x^2 is equal to 5( -1 – 2x + 4x^2) .

6. Factoring

(i) a (a + 3) + 5(a + 3)
(ii) 5a (a – 4) – 7 (a – 4).
(iii) 2x (1 – y) + 3(1 – y).

Solution:

(i) The given expression is a(a + 3) + 5(a + 3).
Factor out the greatest common factor. That is,
(a + 3) (a + 5)

Then,a (a + 3) + 5(a + 3) is equal to (a + 3) (a + 5).

(ii) The given expression is 5a(a – 4) – 7(a – 4).
Factor out the greatest common factor. That is,
(a – 4) (5a – 7).

Then, 5a (a – 4) – 7 (a – 4) is equal to (a – 4) (5a – 7).

(iii) The given expression is 2x(1 – y) + 3(1 – y).
Factor out the greatest common factor. That is,
(1 – y) (2x + 3).

Then, 2x(1 – y) + 3(1- y) is equal to (1 – y) (2x + 3).

7. Factoring the expressions

(i) 6x (x – 2y) + 5y (x – 2y).
(ii) p³ (2x – y) + p²(2x – y).

Solution:

(i) The given expression is 6x (x – 2y) + 5y (x – 2y).
Factor out the greatest common factor. That is,
(x – 2y) (6x + 5y).

Then, 6x (x – 2y) + 5y (x – 2y) is equal to (x – 2y) (6x + 5y).

(ii) The given expression is p^3 (2x – y) + p^2 (2x – y).
Factor out the greatest common factor. That is,
(2x – y)p^2 (p + 1).

Then, p^3 (2x – y) + p^2 (2x – y) is equal to p^2(2x – y) (p + 1).

8. How to factor by grouping polynomials?

(i) 9x (3x – 5y) – 12x²(3x – 5y)
(ii) (a + 5)² – 4 (a + 5)
(iii) 3(x – 2y)² – 5(x – 2y)

Solution:

(i) The given expression is 9x (3x – 5y) – 12x²(3x – 5y).
Factor out the greatest common factor. That is,
3x(3x – 5y)(3 – 4x).

Then, 9x (3x – 5y) – 12x²(3x – 5y) is equal to 3x(3x – 5y)(3 – 4x).

(ii) The given expression is (a + 5)^2 – 4 (a + 5).
Factor out the greatest common factor. That is,
(a + 5) [(a + 5) – 4] = (a + 5) (a + 1).

Then, (a + 5)^2 – 4 (a + 5) is equal to (a + 5) (a + 1).

(iii) The given expression is 3(x – 2y)² – 5(x – 2y).
Factor out the greatest common factor. That is,
(x – 2y) [3(x – 2y) – 5].

Then, 3(x – 2y)² – 5(x – 2y) is equal to (x – 2y) [3(x – 2y) – 5].

9. Factor completely

(i) 2x + 6y – 3(x + 3y)²
(ii) 16(2x – 3y)² – 4(2x – 3y)
(iii) p (x – 3) + q (3 – x)
(iv) 12 (2a – 3b)² – 16 (3b – 2a).
(v) (a + b)(2a + 5) – (a + b)(a + 3).

Solution:

(i) The given expression is 2x + 6y – 3(x + 3y)^2.
Factor out the greatest common factor. That is,
Factor out the greatest common factor. That is,
2(x + 3y) – 3(x + 3y)^2 = (x + 3y) (2 – 3(x + 3y)).

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

(ii) The given expression is 16(2x – 3y)² – 4(2x – 3y).
Factor out the greatest common factor. That is,
4(2x – 3y) [4(2x – 3y) – 1].

Then, 16(2x – 3y)² – 4(2x – 3y) is equal to 4(2x – 3y) [4(2x – 3y) – 1].

(iii) The given expression is
p (x – 3) + q (3 – x).
we can write it as p (x – 3) – q(x – 3).
Factor out the greatest common factor. That is,
(x – 3) (p – q).

Then, p (x – 3) + q (3 – x) is equal to (x – 3) (p – q).

(iv) The given expression is 12 (2a – 3b)² – 16 (3b – 2a).
Expand the above expression. That is,
12 (2a – 3b)^2 + 32a – 48b.
We can write it as 12 (2a – 3b)^2 + 16 (2a – 3b).
Factor out the greatest common factor. That is,
4(2a – 3b) [3(2a – 3b) + 4].

Then, 12 (2a – 3b)² – 16 (3b – 2a) is equal to 4(2a – 3b) [3(2a – 3b) + 4].

(v) The given expression is (a + b)(2a + 5) – (a + b)(a + 3).
Factor out the greatest common factor. That is,
(a + b) [(2a + 5) – (a + 3)] = (a + b) (a + 2).

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

10. How to factor by grouping polynomials?

(i) xp + yp + xq + yq.
(ii) p² – sp – qp + sq.
(iii) xy² – yz² – xy + z²
(iv) a² – ac + ab–bc.
(v) 6xy – y² + 12xz – 2yz.

Solution:

(i) The given expression is xp + yp + xq + yq.
Factor out the greatest common factor. That is,
p(x + y) + q(x + y) = (x + y) (p + q).

Then, xp + yp + xq + yq is equal to (x + y) (p + q).

(ii) The given expression is p2 – sp – qp + sq.
Factor out the greatest common factor. That is,
p(p – s) – q(p – s) = (p – s) (p – q).

Then, p2 – sp – qp + sq is equal to (p – s) (p – q).

(iii) The given expression is xy² – yz² – xy + z²
Factor out the greatest common factor. That is,
xy^2 – xy – yz^2 + z^2 = xy(y – 1) – z(yz – z).

Then, xy² – yz² – xy + z² is equal to xy(y – 1) – z(yz – z).

(iv) : The given expression is a2 – ac + ab – bc.
Factor out the greatest common factor. That is,
a(a – c) + b(a – c) = (a – c) (a + b).

Then, a2 – ac + ab – bc is equal to (a – c) (a + b).

(v) The given expression is 6xy – y² + 12xz – 2yz.
Factor out the greatest common factor. That is,
y(6x – y) + 2z(6x – y) = (6x – y) (y + 2z).

Then, 6xy – y² + 12xz – 2yz is equal to (6x – y) (y + 2z).

11. Solve by factoring

(i) (a – 2b)² + 4a–8b
(ii) b² – ab (1 – a) – a³
(iii) (rp + sq)² + (sp – rq)²
(iv) pq² + (p – 1)q – 1

Solution:

(i) The given expression is (a – 2b)² + 4a–8b.
Factor out the greatest common factor. That is,
(a – 2b)² + 4(a – 2b) = (a – 2b) (a – 2b + 4).

Then, (a – 2b)² + 4a–8b is equal to (a – 2b) (a – 2b + 4).

(ii) The given expression is b² – ab (1 – a) – a³
we can write it as b^2 – ab + a^2b – a^3.
Factor out the greatest common factor. That is,
b(b – a) + a^2(b – a) = (b – a)(b + a^2).

Then, b² – ab (1 – a) – a³ is equal to (b – a)(b + a^2).

(iii) The given expression is (rp + sq)² + (sp – rq)²
We can write it as (rp)^2 + (sq)^2 + 2rpsq + (sp)^2 + (rq)^2 – 2rpsq
it is equal to (rp)^2 + (sq)^2 + (sp)^2 + (rq)^2
Factor out the greatest common factor. That is,
r^2(p^2 + q^2) + s^2(p^2 + q^2) = (p^2 + q^2) (r^2 + s^2)

Then, (rp + sq)² + (sp – rq)² is equal to (p^2 + q^2) (r^2 + s^2)

(iv) The given expression is pq² + (p – 1)q – 1.
we can write it as pq^2 + pq – q – 1= pq(q +1) – (q + 1)
Factor out the greatest common factor. That is,
(q + 1) (pq – 1).

Then, pq² + (p – 1)q – 1 is equal to (q + 1) (pq – 1).

12. Factoring Algebraic Expressions

(i) p³ – 3p² + p – 3
(ii) xy (p² + q²) – pq (x² + y²)
(iii) p² – p(x + 2y) + 2xy

Solution:

(i) The given expression is p3 – 3p2 + p – 3.
We can write it as p^2(p – 3) + (p – 3)
Factor out the greatest common factor. That is, (p – 3) (p^2 + 1).

Then, p3 – 3p2 + p – 3is equal to (p – 3) (p^2 + 1).

(ii) The given expression is xy (p2 + q2) – pq (x2 + y2).
We can write it as xyp^2 + xyq^2 – pqx^2 – pqy^2.
Factor out the greatest common factor. That is, xp (yp –qx) – yq (yp–qx) = (yp – qx) (xp – yq).

Then, xy (p2 + q2) – pq (x2 + y2)is equal to (yp – qx) (xp – yq).

(iii) The given expression is p2 – p(x + 2y) + 2xy.
We can write it as p^2 – px – 2py + 2xy.
Factor out the greatest common factor. That is, p(p -2y) – x(p – 2y) = (p – 2y) (p – x).

Then, p2 – p(x + 2y) + 2xy is equal to (p – 2y) (p – x).

## Worksheet on Factorization by Regrouping | Factoring by Regrouping Worksheet

Are you looking for the help of the Factorization by Regrouping Concept? Then, you are in the correct place. Use the provided Worksheet on Factorization by Regrouping free of cost to learn the entire factorization by regrouping concept. To improve your preparation level, we have given solved examples with explanations in Factorization by Regrouping Practice Worksheets. Test your knowledge and cross-check where you went wrong and overcome those errors in your exams.

Practice all problems available in the Factorization Worksheets to get grip on the complete concept.

## How to Solve Factorization Problems by Regrouping?

1. Factorize each of the following by regrouping

(i) a2 + ab + 9a + 9b
(ii) 6ab – 4b + 6 – 9a
(iii) 10xy + 6x + 5y +3
(iv) a3 + a2 + a + 1
(v) m2 – b + mb – m
(vi) a3 – a2b + 5a – 5b
(vii) x (x + 3) – x – 3
(viii) 3mx + 3my – 2nx – 2ny
(ix) a (a + b – c ) – bc
(x) x3 – x2 + xy2 – x2y2

Solution:

(i) Given expression is a2 + ab + 9a + 9b.
Rearrange the terms
a2 + ab + 9a + 9b.
Group the first two terms and last two terms.
The first two terms are a2 + ab and the second two terms are 9a + 9b.
Take a common from the first two terms.
a (a + b)
Take 9 common from the second two terms.
9 (a + b)
a (a + b) + 9 (a + b)
Then, take (a + b) common from the above expression.
(a + b) (a + 9)

The final answer is (a + b) (a + 9).

(ii) Given expression is 6ab – 4b + 6 – 9a.
Rearrange the terms
6ab – 9a – 4b + 6
Group the first two terms and last two terms.
The first two terms are 6ab – 9a and the second two terms are – 4b + 6.
Take 3a common from the first two terms.
3a (2b – 3)
Take -2 common from the second two terms.
-2 (2b  – 3)
3a (2b – 3) – 2 (2b – 3)
Then, take (2b – 3) common from the above expression.
(2b – 3) (3a – 2)

The final answer is (2b – 3) (3a – 2).

(iii) Given expression is 10xy + 6x + 5y +3.
Rearrange the terms
10xy + 5y + 6x +3
Group the first two terms and last two terms.
The first two terms are 10xy + 5y and the second two terms are 6x +3.
Take 5y common from the first two terms.
5y (2x + 1)
Take 3 common from the second two terms.
3 (2x  + 1)
5y (2x + 1) + 3 (2x  + 1)
Then, take (2x  + 1) common from the above expression.
(2x  + 1) (5y + 3)

The final answer is (2x  + 1) (5y + 3).

(iv) Given expression is a3 + a2 + a + 1.
Group the first two terms and last two terms.
The first two terms are a3 + a2 and the second two terms are a + 1.
Take a2 common from the first two terms.
a2 (a + 1)
Take 1 common from the second two terms.
1 (a + 1)
a2 (a + 1) + 1 (a + 1)
Then, take (a + 1) common from the above expression.
(a + 1) (a2 + 1)

The final answer is (a + 1) (a2 + 1).

(v) Given expression is m2 – b + mb – m.
Rearrange the terms
m2 + mb – b  – m.
Group the first two terms and last two terms.
The first two terms are m2 + mb and the second two terms are – b – m.
Take m common from the first two terms.
m (m + b)
Take -1 common from the second two terms.
-1 (m + b)
m (m + b) -1 (m + b)
Then, take (m + b) common from the above expression.
(m + b) (m – 1)

The final answer is (m + b) (m – 1).

(vi) Given expression is a3 – a2b + 5a – 5b.
Rearrange the terms
a3 + 5a – a2b – 5b.
Group the first two terms and last two terms.
The first two terms are a3 + 5a and the second two terms are – a2b – 5b.
Take a common from the first two terms.
a (a2 + 5)
Take -b common from the second two terms.
– b (a2 + 5)
a (a2 + 5) – b (a2 + 5)
Then, take (m + b) common from the above expression.
(a2 + 5) (a – b)

The final answer is (a2 + 5) (a – b).

(vii) Given expression is x (x + 3) – x – 3.
Rearrange the terms
x2  – x + 3x – 3.
Group the first two terms and last two terms.
The first two terms are x2  – x  and the second two terms are 3x – 3.
Take x common from the first two terms.
x (x – 1)
Take 3 common from the second two terms.
3 (x – 1)
x (x – 1) + 3 (x – 1)
Then, take (x – 1) common from the above expression.
(x – 1) (x + 3)

The final answer is (x – 1) (x + 3).

(viii) Given expression is 3mx + 3my – 2nx – 2ny.
Rearrange the terms
3mx – 2nx + 3my – 2ny.
Group the first two terms and last two terms.
The first two terms are 3mx – 2nx  and the second two terms are 3my – 2ny.
Take x common from the first two terms.
x (3m – 2n)
Take y common from the second two terms.
y (3m – 2n)
x (3m – 2n) + y (3m – 2n)
Then, take (3m – 2n) common from the above expression.
(3m – 2n) (x + y)

The final answer is (3m – 2n) (x + y).

(ix) Given expression is a (a + b – c ) – bc.
Rearrange the terms
a² – ac + ab – bc.
Group the first two terms and last two terms.
The first two terms are a² – ac and the second two terms are ab – bc.
Take a common from the first two terms.
a (a – c)
Take b common from the second two terms.
b (a – c)
a (a – c) + b (a – c)
Then, take (a – c) common from the above expression.
(a – c) (a + b)

The final answer is (a – c) (a + b).

(x) Given expression is x3 – x2 + xy2 – x2y2.
Rearrange the terms
x3 – x2 – x2y2 + xy2
Group the first two terms and last two terms.
The first two terms are x3 – x2 and the second two terms are – x2y2 + xy2.
Take x2 common from the first two terms.
x2 (x – 1)
Take – xy2 common from the second two terms.
– xy2 (x – 1)
x2 (x – 1) – xy2 (x – 1)
Then, take (x – 1) common from the above expression.
(x – 1) (x2 – xy2)

The final answer is (x – 1) (x2 – xy2).

2. Factor grouping the algebraic expressions

(i) 4ab – 7b + 12a – 21
(ii) 7xy – 5x – 28y + 20
(iii) 5mn – 2m – 5n2 + 2n
(iv) 6a2 – 15ac – 8ba + 20bc
(v) 4mx + 5nx – 12my – 15ny

Solution:

(i) Given expression is 4ab – 7b + 12a – 21.
Rearrange the terms
4ab + 12a – 7b – 21
Group the first two terms and last two terms.
The first two terms are 4ab + 12a and the second two terms are – 7b – 21.
Take 4a common from the first two terms.
4a (b + 3)
Take – 7 common from the second two terms.
– 7 (b + 3)
4a (b + 3) – 7 (b + 3)
Then, take (b + 3) common from the above expression.
(b + 3) (4a – 7)

The final answer is (b + 3) (4a – 7).

(ii) Given expression is 7xy – 5x – 28y + 20.
Rearrange the terms
7xy – 28y – 5x + 20
Group the first two terms and last two terms.
The first two terms are 7xy – 28y and the second two terms are – 5x + 20.
Take 7y common from the first two terms.
7y (x – 4)
Take – 5 common from the second two terms.
– 5 (x – 4)
7y (x – 4) – 5 (x – 4)
Then, take (x – 4) common from the above expression.
(x – 4) (7y – 5)

The final answer is (x – 4) (7y – 5).

(iii) Given expression is 5mn – 2m – 5n2 + 2n.
Rearrange the terms
5mn – 5n2 – 2m + 2n
Group the first two terms and last two terms.
The first two terms are 5mn – 5n2 and the second two terms are – 2m + 2n.
Take n common from the first two terms.
5n (m – n)
Take – 2 common from the second two terms.
– 2 (m – n)
5n (m – n) – 2 (m – n)
Then, take (m – n) common from the above expression.
(m – n) (5n – 2)

The final answer is (m – n) (5n – 2).

(iv) Given expression is 6a2 – 15ac – 8ba + 20bc.
Rearrange the terms
6a2 – 8ba – 15ac + 20bc
Group the first two terms and last two terms.
The first two terms are 6a2 – 8ba and the second two terms are – 15ac + 20bc.
Take 2a common from the first two terms.
2a (3a – 4b)
Take – 5c common from the second two terms.
– 5c (3a – 4b)
2a (3a – 4b) – 5c (3a – 4b)
Then, take (3a – 4b) common from the above expression.
(3a – 4b) (2a – 5c)

The final answer is (3a – 4b) (2a – 5c).

(v) Given expression is 4mx + 5nx – 12my – 15ny.
Rearrange the terms
4mx – 12my + 5nx – 15ny
Group the first two terms and last two terms.
The first two terms are 4mx – 12my and the second two terms are + 5nx – 15ny.
Take 4m common from the first two terms.
4m (x – 3y)
Take 5n common from the second two terms.
5n (x – 3y)
4m (x – 3y) + 5n (x – 3y)
Then, take (x – 3y) common from the above expression.
(x – 3y) (4m + 5n)

The final answer is (x – 3y) (4m + 5n).

3. Factorize by regrouping the terms

(i) 7mn – 21nr – 7mx + 21xr
(ii) a2 + ab(b + 1) + b3
(iii) yx2 – 2x(1 – y) – 4
(iv) m2 – m(a + 4b) + 4ab
(v) a – 9 – (a – 9)2 + ab – 9b

Solution:

(i) Given expression is 7mn – 21nr – 7mx + 21xr
Rearrange the terms
7mn – 7mx – 21nr + 21xr
Group the first two terms and last two terms.
The first two terms are 7mn – 7mx and the second two terms are – 21nr + 21xr.
Take 7m common from the first two terms.
7m (n – x)
Take -21r common from the second two terms.
-21r (n – x)
7m (n – x) – 21r (n – x)
Then, take (n – x) common from the above expression.
(n – x) (7m – 21r)

The final answer is (n – x) (7m – 21r).

(ii) Given expression is a2 + ab(b + 1) + b3
Rearrange the terms
a2 + ab + ab2 + b3
Group the first two terms and last two terms.
The first two terms are a2 + ab and the second two terms are ab2 + b3.
Take a common from the first two terms.
a (a + b)
Take b2 common from the second two terms.
b2 (a + b)
a (a + b) + b2 (a + b)
Then, take (a + b) common from the above expression.
(a + b) (a + b2)

The final answer is (a + b) (a + b2).

(iii) Given expression is yx2 – 2x(1 – y) – 4
Rearrange the terms
yx2 + 2xy – 2x – 4
Group the first two terms and last two terms.
The first two terms are yx2 + 2xy and the second two terms are – 2x – 4.
Take xy common from the first two terms.
xy (x + 2)
Take -2 common from the second two terms.
-2 (x + 2)
xy (x + 2) – 2 (x + 2)
Then, take (x + 2) common from the above expression.
(x + 2) (xy – 2)

The final answer is (x + 2) (xy – 2).

(iv) Given expression is m2 – m(a + 4b) + 4ab
Rearrange the terms
m2 – 4mb – ma + 4ab
Group the first two terms and last two terms.
The first two terms are m2 – 4mb and the second two terms are – ma + 4ab.
Take m common from the first two terms.
m (m – 4b)
Take -a common from the second two terms.
-a (m – 4b)
m (m – 4b) – a (m – 4b)
Then, take (m – 4b) common from the above expression.
(m – 4b) (m – a)

The final answer is (m – 4b) (m – a).

(v) Given expression is a – 9 – (a – 9)2 + ab – 9b
Rearrange the terms
a – 9 – (a – 9)2 + ab – 9b
Group the first two terms and last two terms.
The first two terms are a – 9 – (a – 9)2  and the second two terms are ab – 9b.
Take m common from the first two terms.
(a – 9) (1 – a + 9) = (a – 9) (10 – a)
Take b common from the second two terms.
b (a – 9)
(a – 9) (10 – a) + b (a – 9)
Then, take (a – 9) common from the above expression.
(a – 9) (10 – a + b)

The final answer is (a – 9) (10 – a + b).

4. Factorize by grouping the following expressions

(i) (a – 4) – (a – 4)2 + 12 – 3a
(ii) b (c – d )2 – a (d – c) + 3c – 3d
(iii) (a2 + 2a)2 – 7 (a2 + 2a) – y (a2 + 2a) +7y
(iv) m4x + m3 (2x – y ) – m(2my + z) – 2z
(v) x3 – 2x2y + 3xy2 – 6y3
(vi) m2 + n – mn – m
(vii) 5ab – b2 + 15ca – 3bc
(viii) xy2 – yz2 – xy + z2

Solution:

(i) Given expression is (a – 4) – (a – 4)2 + 12 – 3a
Rearrange the terms
(a – 4) – (a – 4)2 – 3a + 12
Group the first two terms and last two terms.
The first two terms are (a – 4) – (a – 4)2  and the second two terms are – 3a + 12.
Take (a – 4) common from the first two terms.
(a – 4) (1 – a + 4) = (a – 4) (5 – a)
Take -3 common from the second two terms.
-3 (a – 4)
(a – 4) (5 – a) – 3 (a – 4)
Then, take (a – 4) common from the above expression.
(a – 4) (5 – a – 3) = (a – 4) (2 – a)

The final answer is (a – 4) (2 – a).

(ii) Given expression is b (c – d )2 – a (d – c) + 3c – 3d
Rearrange the terms
b (c – d )2 + a (c – d) + 3c – 3d
Group the first two terms and last two terms.
The first two terms are b (c – d )2 + a (c – d) and the second two terms are 3c – 3d.
Take (c – d) common from the first two terms.
(c – d) (b (c – d) + a) = (c – d) (bc – bd + a)
Take 3 common from the second two terms.
3 (c – d)
(c – d) (bc – bd + a) + 3 (c – d)
Then, take (a – 4) common from the above expression.
(c – d) (bc – bd + a + 3)

The final answer is (c – d) (bc – bd + a + 3).

(iii) Given expression is (a2 + 2a)2 – 7 (a2 + 2a) – y (a2 + 2a) +7y
Rearrange the terms
(a2 + 2a)2 – y (a2 + 2a) – 7 (a2 + 2a) + 7y
Group the first two terms and last two terms.
The first two terms are (a2 + 2a)2 – y (a2 + 2a) and the second two terms are – 7 (a2 + 2a) + 7b.
Take (a2 + 2a) common from the first two terms.
(a2 + 2a) (a2 + 2a – y)
Take – 7 common from the second two terms.
– 7 (a2 + 2a – y)
(a2 + 2a) (a2 + 2a – y) – 7 (a2 + 2a – y)
Then, take (a2 + 2a – y) common from the above expression.
(a2 + 2a – y) (a2 + 2a – 7)

The final answer is (a2 + 2a – y) (a2 + 2a – 7).

(iv) Given expression is m4x + m3 (2x – y ) – m(2my + z) – 2z
Rearrange the terms
m4x + 2xm3 – ym3 –  2m2y – zm – 2z
Group the first two terms, middle terms, and last two terms.
The first two terms are m4x + 2xm3, the middle terms are – ym3 –  2m2y, and the second two terms are – zm – 2z.
Take xm3 common from the first two terms.
xm3 (m + 2)
Take – m2y common from the middle two terms.
– m2y (m + 2)
Take – z common from the middle two terms.
– z (m + 2)
xm3 (m + 2) – m2y (m + 2) – z (m + 2)
Then, take (m + 2) common from the above expression.
(m + 2) (xm3 – m2y – z)

The final answer is (m + 2) (xm3 – m2y – z).

(v) Given expression is x3 – 2x2y + 3xy2 – 6y3
Rearrange the terms
x3 – 2x2y + 3xy2 – 6y3
Group the first two terms, and last two terms.
The first two terms are x3 – 2x2y, and the second two terms are 3xy2 – 6y3.
Take x2 common from the first two terms.
x2 (x – 2y)
Take 3y2 common from the middle two terms.
3y2 (x – 2y)
x2 (x – 2y) + 3y2 (x – 2y)
Then, take (x – 2y) common from the above expression.
(x – 2y) (x2 + 3y2)

The final answer is (x – 2y) (x2 + 3y2).

(vi) Given expression is m2 + n – mn – m
Rearrange the terms
m2 – m + n – mn
Group the first two terms, and last two terms.
The first two terms are m2 – m, and the second two terms are n – mn.
Take m common from the first two terms.
m (m – 1)
Take -n common from the middle two terms.
-n (m – 1)
m (m – 1) – n (m – 1)
Then, take (m – 1) common from the above expression.
(m – 1) (m – n)

The final answer is (m – 1) (m – n).

(vii) Given expression is 5ab – b2 + 15ca – 3bc
Rearrange the terms
5ab + 15ca – b2 – 3bc
Group the first two terms, and last two terms.
The first two terms are 5ab + 15ca, and the second two terms are – b2 – 3bc.
Take 5a common from the first two terms.
5a (b + 3c)
Take -b common from the middle two terms.
-b (b + 3c)
5a (b + 3c) – b (b + 3c)
Then, take (b + 3c) common from the above expression.
(b + 3c) (5a – b)

The final answer is (b + 3c) (5a – b).

(viii) Given expression is xy2 – yz2 – xy + z2
Rearrange the terms
xy2 – xy – yz2 + z2
Group the first two terms, and last two terms.
The first two terms are xy2 – xy, and the second two terms are – yz2 + z2.
Take xy common from the first two terms.
xy (y – 1)
Take -z2 common from the middle two terms.
-z2 (y – 1)
xy (y – 1) – z2 (y – 1)
Then, take (y – 1) common from the above expression.
(y – 1) (xy – z2)

The final answer is (y – 1) (xy – z2).

## Worksheet on Factoring the Differences of Two Squares | Factoring Difference of Squares Worksheet

Are you looking for guidance on the concept of Differences of Two Squares Factorization? Then, check out the Worksheet on Factoring the Differences of Two Squares now. Practice as much as you can using our Factoring Difference of Squares Worksheet and get a good grip on the topic. Solve all problems available on the Difference Between Two Squares Worksheet and test your subject knowledge.

Improve your problem-solving skills by solving the Differences of Two Squares Factorization Worksheet frequently and assess your preparation standards. All the questions are framed by the subject experts to help the students to have perfect knowledge of the concept.

## How to Find Factorization of the Differences of Two Squares?

Factorize given algebraic expression using the following identity a2 – b2 = (a + b) (a – b).

1. Factorize the following by taking the difference of squares

(i) a2 – 9
(ii) x2 – 1
(iii) 49 – a2
(iv) 4a2 – 25
(v) x2y2 – 16
(vi) m4 – n4

Solution:

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

The final answer is [a + 3] [a – 3]

(ii) Given expression is x2 – 1
Rewrite the given expression in the form of a2 – b2.
(x)2 – (1)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = x and b = 1
[x + 1] [x – 1]

The final answer is [x + 1] [x – 1]

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

The final answer is [7 + a] [7 – a]

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

The final answer is [2a + 5] [2a – 5]

(v) Given expression is x2y2 – 16
Rewrite the given expression in the form of a2 – b2.
(xy)2 – (4)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = xy and b = 4
[xy + 4] [xy – 4]

The final answer is [xy + 4] [xy – 4]

(vi) Given expression is m4 – n4
Rewrite the given expression in the form of a2 – b2.
(m2)2 – (n2)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = m2 and b = n2
[m2 + n2] [m2 – n2]
From the above equation, [m2 – n2] is in the form of a2 – b2.
[(m)2 – (n)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = m and b = n
[m + n] [m – n]
Now, [m2 + n2] [m2 – n2]
[m2 + n2] [m + n] [m – n]

The final answer is [m2 + n2] [m + n] [m – n]

2. Factoring by the Difference of Two Perfect Squares

(i) 144x2 – 169y2
(ii) 1 – 0.09x2
(iii) 16m2 – 121
(iv) – 64x2 + (9/25) y2
(v) m4 – 256
(vi) (a + b)4 – c4

Solution:

(i) Given expression is 144x2 – 169y2
Rewrite the given expression in the form of a2 – b2.
(12x)2 – (13y)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 12x and b = 13y
[12x + 13y] [12x – 13y]

The final answer is [12x + 13y] [12x – 13y]

(ii) Given expression is 1 – 0.09x2
Rewrite the given expression in the form of a2 – b2.
(1)2 – (0.3x)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 1 and b = 0.3x
[1 + 0.3x] [1 – 0.3x]

The final answer is [1 + 0.3x] [1 – 0.3x]

(iii) Given expression is 16m2 – 121
Rewrite the given expression in the form of a2 – b2.
(4m)2 – (11)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 4m and b = 11
[4m + 11] [4m – 11]

The final answer is [4m + 11] [4m – 11]

(iv) Given expression is – 64x2 + (9/25) y2
Rewrite the given expression in the form of a2 – b2.
(9/25) y2 – 64x2 = (3/5y)2 – (8x)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 3/5y and b = 8x
[3/5y + 8x] [3/5y – 8x]

The final answer is [3/5y + 8x] [3/5y – 8x]

(v) Given expression is m4 – 256
Rewrite the given expression in the form of a2 – b2.
(m2)2 – ( (4)2)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = m2 and b = (4)2
[m2 + (4)2] [m2 – (4)2]
[m2 + 16] [(m)2 – (4)2]
From the above equation, [(m)2 – (4)2] is in the form of a2 – b2.
[(m)2 – (4)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = m and b = (4)
[m + (4)] [m – (4)]
Now, [m2 + 16] [(m)2 – (4)2]
[m2 + 16] [m + (4)] [m – (4)]
[m2 + 16] [m + 4] [m – 4]

The final answer is [m2 + 16] [m + 4] [m – 4]

(vi) Given expression is (a + b)4 – c4
Rewrite the given expression in the form of a2 – b2.
((a + b)2)2 – ( (c)2)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = (a + b)2 and b = ((c)2)2
[(a + b)2 + (c)2] [(a + b)2 – (c)2]
[(a + b)2 + (c)2] [(a + b)2 – (c)2]
From the above equation, [(a + b)2 – (c)2] is in the form of a2 – 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]
Now, [(a + b)2 + (c)2] [(a + b)2 – (c)2]
[(a + b)2 + (c)2] [a+ b + c] [a + b – c]

The final answer is [(a + b)2 + (c)2] [a+ b + c] [a + b – c]

3. Factorize using the formula of differences of two squares

(i) 36x2 – y2
(ii) a2b2 – 16
(iii) 9x4y4 – 25m4n4
(iv) a4 – 256
(v) 81m2 – 49n2
(vi) a2 – (b – c)2

Solution:

(i) Given expression is 36x2 – y2
Rewrite the given expression in the form of a2 – b2.
(6x)2 – (y)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 6x and b = (y)
[6x + y] [6x – y]

The final answer is [6x + y] [6x – y]

(ii) Given expression is a2b2 – 16
Rewrite the given expression in the form of a2 – b2.
(ab)2 – (4)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = ab and b = 4
[ab + 4] [ab – 4]

The final answer is [ab + 4] [ab – 4]

(iii) Given expression is 9x4y4 – 25m4n4
Rewrite the given expression in the form of a2 – b2.
(3x2y2)2 – (5m2n2)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 3x2y2 and b = 5m2n2
[3x2y2 + 5m2n2] [3x2y2 – 5m2n2]

The final answer is [3x2y2 + 5m2n2] [3x2y2 – 5m2n2]

(iv) Given expression is a4 – 256
Rewrite the given expression in the form of a2 – b2.
(a2)2 – ((4)2)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a2 and b = (4)2
[a2 + (4)2] [a2 – (4)2]
[a2 + 16] [a2 – (4)2]
From the above equation, [a2 – (4)2] is in the form of a2 – b2.
[(a)2 – (4)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a and b = 4
[a + 4] [a – 4]
Now, [a2 + 16] [a2 – (4)2]
[a2 + 16] [a + 4] [a – 4]

The final answer is [a2 + 16] [a + 4] [a – 4]

(v) Given expression is 81m2 – 49n2
Rewrite the given expression in the form of a2 – b2.
(9m)2 – (7n)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 9m and b = 7n
[9m + 7n] [9m – 7n]

The final answer is [9m + 7n] [9m – 7n]

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

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

4. Factor the difference of two perfect squares

(i) 16a2 – (3b + 2y)2
(ii) (3m + 4n)2 – (4n + 5n)2
(iii) (a + b)2 – (a – b)2
(iv) 50x2 – 72y2
(v) x4 – (y + z)4
(vi) x2 – 1/169

Solution:

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

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

(ii) Given expression is (3m + 4n)2 – (4n + 5n)2
Rewrite the given expression in the form of a2 – b2.
(3m + 4n)2 – (9n)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 3m + 4n and b = 9n
[3m + 4n + 9n] [3m + 4n – 9n]
[3m + 13n] [3m – 5n]

The final answer is [3m + 13n] [3m – 5n]

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

(iv) Given expression is 50x2 – 72y2
Rewrite the given expression in the form of a2 – b2.
2[(5x)2 – (6y)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 5x and b = 6y
2{[5x + 6y] [5x – 6y]}

The final answer is 2{[5x + 6y] [5x – 6y]}

(v) Given expression is x4 – (y + z)4
Rewrite the given expression in the form of a2 – b2.
[(x2)2 – ((y + z)2)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = x2 and b = (y + z)2
{[x2 + (y + z)2] [x2 – (y + z)2]}
{[x2 + y2 + z2+ 2yz] [x2 – (y + z)2]}
From the above equation, [x2 – (y + z)2] is in the form of a2 – b2.
[x2 – (y + z)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = x and b = y + z
[x + y + z] [x – (y + z)] = [x + y + z] [x – y – z]
Now, {[x2 + y2 + z2+ 2yz] [x2 – (y + z)2]}
{[x2 + y2 + z2+ 2yz] [x + y + z] [x – y – z]}

The final answer is {[x2 + y2 + z2+ 2yz] [x + y + z] [x – y – z]}

(vi) Given expression is x2 – 1/169
Rewrite the given expression in the form of a2 – b2.
(x)2 – (1/13)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = x and b = 1/13
[x + 1/13] [x – 1/13]

The final answer is [x + 1/13] [x – 1/13]

5. Factor each expression as a difference between two squares

(i) 9 (a + b)2 – 4 (a – b)2
(ii) 16/49 – 25m2
(iii) 9ab2 – a3
(iv) 4 (3a + 1)2 – 9 (a – 2)2
(v) 1 – 121m2
(vi) 169x2 – 1

Solution:

(i) Given expression is9 (a + b)2 – 4 (a – b)2
Rewrite the given expression in the form of a2 – b2.
(a + b)2 – (2 (a – b))2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a + b and b = 2 (a – b)
[a + b + 2 (a – b)] [a + b – 2 (a – b)]
[a + b + 2a – 2b] [a + b – 2a + 2b]
[3a – b] [3b – a]

The final answer is [3a – b] [3b – a]

(ii) Given expression is 16/49 – 25m2
Rewrite the given expression in the form of a2 – b2.
(4/7)2 – (5m)2
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 4/7 and b = 5m
[4/7 + 5m] [4/7 – 5m]

The final answer is [4/7 + 5m] [4/7 – 5m]

(iii) Given expression is 9ab2 – a3
Rewrite the given expression in the form of a2 – b2.
a[(3b)2 – (a)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 3b and b = a
a{[3b + a] [3b – a]}

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

(iv) Given expression is 4 (3a + 1)2 – 9 (a – 2)2
Rewrite the given expression in the form of a2 – b2.
4 (9a2 + 1 + 6a) – 9 (a2 + 4 – 4a) = 36a2 + 4 + 24a – 9a2 – 36 + 36a = 27a2 + 60a – 32 = (9a – 4) (3a + 8)
Now, apply the formula of a2 – b2 = (a + b) (a – b),
(9a – 4) (3a + 8)

The final answer is (9a – 4) (3a + 8)

(v) Given expression is 1 – 121m2
Rewrite the given expression in the form of a2 – b2.
[(1)2 – (11m)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 1 and b = 11m
{[1 + 11m] [1 – 11m]}

The final answer is {[1 + 11m] [1 – 11m]}

(vi) Given expression is 169x2 – 1
Rewrite the given expression in the form of a2 – b2.
[(13x)2 – (1)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 13x and b = 1
{[13x + 1] [13x – 1]}

The final answer is {[13x + 1] [13x – 1]}

6. Factor using the identity

(i) 1 – (m + n)2
(ii) a2b2 – 25/c2
(iii) a12b4 – a4b12
(iv) 100 (m – n)2 – 121 (x + y)2
(v) 2a – 50a3
(vi) 25/a2 – (4a2)/9
(vii) a4 – 1/(b4)
(viii) 75a3b2 – 108ab4

Solution:

(i) 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]}

(ii) Given expression is a2b2 – 25/c2
Rewrite the given expression in the form of a2 – b2.
[(ab)2 – (5/c)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = ab and b = 5/c
{[ab + 5/c] [ab – (5/c)]}
{[ab + 5/c] [ab – 5/c]}

The final answer is {[ab + 5/c] [ab – 5/c]}

(iii) Given expression is a12b4 – a4b12
Rewrite the given expression in the form of a2 – b2.
[(a6b2)2 – (a2b6)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a6b2 and b = a2b6
{[a6b2 + a2b6] [a6b2 – a2b6]} = a2b2{[a4 + b4] a2b2[a4 – b4]} = a4b4 [a4 + b4] [a4 – b4]
From the above equation, [(a2)2 – (b2)2] is in the form of a2 – b2.
[(a2)2 – (b2)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a2 and b = b2
[a2 + b2] [a2 – b2
Now, a4b4 [a4 + b4] [a4 – b4]
a4b4 [a4 + b4] [a2 + b2] [a2 – b2]
From the above equation, [a2 – b2] is in the form of a2 – b2.
[a2 – b2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a and b = b
[a + b] [a – b]
Now, a4b4 [a4 + b4] [a2 + b2] [a2 – b2]
a4b4 [a4 + b4] [a2 + b2] [a + b] [a – b]

The final answer is a4b4 [a4 + b4] [a2 + b2] [a + b] [a – b]

(iv) Given expression is 100 (m – n)2 – 121 (x + y)2
Rewrite the given expression in the form of a2 – b2.
[(10(m – n))2 – (11 (x + y))2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 10(m – n) and b = 11 (x + y)
{[10(m – n) + 11 (x + y)] [10(m – n) – 11 (x + y)]}
{[10m – 10n + 11x + 11y] [10m – 10n – 11x – 11y]}

The final answer is {[10m – 10n + 11x + 11y] [10m – 10n – 11x – 11y]}

(v) Given expression is 2a – 50a3
Rewrite the given expression in the form of a2 – b2.
2a[(1)2 – (5a)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 1 and b = 5a
2a {[1 + 5a] [1 – 5a]}

The final answer is 2a {[1 + 5a] [1 – 5a]}

(vi) Given expression is 25/a2 – (4a2)/9
Rewrite the given expression in the form of a2 – b2.
[(5/a)2 – (2a/3)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 5/a and b = 2a/3
{[5/a + 2a/3] [5/a – 2a/3]}

The final answer is {[5/a + 2a/3] [5/a – 2a/3]}

(vii) Given expression is a4 – 1/(b4)
Rewrite the given expression in the form of a2 – b2.
[(a2)2 – (1/b2)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a2 and b = 1/b2
{[a2 + 1/b2] [a2 – 1/b2]}
From the above equation, [a2 – 1/b2] is in the form of a2 – b2.
[a2 – (1/b)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = a and b = 1/b
[a + 1/b] [a – 1/b]
Now, {[a2 + 1/b2] [a2 – 1/b2]}
{[a2 + 1/b2] [a + 1/b] [a – 1/b] }

The final answer is {[a2 + 1/b2] [a + 1/b] [a – 1/b]}

(vi) Given expression is 75a3b2 – 108ab4
Rewrite the given expression in the form of a2 – b2.
3ab2[(5a)2 – (6b)2]
Now, apply the formula of a2 – b2 = (a + b) (a – b), where a = 5a and b = 6b
3ab2{[5a + 6b] [5a – 6b]}

The final answer is 3ab2{[5a + 6b] [5a – 6b]}

## Worksheet on Factoring Binomials | Factoring Binomials Algebra Worksheets

Do you feel difficult to solve factorization problems when Binomial is a Common Factor? Don’t worry!! We have given a Worksheet on Factoring Binomials for your practice. Solve all the questions available in the Factoring Binomials Worksheets and cross-check answers to test your preparation level.

Most of the questions given in this Common Binomial Factor Worksheet impose in the exam. Therefore, students can practice and get good scores easily by practicing all the methods available in the Binomial Factorization Worksheets. Have a look at the Factorization Worksheets if you want to get a complete grip on the entire factorization concept.

## How to do Factorisation when a Binomial is a Common Factor?

1. Factorize the following binomials

(i) 3x + 21
(ii) 7a – 14
(iii) b3 + 3b
(iv) 20a + 5a2
(v) – 16m + 20m3
(vi) 5a2b + 15ab2
(vii) 9m2 + 5m
(viii) 19x – 57y
(ix) 25x2y2z3 – 15xy3z

Solution:

(i) The given expression is 3x + 21
Here, the first term is 3x and the second term is 21
By comparing the above two terms, we can observe the greatest common factor and that is 3
Now, factor out the greatest common factor from the expression
That is, 3 [x + 7]
3 [x + 7]

Therefore, the resultant value for the expression 3x + 21 is 3 [x + 7]

(ii) The given expression is 7a – 14
Here, the first term is 7a and the second term is 14
By comparing the above two terms, we can observe the greatest common factor and that is 7
Now, factor out the greatest common factor from the expression
That is, 7 [a – 2]
7 [a – 2]

Therefore, the resultant value for the expression 7a – 14 is 7 [a – 2]

(iii) The given expression is b3 + 3b
Here, the first term is b3 and the second term is 3b
By comparing the above two terms, we can observe the greatest common factor and that is b
Now, factor out the greatest common factor from the expression
That is, b [b² + 3]
b [b² + 3]

Therefore, the resultant value for the expression b3 + 3b is b [b² + 3]

(iv) The given expression is 20a + 5a2
Here, the first term is 20a and the second term is 5a2
By comparing the above two terms, we can observe the greatest common factor and that is 5a
Now, factor out the greatest common factor from the expression
That is, 5a [4 + a]
5a [4 + a]

Therefore, the resultant value for the expression 20a + 5a2 is 5a [4 + a]

(v) The given expression is – 16m + 20m3
Here, the first term is – 16m, and the second term is 20m3
By comparing the above two terms, we can observe the greatest common factor and that is 4m
Now, factor out the greatest common factor from the expression
That is, 4m [-4 + 5m²]
4m [-4 + 5m²]

Therefore, the resultant value for the expression – 16m + 20m3 is 4m [-4 + 5m²]

(vi) The given expression is 5a2b + 15ab2
Here, the first term is 5a2b and the second term is 15ab2
By comparing the above two terms, we can observe the greatest common factor and that is 5ab
Now, factor out the greatest common factor from the expression
That is, 5ab [a + 3b]
5ab [a + 3b]

Therefore, the resultant value for the expression 5a2b + 15ab2 is 5ab [a + 3b]

(vii) The given expression is 9m2 + 5m
Here, the first term is 9m2 and the second term is 5m
By comparing the above two terms, we can observe the greatest common factor and that is m
Now, factor out the greatest common factor from the expression
That is, m [9m + 5]
m [9m + 5]

Therefore, the resultant value for the expression 9m2 + 5m is m [9m + 5]

(viii) The given expression is 19x – 57y
Here, the first term is 19x and the second term is – 57y
By comparing the above two terms, we can observe the greatest common factor and that is 19
Now, factor out the greatest common factor from the expression
That is, 19 [x – 3y]
19 [x – 3y]

Therefore, the resultant value for the expression 19x – 57y is 19 [x – 3y]

(ix) The given expression is 25x2y2z3 – 15xy3z
Here, the first term is 25x2y2z3 and the second term is – 15xy3z
By comparing the above two terms, we can observe the greatest common factor and that is 5xy2z
Now, factor out the greatest common factor from the expression
That is, 5xy2z [5xz2 – 3y]
5xy2z [5xz2 – 3y]

Therefore, the resultant value for the expression 25x2y2z3 – 15xy3z is 5xy2z [5xz2 – 3y]

2. Factor each of the following algebraic expression

(i) 13x + 39
(ii) 19a – 57b
(iii) 21ab + 49abc
(iv) – 16x + 20x3
(v) 12a2b – 42abc
(vi) 27m3n3 + 36m4n2

Solution:

(i) The given expression is 13x + 39
Here, the first term is 13x and the second term is 39
By comparing the above two terms, we can observe the greatest common factor and that is 13
Now, factor out the greatest common factor from the expression
That is, 13 [x + 3]
13 [x + 3]

Therefore, the resultant value for the expression 13x + 39 is 13 [x + 3]

(ii) The given expression is 19a – 57b
Here, the first term is 19a and the second term is – 57b
By comparing the above two terms, we can observe the greatest common factor and that is 19
Now, factor out the greatest common factor from the expression
That is, 19 [a – 3b]
19 [a – 3b]

Therefore, the resultant value for the expression 19a – 57b is 19 [a – 3b]

(iii) The given expression is 21ab + 49abc
Here, the first term is 21ab and the second term is 49abc
By comparing the above two terms, we can observe the greatest common factor and that is 7ab
Now, factor out the greatest common factor from the expression
That is, 7ab [3 + 7c]
7ab [3 + 7c]

Therefore, the resultant value for the expression 21ab + 49abc is 7ab [3 + 7c]

(iv) The given expression is – 16x + 20x3
Here, the first term is – 16x and the second term is 20x3
By comparing the above two terms, we can observe the greatest common factor and that is 4x
Now, factor out the greatest common factor from the expression
That is, 4x [-4 + 5x²]
4x [-4 + 5x²]

Therefore, the resultant value for the expression – 16x + 20x3 is 4x [-4 + 5x²]

(v) The given expression is 12a2b – 42abc
Here, the first term is 12a2b and the second term is – 42abc
By comparing the above two terms, we can observe the greatest common factor and that is 6ab
Now, factor out the greatest common factor from the expression
That is, 6ab [2a – 7bc]
6ab [2a – 7bc]

Therefore, the resultant value for the expression 12a2b – 42abc is 6ab [2a – 7bc]

(vi) The given expression is 27m3n3 + 36m4n2
Here, the first term is 27m3n3 and the second term is 36m4n2
By comparing the above two terms, we can observe the greatest common factor and that is 9m3n2
Now, factor out the greatest common factor from the expression
That is, 9m3n2 [3n + 4m]
9m3n2 [3n + 4m]

Therefore, the resultant value for the expression 27m3n3 + 36m4n2 is 9m3n2 [3n + 4m]