Mounika Kandukuri

(i) 5x² + √8x + 5 = 0 is in the form of ax² + bx + c = 0. So, it is a quadratic equation.

(ii) x + 6/x = x

(x² + 6)/x = x

x² + 6 = x²

Which is not in the form of ax² + bx + c = 0. So, it is not a quadratic equation.

(iii) x² = 0 is in the form of ax² + bx + c = 0. So, it is a quadratic equation.

(iv) x² – 4x = 0 is in the form of ax² + bx + c = 0. So, it is a quadratic equation.

(v) x² – 25 = 0 is in the form of ax² + bx + c = 0. So, it is a quadratic equation.

(i) x² = 64 can be written as x² – 64 = 0

Which is in the form of ax² + bx + c = 0. So, it is a quadratic equation.

(ii) x² + 4x + 2 = 0 is in the form of ax² + bx + c = 0. So, it is a quadratic equation.

(iii) x² + 3/x² = 8

(x⁴ + 3) /x² = 8

x⁴ + 3 = 8x²

Which is not in the form of ax² + bx + c = 0. So, it is not a quadratic equation.

(iv) x (x + 1) – (x + 2) (x + 2) = -8

x² + x – [x² + 4x + 4] = -8

x² + x – x² – 4x – 4 = -8

-3x – 4 = -8

3x = -4 + 8 = 4

Which is not in the form of ax² + bx + c = 0. So, it is not a quadratic equation.

(v) x² – √x + 4 = 0 is not in the form of ax² + bx + c = 0. So, it is not a quadratic equation.

(i) Given equation is x² – 6√5x + 25 = 0

a = 1, b = -6√5, c = 25

roots (α, β) = [-b ± √(b² – 4ac)] / 2a

= [-(-6√5) ± √((-6√5)² – 4 * 1 * 25)] / (2 * 1)

= [6√5 ± √(180 – 100)] / 2

= [6√5 ± √80] / 2

= [6√5 ± 4√5] / 2

= [6√5 + 4√5] / 2 and [6√5 – 4√5] / 2

= 10√5 / 2 and 2√5 / 2

= 5√5 and √5

So, the roots are satisfying the equation.

(ii) Given quadratic equation is 8x² + 2x – 1 = 0

8x² + 4x – 2x – 1 = 0

= 4x(2x + 1) -1(2x + 1) = 0

(2x + 1)(4x – 1) = 0

2x + 1 = 0 and 4x – 1 = 0

2x = -1 and 4x = 1

x = -1/2 and x = 1/4

Therefore, the given roots are correct.

(iii) Given quadratic equation is x² – 3x = 40

It can also be written as x² – 3x – 40 = 0

By using the factorization method

x² – 8x + 5x – 40 = 0

x(x – 8) + 5(x – 8) = 0

(x + 5)(x – 8) = 0

x – 8 = 0 and x + 5 = 0

x = 8 and x = -5

Therefore, the given roots are correct.

(iv) Given details are 6x² – 13x + 5 = 0; x = 1/2, and x = 5/3

6x² – 10x – 3x + 5 = 0

2x(3x – 5) – 1(3x – 5) = 0

(2x – 1)(3x – 5) = 0

2x – 1 = 0 and 3x – 5 = 0

2x = 1 and 3x = 5

x = 1/2 and x = 5/3

Therefore, the given roots are correct.

(i) Given quadratic equation is x² – (√2 – √3)x – √6

a = 1, b = √3 – √2, c = -√6

Quadratic equation roots formula is given as

roots (α, β) = [-b ± √(b² – 4ac)] / 2a

= [-(√3 – √2) ± √((√3 – √2)² – 4 * 1 * (-√6)] / 2 * 1

= [(√2 – √3) ± √(5 – 2√6 + 4√6)] / 2

= [(√2 – √3) ± √(5 + 2√6)] / 2

= [(√2 – √3) + √(5 + 2√6)] / 2 and [(√2 – √3) – √(5 + 2√6)] / 2

= [1.141 – 1.732 + √(5 + 2(2.44)] / 2 and [1.141 – 1.732 – √(5 + 2(2.44)] / 2

= [-0.591 + √(5 + 4.88)] / 2 and [-0.591 – √(5 + 4.88)] / 2

= [-0.591 + √9.88] / 2 and [-0.591 – √9.88] / 2

= [-0.591 + 3.143] / 2 and [-0.591 – 3.143] / 2

= 2.55 / 2 and -3.734 / 2

= 1.275 and -1.875

= √2 and -√3

(ii) Given equation is 5x² + 49x + 72 = 0

5x² + 40x + 9x + 72 = 0

5x(x + 8) +9(x + 8) = 0

(5x + 9) (x + 8) = 0

(5x + 9) = 0 and (x + 8) = 0

5x = -9 and x = -8

x = -9/5

The roots are (-8, -9/5)

(iii) Given equation is x² + 2x – 24 = 0

x² + 6x – 4x – 24 = 0

x(x + 6) -4(x + 6) = 0

(x – 4)(x + 6) = 0

(x – 4) = 0 and (x + 6) = 0

x = 4 and x = -6

The roots are (4, -6).

(i) Given quadratic equation is 50x² + 110x + 48 = 0

50x² + 80x + 30x + 48 = 0

10x(5x + 8) + 6(5x + 8) = 0

(10x + 6) (5x + 8) = 0

(10x + 6) = 0 and (5x + 8) = 0

10x = -6 and 5x = -8

x = -6/10 and x = -8/5

Roots are (-3/5, -8/5)

(ii) Given quadratic equation is 4x² + 12x + 9 = 0

4x² + 6x + 6x + 9 = 0

2x(2x + 3) +3(2x + 3) = 0

(2x + 3) (2x + 3) = 0

(2x + 3) = 0 and (2x + 3) = 0

2x = -3 and 2x = -3

x = -3/2 and x = -3/2

The roots are (-3/2, -3/2)

(iii) Given quadratic equation is 16x² – 49 = 0

(4x)² – 7² = 0

(4x)² = 7²

Applying square root on both sides

√(4x)² = √7²

4x = ±7

x = 7/ 4 and x = -7/4

The roots are (-7/4, 7/4).

(a) x ≥ 8

The variable x is greater than equal to 8. The possible values of x are 8 and more than 8.

(b) x < -2

The variable x is less than -2. The possible values of x are -1 and less than it.

(c) x > 6

The variable x is greater than 6. The possible values of x are 7, 8, and so on.

(d) x ≤ 5

The variable x is less than and equal to 5. The possible values of x are 5, 4, 3, 2, 1 and so on.

(a) x < 2 increased by 2.

Add 2 to both sides of inequation.

x + 2 < 2 + 2

x + 2 < 4.

(b) x > 4 decreased by 7.

Subtract 7 from both sides of the inequation

x – 7 > 4 – 7

x – 7 > -3

(c) x ≤ 6 multiplied by 4.

x x 4 ≤ 6 x 4

4x ≤ 24

(d) x ≥ 25 divided by 5.

x/5 ≥ 25/5

x/5 ≥ 5

(a) x ≤ -4

The variable x is less than and equal to -4. The possible values of x are -4, -5, -6, and so on.

(b) x > 9

The variable x is greater than 9. The possible values of x are 10, 11, 12, and so on.

(c) x < -1

The variable x is less than -1. The possible values of x are -2, -3, -4, -5, -6, and so on.

(a) x < 5 multiplied by 2.

Multiply 2 by both sides.

x x 2 < 5 x 2

2x < 10.

(b) x > 6 divided by 3.

Divide 3 by each side

x/3 > 6/3

x/3 > 2.

(c) x ≥ 10 increased by 5.

Add 5 to each side

x + 5 ≥ 10 + 5

x + 5 ≥ 15.

(a) x < -9, x ∈ I

Replacement set = {. . ., -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, . . .}

Solution set for given inequation S = {. . . -12, -11, -10}

(b) x ≥ 7, x ∈ N

Replacement set = {1, 2, 3, 4, 5, . . .}

Solution set for given inequation S = {7, 8, 9, . . . }

(c) x > 8, x ∈ W

Replacement set = {0, 1, 2, 3, 4, 5, 6, . . .}

Solution set for given inequation S = {9, 10, 11, 12, 13, . . }

Given linear inequation is -5 < x ≤ 5

It has two cases.

Case I: -5 < x

Case II: x ≤ 5

(i) x ∈ I

Replacement set = {. . ., -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, . . .}

Solution set for -5 < x is {-4, -3, -2, -1, 0, . . .} = P

Solution set for x ≤ 5 is {. . . 1, 2, 3, 4, 5} = Q

Therefore, required solution set S = P ∩ Q

= {-4, -3, -2, -1, 0, . . .} ∩ {. . . 1, 2, 3, 4, 5}

= {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

(ii) x ∈ W

Replacement set = {0, 1, 2, 3, 4, 5, 6, . . .}

Solution set for -5 < x is {0, 1, 2, 3, 4, 5, 6, . . .} = P

Solution set for x ≤ 5 is {0, 1, 2, 3, 4, 5} = Q

Therefore, required solution set S = P ∩ Q

= {0, 1, 2, 3, 4, 5}

(iii) x ∈ N

Replacement set = {1, 2, 3, 4, 5, . . .}

Solution set for -5 < x is {1, 2, 3, 4, 5, . . .} = P

Solution set for x ≤ 5 is {1, 2, 3, 4, 5} = Q

Therefore, required solution set S = P ∩ Q

= {1, 2, 3, 4, 5}

(a) x – 6 < 4, x ∈ W

Add 6 to both sides of the linear inequation.

x – 6 + 6 < 4 + 6

x < 10

Replacement set = {0, 1, 2, 3, 4, 5, 6, . . .}

Solution set for x < 10 is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(b) 4x + 7 > 15, x ∈ N

Subtract 7 from both sides of the inequation.

4x + 7 – 7 > 15 – 7

4x > 8

Divide each side of the inequation by 4.

4x/4 > 8/4

x > 2.

Replacement set = {1, 2, 3, 4, 5, . . .}

Solution set for x > 2 is {3, 4, 5, . . .}

(c) x/2 + 5 ≤ 6, x ∈ I

Subtract 5 from both sides of the inequation.

x/2 + 5 – 5 ≤ 6 – 5

x/2 ≤ 1

Multiply each side of the inequation by 2.

x/2 x 2 ≤ 1 x 2

x ≤ 2

Replacement set = {. . ., -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, . . .}

Solution set for x ≤ 2 is {. . . -1, 0, 1, 2}

(a) -x/3 > 5

Multiply each side of the inequation by -3.

-x/3 x (-3) < 5 x (-3)

x < -15

Replacement set = {. . ., -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, . . .}

Therefore, solution set for x > -15 is {. . . -17, -16}

(b) 6x – 7 ≥ 17

Add 7 to both sides.

6x – 7 + 7 ≥ 17 + 7

6x ≥ 24

Divide each side by 6.

6x/6 ≥ 24/6

x ≥ 4

Replacement set = {. . ., -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, . . .}

Therefore, solution set for x ≥ 4 is {4, 5, 6, 7, . . .}

(c) -3x > 12

Divide each side by -3.

-3x/(-3) > 12/(-3)

x < -4

Replacement set = {. . ., -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, . . .}

Therefore, solution set for x < -4 is {. . . -8, -7, -6, -5}

(a) x/2 > x/3 + 1, x ∈ W

Move variable x to one side and constants to other side.

x/2 – x/3 > 1

L.C.M of 3, 2 is 6.

(3x – 2x) / 6 > 1

x/6 > 1

Divide each side by (6)

x/6 x (6) > 1 x (6)

x > 6

Replacement set = {0, 1, 2, 3, 4, 5, 6, . . .}

Therefore, solution set for x > 6 is {7, 8, 9, . .}

(b) 3x + 5 < 4x – 6, x ∈ W

Move variable x to one side and constants to other side.

5 – 6 < 4x – 3x

-1 < x

Replacement set = {0, 1, 2, 3, 4, 5, 6, . . .}

Therefore, solution set for -1 < x is {0, 1, 2, 3, 4, 5, 6, . . .}

(c) x/6 – 1 ≥ 3, x ∈ W

Add 1 to both sides.

x/6 ≥ 3 + 1

x/6 ≥ 4

Multiply each side by 6.

x/6 x 6 ≥ 4 x 6

x ≥ 24

Replacement set = {0, 1, 2, 3, 4, 5, 6, . . .}

Therefore, solution set for x ≥ 24 is {24, 25, 26, . . .}

(d) 8x – 9 > 9x + 10, x ∈ W

Move variable x to one side and constants to other side.

-9 – 10 > 9x – 8x

-19 > x

Replacement set = {0, 1, 2, 3, 4, 5, 6, . . .}

Therefore, solution set for -19 > x is {0, 1, 2, 3, 4, 5, 6, . . .}

(a) 7 – 3x < – 2, x ∈ N

Add 3x to each side

7 – 3x + 3x < -2 + 3x

7 < -2 + 3x

Add 2 to both sides

7 + 2 < -2 + 2 + 3x

9 < 3x

Divide each side by 3.

9/3 < 3x/3

3 < x

Replacement set = {1, 2, 3, 4, 5, . . .}

Therefore, solution set for 3 < x is {4, 5, 6, 7. . }

(b) 8x – 5 ≥ 18, x ∈ N

Add 5 to each side

8x – 5 + 5 ≥ 18 + 5

8x ≥ 23

Divide each side by 8.

8x/8 ≥ 23/8

x ≥ 2.87

Replacement set = {1, 2, 3, 4, 5, . . .}

Therefore, solution set for x ≥ 2.87 is {2.87, 3, 4, 5, . . }

(c) 5x/2 > 1, x ∈ N

Multiply each side by 2.

5x/2 x 2 > 1 x 2

5x > 2

Divide each side by 5.

5x/5 > 2/5

x > 0.4

Replacement set = {1, 2, 3, 4, 5, . . .}

Therefore, solution set for x > 0.4 is {1, 2, 3, 4, 5, . . .}

(a) [(5a – 1) / 8] – [(3a – 2) / 7] + [(a – 5) / 4]

L.C.M of denominators 8, 7, 4 is 56

= [7(5a – 1) / 56] – [8(3a – 2) / 56] + [14(a – 5) / 56]

= [(35a – 7) / 56] – [(24a – 16) / 56] + [(14a – 70) / 56]

= (35a – 7 – (24a – 16) + 14a – 70) / 56

= (35a – 7 – 24a + 16 + 14a – 70) / 56

= (25a – 61) / 56

(b) [5a / (a – 5)] – [a² / (a² – 25)]

The lowest common multiple of denominators is (a – 5), (a² – 25) = (a – 5), (a + 5)(a – 5) = (a + 5)(a – 5)

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by (a + 5) / (a + 5) in case of [5a / (a – 5)], 1/ 1 in case of [a² / (a² – 25)].

Therefore, [5a / (a – 5)] – [a² / (a² – 25)]

= [5a(a + 5) / (a + 5)(a – 5)] – [a² / (a + 5)(a – 5)]

= [(5a² + 25a – a²] / (a + 5)(a – 5)

= (4a² + 25a) / (a + 5)(a – 5)

(c) [2(x – 3) / (x² -5x +6)] + [3(x – 1) / (x² – 4x + 3)] + [5(x – 2) / (x² – 3x + 2)]

The least common multiple of (x² -5x +6), (x² – 4x + 3), (x² – 3x + 2) is calculated by finding the factors.

Factors of (x² -5x +6), (x² – 4x + 3), (x² – 3x + 2) are (x² – 3x – 2x + 6), (x² – 3x – x + 3), (x² – 2x – x + 2)

= (x(x – 3) – 2(x – 3)), (x(x – 3) – 1(x – 3)), (x(x – 2) -1(x – 2))

= (x – 3)(x – 2), (x – 3)(x – 1), (x – 2)(x – 1)

The L.C.M of (x² -5x +6), (x² – 4x + 3), (x² – 3x + 2) is (x – 3)(x – 2)(x – 1).

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by (x – 1) / (x – 1) in case of [2(x – 3) / (x² -5x +6)], (x – 2) / (x – 2) in case of [3(x – 1) / (x² – 4x + 3)], (x – 3) / (x – 3) in case of [5(x – 2) / (x² – 3x + 2)].

Therefore, [2(x – 3) / (x² -5x +6)] + [3(x – 1) / (x² – 4x + 3)] + [5(x – 2) / (x² – 3x + 2)]

= [2(x – 3)(x – 1) / (x – 3)(x – 2)(x – 1)] + [3(x – 1)(x – 2) / (x – 3)(x – 2)(x – 1)] + [5(x – 2)(x – 3) / (x – 3)(x – 2)(x – 1)]

= [2(x – 3)(x – 1) + 3(x – 1)(x – 2) + 5(x – 2)(x – 3)] / (x – 3)(x – 2)(x – 1)

= [2(x² – x – 3x + 3) + 3(x² -x – 2x +2) +5(x² – 2x – 3x + 6)] / (x – 3)(x – 2)(x – 1)

= [2(x² – 4x + 3) + 3(x² – 3x + 2) + 5(x² – 5x + 6)] / (x – 3)(x – 2)(x – 1)

= [2x² – 8x + 6 + 3x² – 9x + 6 + 5x² – 25x + 30] / (x – 3)(x – 2)(x – 1)

= [10x² – 42x + 42] / (x – 3)(x – 2)(x – 1)

= 2[5x² – 21x + 21] / (x – 3)(x – 2)(x – 1)

(a) [(x + y) / (2x – 3)] x [(x – y) / (2x + 3)]

Multiply the numerator and denominator separately.

= (x + y)(x – y) / (2x – 3)(2x + 3)

= (x² – y²) / (4x² – 9)

(b) [(14a² – 7a) / (12a³ + 24a²)] ÷ [(2a – 1) / (a² + 2a)]

= [(14a² – 7a) / (12a³ + 24a²)] x [(a² + 2a) / (2a – 1)]

= (14a² – 7a) (a² + 2a) / (12a³ + 24a²) (2a – 1)

= 7a²(2a – 1) (a + 2) / 12a²(a + 2) (2a – 1)

Cancel the common terms a², (2a – 1), and (a + 2) in both numerator and denominator.

= 7 / 12

(c) [(m² – m – 12) / 5m] x [(m³ – m) / (m² – 9)]

= (m² – m – 12) (m³ – m) / 5m(m² – 9)

= m(m² – 4m + 3m – 12) (m² – 1) / 5m (m² – 3²)

Cancel the common term m in both numerator and denominator.

= (m(m – 4) + 3(m – 4) (m² – 1) / 5(m + 3) (m – 3)

= (m + 4) (m – 3) (m² – 1) / 5(m + 3) (m – 3)

Cancel the common term (m – 3) in both numerator and denominator.

= (m + 4) (m² – 1) / 5(m + 3).

(a) (abc + bc²) / (acd + dc²)

= bc(a + c) / dc(a + c)

Cancel the common term c(a + c) in both numerator and denominator.

= b / d.

(b) (4a² – 9b²) / (4a² + 6ab)

= [(2a)² – (3b)²] / (4a² + 6ab)

= (2a + 3b) (2a – 3b) / 2a(2a + 3b)

Cancel the common term (2a + 3b) in both numerator and denominator of the algebraic fraction.

= (2a – 3b) / 2a.

(c) [(a + 1)³ – (a – 1)³] / (3a³ + a)

= [(a³ + 3a² + 3a + 1) – (a³ – 3a² + 3a – 1)] / a(3a² + 1)

= [a³ + 3a² + 3a + 1 – a³ + 3a² – 3a + 1] / a(3a² + 1)

= (6a² + 2) / a(3a² + 1)

= 2(3a² + 1) / a(3a² + 1)

Cancel the common term (3a² + 1) in both numerator and denominator.

= 2 / a.

(a) [(a – b) / b)] + [(a + b) / b] – [(a² – b²) / 2ab]

The least common multiple of denominators b, b, 2ab is 2ab.

To make the fractions having the common denominator both numerator and denominator of these are to be multiplied by 2a / 2a in case of [(a – b) / b)] and [(a + b) / b], 1 / 1 in case of [(a² – b²) / 2ab].

Therefore, [(a – b) / b)] + [(a + b) / b] – [(a² – b²) / 2ab]

= [2a(a – b) / 2ab] + [2a(a + b) / 2ab] – [(a² – b²) / 2ab]

= [2a(a – b) + 2a(a + b) – (a² – b²)] / 2ab

= [2a² – 2ab + 2a² + 2ab – a² + b²] / 2ab

= [3a² + b²] / 2ab.

(b) (ab – 3b²)² / (a²b² – 27b⁵)

= (a²b² – 6ab³ + 9b⁴) / (a²b² – 27b⁵)

= b²(a² – 6ab + 9b²) / b²(a² – 27b³)

Cancel the common term b² in both numerator and denominator.

= (a – 3b)² / (a² – (3b)³)

(c) [(a³ – 2a)² – (a² – 2)²] / [(a – 1)(a + 1) (a² – 2)]

= [(a⁶ – 4a⁴ + 4a²) – (a⁴ – 4a² + 4)] / [(a – 1)(a + 1) (a² – 2)]

= [a⁶ – 4a⁴ + 4a² – a⁴ + 4a² – 4] / [(a – 1)(a + 1) (a² – 2)]

= [a⁶ – 5a⁴ + 8a² – 4] / [(a – 1)(a + 1) (a² – 2)].

1 / (a² – 5a + 6) – 1 / (a² – 4a + 3)

Get the factors of the denominators (a² – 5a + 6), (a² – 4a + 3)

= (a² – 3a – 2a + 6), (a² – 3a – a + 3)

= (a(a – 3) – 2(a – 3)), (a(a – 3) -1(a – 3))

=(a – 3) (a – 2), (a – 1)(a – 3)

The L.C.M of denominators is (a – 3) (a – 2) (a – 1).

= (a – 1) / (a – 3) (a – 2) (a – 1) – (a – 2) / (a – 3) (a – 2) (a – 1)

= [(a – 1) – (a – 2)] / (a – 3) (a – 2) (a – 1)

= (a – 1 – a + 2) / (a – 3) (a – 2) (a – 1)

= 1 / (a – 3) (a – 2) (a – 1)

a – (a – 1) / 2 + (a – 2) / 6

The lowest common multiple of 2, 6 is 6.

= a/6 – (a-1)/6 + (a-2)/6

=[a – (a – 1) + a – 2] / 6

= [a – a + 1 + a – 2] / 6

= (a – 1) / 6.

[(25x² – 16a²) / (ax + 2a²) x [2a / (x – 2a)] x [(x + 2a) / (5x – 4a)]

Multiply the numerators and denominators together.

= [((5x)² – (4a)²) / a(x + 2a)] x [2a / (x – 2a)] x [(x + 2a) / (5x – 4a)]

= [(5x – 4a) (5x + 4a) * 2a * (x + 2a)] / [a(x + 2a) (x – 2a) (5x – 4a)]

Cancel the common terms.

[5x / (x² – 25)] + [(x² + x – 20) / (x² + 2x – 15)]

Denominators of the two algebraic fractions are (x² – 25), (x² + 2x – 15)

The factors of denominators are (x² – 5²), (x² + 5x – 3x – 15)

= (x – 5) (x + 5), (x(x + 5) -3(x + 5))

= (x – 5) (x + 5), (x + 5) (x – 3)

L.C.M of denominators are (x – 5) (x + 5) (x – 3).

= [5x(x – 3)/ (x – 5) (x + 5) (x – 3)] + [(x² + 5x – 4x – 20) / (x – 5) (x + 5) (x – 3)]

= [(5x² – 3x) / (x – 5) (x + 5) (x – 3)] + [x² + x – 20 / (x – 5) (x + 5) (x – 3)]

= [5x² – 3x + x² + x – 20] / (x – 5) (x + 5) (x – 3)

= [6x² – 2x – 20] / (x – 5) (x + 5) (x – 3)

= 2(3x² – x – 10] / (x – 5) (x + 5) (x – 3)

= 2(3x² – 6x +5x – 10) / (x – 5) (x + 5) (x – 3)

= 2(3x(x – 2) + 2(x – 2)) / (x – 5) (x + 5) (x – 3)

= 2(3x + 2) (x – 2) / (x – 5) (x + 5) (x – 3)

[3x / 4a²b] – [7 / 6ab⁵] – [5x / 2ab²]

L.C.M of 4a²b, 6ab⁵, and 2ab² is 12a²b⁵.

= [3x . 3 . b⁴ / 12a²b⁵] – [7 . 2. a / 12a²b⁵] – [5x . 6ab³ / 12a²b⁵]

= (9xb⁴ – 14a – 30ab³x) / 12a²b⁵

[(2a – b) / 10a] – [b / 2a] + [(2b – a) / 15a]

The least common multiple of 10a, 2a, 15a is 30a.

= [3(2a – b) / 30a] – [15b / 30a] + [2(2b – a) / 30a]

= [6a – b – 15b + 4b – a] / 30a

= (5a – 12b) / 30a.

[(3m² – 9m) / (16m² – 1)] ÷ [(4m² – m) / (m – 2)]

Reverse the second algebraic expression and multiply it with the first one.

= [(3m² – 9m) / (16m² – 1)] x [(m – 2) / (4m² – m)]

= [(3m² – 9m) / ((4m)² – 1²) / (3m² – 9m)] x [(m – 2) / (4m² – m)]

= [(3m² – 9m) (m – 2)] / [(4m – 1) (4m + 1) (4m² – m)]

= [m(3m – 9) (m – 2)] / [(4m – 1) (4m + 1) m (4m – 1)]

Cancel the common term m in both the numerator and denominator.

= [(3m – 9) (m – 2)] / [(4m – 1)² (4m + 1)]

(a) [3x + 7] / [6x² – 25x – 91]

Get the factors of the denominator.

= (3x + 7) / (6x² + 14x – 39x – 91)

= (3x + 7) / (2x(3x + 7) – 13(3x + 7))

= (3x + 7) / (2x – 13) (3x + 7)

Cancel the common factor (3x + 7) in the numerator and denominator to get the reduced form,

= 1 / (2x – 13)

(b) [5a³ + 8a²b + ab² – 2b³] / [2a³ + 9a²b- 8ab² – 15b³]

Find the factors of numerator and denominator of the algebraic expression.

= [(5a – b) (a² + 2ab + b²)] / [(2a – 3b) (a² + 5b² + 6ab)]

= [(5a – b) (a² + ab + ab + b²)] / [(2a – 3b) (a² + 5ab + ab + 5b²)]

= [(5a – b) (a(a + b) +b(a + b)] / [(2a – 3b) (a(a + 5b) + b(a + 5b)]

= [(5a – b) (a + b) (a + b)] / [(2a – 3b) (a + b) (a + 5b)]

Cancel the common factor (a + b) in both the numerator and denominator of the fraction.

= [(5a – b) (a + b)] / [(2a – 3b) (a + 5b)]

Get the factors of each polynomial.

= [(c² – 5c – 5c + 25) / (4c² + 10c + 2c + 5)] ÷ [4c² + 6c + 2c + 3) / (c² + 3c + c+ 3)]

= [(c(c – 5) – 5(c – 5)) / (2c(c + 5) + 1(2c + 5)] ÷ [2c(2c + 3) +1(2c + 3)) / (c(c + 3) +1(c + 3))]

= [(c – 5) (c – 5) / (2c + 5) (c + 5)] ÷ [(2c + 1)(2c + 3) / (c + 1)(c + 3)]

= [(c – 5) (c – 5) / (2c + 5) (c + 5)] x [(c + 1)(c + 3) / (2c + 1)(2c + 3)]

= [(c – 5)² (c + 1)(c + 3) / (2c + 5) (c + 5) (2c + 1)(2c + 3)]

(a) [1/(6a-2)] – [1/2(a-⅓)] + [1/(1-3a)]

= [1/(6a-2)] – [1/2(3a-1) / 3)] + [1/(1-3a)]

= [1/(6a-2)] – [3/2(3a – 1)] + [1/(1-3a)]

= [1/(6a-2)] – [3/2(3a – 1)] – [1/(3a)-1]

L.C.M of (6a – 2), (3a – 1), (3a – 1) is 2(3a – 1).

= [1 / 2(3a – 1)] – [3 / 2(3a – 1)] – [2/ 2(3a – 1)]

= [1 – 3 – 2] / 2(3a – 1)

= -4/2(3a – 1)

= -1/ (3a – 1)

(b) [(a⁴ – b⁴) / a²b²] : [(1 + b²/a²)(1 – 2a/b + a²/b²)]

(1 – 2a/b + a²/b²)

L.C.M of b, b² is b².

(b² – 2ab + a²) / b²

= (a – b)² / b²

(1 + b²/a²)

= (a² + b²) / a²

(1 + b²/a²)(1 – 2a/b + a²/b²) = ((a² + b²) / a²) ((a – b)² / b²)

Multiply numerators and denominators.

= (a² + b²) (a – b)² / a²b²

[(a⁴ – b⁴) / a²b²] : [(1 + b²/a²)(1 – 2a/b + a²/b²)]

= [(a⁴ – b⁴) / a²b²] : [(a² + b²) (a – b)² / a²b²]

= [(a⁴ – b⁴) (a² + b²) (a – b)²] / [a²b² a²b²]

= [(a⁴ – b⁴) (a² + b²) (a – b)²] / [a⁴b⁴]

(a) [2(x – 3) / x² – 5x + 6] + [3(x – 1) / x² – 4x + 3] + [5(x – 2) / x² – 3x + 2]

We observe that the denominators of three fractions are (x² – 5x + 6), (x² – 4x + 3), and (x² – 3x + 2)

The factors of denominators are

= (x² – 3x – 2x + 6), (x² – 3x – x + 3), (x² – 2x – x + 2)

= (x(x – 3) -2(x – 3)), (x(x – 3) – 1(x – 3)), (x(x – 2)-1(x – 2))

= (x-3)(x-2), (x-1)(x-3), (x-1)(x-2)

Therefore, required lowest common multiple of denominators is (x-3)(x-2)(x-1).

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by (x-1) ÷ (x-1) in case of [2(x – 3) / x² – 5x + 6], (x – 2) ÷ (x – 2) in case of [3(x – 1) / x² – 4x + 3], (x -3) ÷ (x – 3) in case of [5(x – 2) / x² – 3x + 2].

Therefore, [2(x – 3) / x² – 5x + 6] + [3(x – 1) / x² – 4x + 3] + [5(x – 2) / x² – 3x + 2]

= [2(x – 3)(x – 1) / (x-3)(x-2)(x-1)] + [3(x – 1)(x – 2) / (x-3)(x-2)(x-1)] + [5(x – 2)(x – 3) / (x-3)(x-2)(x-1)]

= [2(x – 3) (x – 1) + 3(x-1)(x – 2) + 5(x – 2)(x – 3)] / (x-3)(x-2)(x-1)

= [2x²- 8x+ 6 + 3x²- 9x+ 6 + 5x²- 25x+ 30] / (x-3)(x-2)(x-1)

= [10x² – 42x + 42] / (x-3)(x-2)(x-1)

= 2(5x² – 21x +21)/ (x-3)(x-2)(x-1)

(b) [a / (a² – b²)] – [1 / (a – b)] + 1 / (a + b) + 1/a – 1/b + [(a² – ab + b²) / (ab(a – b))]

We observe that the denominators of algebraic fractions are (a² – b²), (a – b), (a + b), a, b, (ab(a – b))

L.C.M of denominators is ab (a² – b²)².

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by ab(a² – b²) ÷ ab (a² – b²) in case of a / (a² – b²), ab(a + b)(a² – b²) ÷ ab(a + b)(a² – b²) in case of 1 / (a – b), ab(a – b)(a² – b²) ÷ ab(a – b)(a² – b²) in case of 1 / (a + b), b(a² – b²)² ÷ b(a² – b²)² in case of 1/a, a(a² – b²)² ÷ a(a² – b²)² in case of 1/b, (a+b) (a² – b²) ÷ (a + b)(a² – b² in case of [(a² – ab + b²) / (ab(a – b)).

Therefore, [a / (a² – b²)] – [1 / (a – b)] + 1 / (a + b) + 1/a – 1/b + [(a² – ab + b²) / (ab(a – b))]

= [a²b(a² – b²) / ab (a² – b²)²] – [ab(a + b)(a² – b²) / ab (a² – b²)²] + [ab(a – b)(a² – b²) / ab (a² – b²)²] + [b(a² – b²)² / ab (a² – b²)²] – [a(a² – b²)²/ ab (a² – b²)²] + [(a+b) (a² – b²)(a² – ab + b²) / ab (a² – b²)²]

= [a²b(a² – b²) – ab(a + b)(a² – b²) + ab(a – b)(a² – b²) + b(a² – b²)² + a(a² – b²)² + (a+b) (a² – b²)(a² – ab + b²)] / ab (a² – b²)²

= [a⁴b-a²b³ – (a⁴b-a²b³+a³b²-ab⁴) + a⁴b-a²b³-a³b²+ab⁴ + ba⁴-2b³a²+b⁵ + a⁵-2a³b²+ab⁴ + a⁵-a³b²+2a²b³-b³a²-b⁵] /ab (a² – b²)²

= [a⁴b-a²b³ – a⁴b+a²b³-a³b²+ab⁴ + a⁴b-a²b³-a³b²+ab⁴ + ba⁴-2b³a²+b⁵ + a⁵-2a³b²+ab⁴ + a⁵-a³b²+2a²b³-b³a²-b⁵] /ab (a² – b²)²

= [2a⁴b – 5a³b² + 3ab⁴ – 2b³a² + 2a⁵] / ab (a² – b²)²

(c) (x – y) / xy + (y – z) / yz + (z – x) / zx

We observe that the denominators of algebraic fractions are xy, yz, zx.

The L.C.M of denominators is xyz.

To make the fractions having a common denominator both the numerator and denominator of these are to be multiplied by z ÷ z in case of (x – y) / xy, x ÷ x in case of (y – z) / yz, y ÷ y in case of (z – x) / zx.

Therefore, (x – y) / xy + (y – z) / yz + (z – x) / zx

= (x – y)z / xyz + (y – z)x / xyz + (z – x)y / xyz

= (xz – yz) /xyz + (xy – xz) / xyz + (yz -xz) / xyz

= (xz – yz + xy – xz + yz – xz) / xyz

= 1 / xyz.

(a) [x² + 2xy + y² / (x – y)] – [x² – 2xy + y²/ (x + y)] – 1 / (x² – y²)

We observe that the denominators of three fractions are (x – y), (x + y), and (x² – y²)

Therefore, L.C.M of denominators = (x² – y²)

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by (x + y) ÷ (x + y) in case of [x² + 2x + 1 / (x – y)], (x – y) ÷ (x – y) in case of [x² – 2x + 1/ (x + y)].

Therefore, [x² + 2xy + y² / (x – y)] – [x² – 2xy + y²/ (x + y)] – 1 / (x² – y²)

= [(x² + xy + xy + 1) / (x – y)] – [(x² – xy – xy + 1) / (x + y)] – 1 / (x²- y²)

= [(x+y) (x + y) (x + y) / (x – y) (x + y)] – [(x – y) (x – y) (x – y) / (x – y) (x + y)] – 1 / (x² – y²)

= [(x+y) (x + y) (x + y) – (x – y) (x – y) (x – y) – 1] / [(x – y) (x + y)]

= [(x + y)³ – (x – y)³ – 1] / [(x – y) (x + y)]

= [(x³ + y³ + 3x²y + 3xy²) – (x³ – y³ – 3x²y + 3xy²) – 1] / [(x – y) (x + y)]

= [x³ + y³ + 3x²y + 3xy² – x³ + y³ + 3x²y – 3xy² – 1] / [(x – y) (x + y)]

= [2y³ + 6x²y – 1] / [(x – y) (x + y)]

(b) 5a / (a² – 25) + [(a + 2) / (a² + 10a + 25)] – [(a³ + 2a² + a) / (a² – 10a + 25)]

We observe that the denominators of three fractions are (a² – 25), (a² + 10a + 25), and (a² – 10a + 25)

The factors of denominators are

= (a² – 5²), (a² + 5a + 5a + 25), (a² – 5a – 5a + 25)

= (a + 5) (a – 5), (a + 5) (a + 5), (a – 5) (a – 5)

So, L.C.M of denominators is (a – 5)(a – 5)(a + 5)(a + 5)

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by (a – 5)(a + 5) ÷ (a – 5)(a + 5) in case of 5a / (a² – 25), (a – 5)(a – 5) ÷ (a – 5)(a – 5) in case of [(a + 2) / (a² + 10a + 25)], (a + 5)(a + 5) ÷(a + 5)(a + 5) in case of [(a³ + 2a² + a) / (a² – 10a + 25)].

Therefore, 5a / (a² – 25) + [(a + 2) / (a² + 10a + 25)] – [(a³ + 2a² + a) / (a² – 10a + 25)]

= 5a (a – 5)(a + 5)/ (a – 5)(a – 5)(a + 5)(a + 5) + [(a + 2)(a – 5)(a – 5) / (a – 5)(a – 5)(a + 5)(a + 5)] – [(a³ + 2a² + a) (a + 5)(a + 5) / (a – 5)(a – 5)(a + 5)(a + 5)]

= [5a (a – 5)(a + 5) + (a + 2)(a – 5)(a – 5) – (a³ + 2a² + a) (a + 5)(a + 5)] / (a – 5)(a – 5)(a + 5)(a + 5)

= [5a (a – 5)(a + 5) + (a + 2)(a – 5)(a – 5) – a(a² + 2a + 1) (a + 5)(a + 5)] / (a – 5)(a – 5)(a + 5)(a + 5)

= [5a (a – 5)(a + 5) + (a + 2)(a – 5)(a – 5) – a (a + 1) (a + 1) (a + 5)(a + 5)] / (a – 5)(a – 5)(a + 5)(a + 5)

= [5a³ – 125a + a³ -8a² + 5a + 50 – (a⁵ +12a⁴ +46a³ +60a² + 25a)] / (a – 5)(a – 5)(a + 5)(a + 5)

= [6a³ – 8a² – 120a + 50 – a⁵ -12a⁴ -46a³ -60a² – 25a] / (a – 5)(a – 5)(a + 5)(a + 5)

= [- a⁵ -12a – 40a³ – 68a² – 140a + 50] / (a – 5)(a – 5)(a + 5)(a + 5)

(c) [5x / (2x – 1)] – [(x – 2) / x] + [2x / (x + 2)]

We observe that the denominators of three fractions are (2x – 1), x, and (x + 2)

The L.C.M of denominators is x(2x – 1) (x + 2).

To make the fractions having common denominator both numerator and denominator of these are to be multiplied by x(x + 2) ÷ x(x + 2) in case of 5x / (2x – 1), (x + 2) (2x – 1) ÷ (x + 2) (2x – 1) in case of [(x – 2) / x], x(2x – 1) ÷x(2x – 1) in case of [2x / (x + 2)].

Therefore, [5x / (2x – 1)] – [(x – 2) / x] + [2x / (x + 2)]

= [5x²(x + 2) / x(2x – 1) (x + 2)] – [(x + 2) (2x – 1)(x – 2)/ x(2x – 1) (x + 2)] + [2x²(2x – 1)/ x(2x – 1) (x + 2)]

= [5x²(x + 2) – (x + 2) (2x – 1)(x – 2) + 2x²(2x – 1)] / x(2x – 1) (x + 2)

= [5x³ + 10x² – 2x³ – x² – 8x + 4 + 4x³ – 2x²] / x(2x – 1) (x + 2)

= [7x³ + 7x² – 8x + 4] / x(2x – 1) (x + 2)

(a) [a(a + b) / (a – b)] x [(a – b) / b(a + b)] x a/b

Multiply the numerators together and denominators together.

= [a(a + b) (a – b) a] / [(a – b) (a + b) b b]

= [a² (a + b) (a – b)] / [b² (a + b) (a – b)

Cancel the common terms ( a+ b), (a – b) in both numerator and denominators.

= a² / b², which is the reduced form of the algebraic expression.

(b) [(x² – 2x) / (x² + 3x – 10)] ÷ [(x² + 4x – 21) / (x² + 2x – 15)]

Find the factors of the numerator and denominator of the algebraic expressions.

= [x(x – 2) / (x² + 5x – 2x – 10)] ÷ [(x² + 7x – 3x – 2) / (x² +5x – 3x – 15)]

= [x(x – 2) / (x(x + 5) – 2(x + 5)] ÷ [(x(x + 7) – 3 (x + 7)) / (x (x + 5) – 3 (x + 5))]

= [x(x – 2)/ (x – 2) (x + 5)] ÷ [(x + 7) (x – 3) / (x + 5) (x – 3)]

Cancel the common terms in the numerator and denominators.

= [x / (x + 5)] ÷ [(x + 7) / (x + 5)]

Reverse the second fraction and multiply with the first one.

= [x / (x + 5)] x [(x + 5) / (x + 7)]

= [x(x + 5) / (x + 5) (x + 7)]

Cancel the common term (x + 5) in both numerator and denominator.

= x / (x + 7), which is the lowest term of the algebraic expression.

(c) [(a² – 15a + 4)/(a² – 7a + 10)] x [(a² – a – 2) / (a² + 2a – 3)] ÷ [(a² – 5a + 4) / (a² + 8a + 15)]

Find the factors of the numerator and denominator of the algebraic expressions.

= [(a² – 15a + 4) / (a² – 5a – 2a + 10)] x [(a² – 2a + a – 2) / (a² + 3a – a  – 3)] ÷ [(a² – 4a – a + 4) / (a² + 5a + 3a + 15)]

= [(a² – 15a + 4) / (a(a – 5) – 2(a – 5))] x [(a(a – 2) + 1(a – 2) / (a(a+3) – 1(a +3)] ÷ [(a(a – 4) -1(a – 4) / (a(a + 5) + 3(a + 5)]

= [(a² – 15a + 4) / (a – 5)(a – 2)] x [(a – 2)(a + 1) / (a – 1) (a + 3)] ÷ [(a -1) (a -4) / (a + 3) (a + 5)]

= [(a² – 15a + 4) (a – 2)(a + 1)/ (a – 5)(a – 2)(a – 1) (a + 3)] ÷ [(a -1) (a -4) / (a + 3) (a + 5)]

Cancel the common terms in the numerator and denominator.

= [(a² – 15a + 4) (a + 1) / (a – 5) (a + 3)] x [(a + 3) (a + 5) / (a -1) (a -4)]

= [(a² – 15a + 4) (a + 1) (a + 3) (a + 5) / (a – 5) (a + 3)(a -1) (a -4)]

= [(a² – 15a + 4) (a + 1) (a + 5) / (a – 5) (a – 4)]

(a) [x² – 5x] / [x – 5]

The factors of numerator and denominator are

= x(x – 5) / (x – 5)

Cancel the common term (x – 5) in both denominator and numerator.

= x, which is the lowest form of the given algebraic fraction.

(b) [a² – 49] / [a² – 7a]

Factorize the numerator and denominator of the given algebraic fraction,

= [a² – 7²] / [a² – 7a]

= (a + 7) (a – 7) / a(a – 7)

Cancel the common term (a – 7) in both numerator and denominator.

= (a + 7) / a, which is the lowest form of the given algebraic fraction.

(c) [a² + 6a + 5] / [a² + 8a + 15]

By factorizing the numerator, denominator of the given algebraic fraction

= [a² + 5a + a + 5] / [a² + 5a + 3a + 15]

= [a(a + 5) + 1(a + 5)] / [a(a + 5) + 3(a + 5)]

= (a + 5) (a + 1) / (a + 5) (a + 3)

We observed that in the numerator and denominator of the algebraic fraction, (a + 5) is the common factor.

Now, when the numerator and denominator of the algebraic fraction is divided by this common factor or their H.C.F. the algebraic fraction becomes,

= [(a + 5) (a + 1)] / (a + 5) / [(a + 5) (a + 3)] / (a + 5)

= (a + 1) / (1 + 3), which is the lowest form of the given algebraic fraction.

(a) ab / [a²x² – ax]

Factorize the numerator and denominator of the algebraic fraction

= ab / ax[ax – 1]

We can observe that in the denominator and numerator of the algebraic fraction, a is the common factor. Now cancel the common factor and write the remaining terms as it is.

= b / x [ax – 1], which is the lowest form of the given algebraic fraction.

(b) 20x²y²z² / 25(y² – x²y²)

Get the factors of the numerator and denominator of the algebraic fraction.

= [5 x 4 x x²y²z²] / [5 x 5 x y²(1 – x²)]

= [5 x 4 x x²y²z²] / [5 x 5 x y²(1² – x²)]

= [5 x 4 x x²y²z²] / [5 x 5 x y² (1 – x) (1 + x)]

We observed that in the numerator and denominator of the algebraic fraction, 5y² is the common factor. Now cancel the common factor in both and write the remaining factor as it is.

= [4 x x²z²] / [5 (1 – x) (1 + x)]

= 4x²z² / 5(1 – x) (1 + x), which is the lowest form of the given algebraic fraction.

(c) [25p² – 16q²] / [5p² + 50p + 4pq+ 40q]

Factorize the numerator, denominator of the algebraic fraction.

= [(5p)² – (4q)²] / [5p(p + 10) + 4q(P + 10)]

= (5p – 4q) (5p + 4q) / [(5p + 4q) (p + 10)]

We observed that in the numerator and denominator of the algebraic fraction, (5p + 4q) is the common factor. Now cancel the common factor in both and write the remaining factor as it is.

= (5p – 4q) / (p + 10), which is the lowest form of the given algebraic fraction.

(a) [x(2a² – 3ax)] / [a(4a²x – 9x³)]

Get the factors of the numerator and denominator of the algebraic fraction.

= [x(2a² – 3ax)] / [ax(4a² – 9x²)]

= [x(2a² – 3ax)] / [ax((2a)² – (3x)²)]

= [ax(2a – 3x)] / [ax(2a – 3x) (2a + 3x)]

We observed that in the numerator and denominator of the algebraic fraction, (2a + 3x), a, x are the common factors. Now cancel the common factors in both and write the remaining factor as it is.

= 1 / (2a + 3x), which is the lowest form of the given algebraic fraction.

(b) [x³ + 9x² + 20x] / [x² + 2x – 15]

Factorize the algebraic expressions.

= [x(x² + 9x + 20)] / [x² + 5x – 3x – 15]

= [x(x² + 5x + 4x + 20)] / [x(x + 5) – 3(x + 5)]

= [x(x (x + 5) + 4(x + 5)] / (x – 3) (x + 5)

= [x(x + 5) ( x + 4)] / (x – 3) (x + 5)

We observed that in the numerator and denominator of the algebraic fraction, (x + 5) are the common factor. Now cancel the common factor in both and write the remaining factor as it is.

= x(x + 4) / (x – 3), which is the lowest form of the given algebraic fraction.

(c) [x⁴ + 20x³ + 150x² + 500x + 625] / [x² – x – 30]

Get the factors of the numerator and denominator of the given algebraic fraction.

= (x + 5)⁴ / [x² – 6x + 5x – 30]

= (x + 5)⁴ / [x(x – 6) + 5(x – 6)]

= (x + 5)⁴ / (x – 6) (x + 5)

We observed that in the numerator and denominator of the algebraic fraction, (x + 5) are the common factor. Now cancel the common factor in both and write the remaining factor as it is.

= (x + 5)³ / (x – 6), which is the lowest form of the given algebraic fraction.

(a) [3a² – 6ab] / [2a²b – 4ab²]

Get the factors of the numerator and denominator of the given algebraic fraction.

= [3a(a – 2b)] / [2ab(a – 2b)]

We observed that in the numerator and denominator of the algebraic fraction, a(a – 2b) are the common factor. Now cancel the common factor in both and write the remaining factor as it is.

= 3 / 2b, which is the lowest form of the given algebraic fraction.

(b) [x + 2] / [x² – 4]

Find the factors of the numerator and denominator of the fraction.

= (x + 2) / (x² – 2²)

= (x + 2) / (x + 2) (x – 2)

We can observe that in the denominator and numerator of the algebraic fraction, (x + 2) is the common factor. Now cancel the common factor and write the remaining terms as it is.

= 1 / (x – 2), which is the lowest form of the given algebraic fraction.

(a) x² – 2x – 35 and x² + 14x + 45

First polynomial = x² – 2x – 35

= x² – 7x + 5x – 35 = x(x – 7) +5(x – 7)

= (x – 7) (x + 5)

Second Polynomial = x² + 14x + 45

= x² + 9x + 5x + 45 = x(x + 9) + 5(x + 9)

= (x + 9) (x + 5)

In both polynomials, the common factor is (x + 5) and extra factors are (x – 7) and (x + 9).

Therefore, required H.C.F = (x + 5)

Required L.C.M = (x + 5) (x – 7) (x + 9).

(b) x² + 6x+ 9, x² + 3x

First polynomial = x² + 6x+ 9

= x² + 3x + 3x + 9 = x(x + 3) + 3(x + 3)

= (x + 3) (x + 3)

Second Polynomial =x² + 3x

= x(x + 3).

In both polynomials, the common factor is (x + 3), extra common factorsare x, (x + 3).

Therefore, the required highest common factor is (x + 3).

Required least common factor is x (x + 3)².

(c) x³ – 9x² + 14x, x³ – 49x

First polynomial = x³ – 9x² + 14x

= x(x² – 9x + 14) = x(x² -7x -2x +14)

= x(x(x – 7) -2(x – 7)) = x (x – 2) (x – 7)

Second Polynomial = x³ – 49x

= x(x² – 49) = x(x² – 7²)

= x(x + 7) (x – 7)

In both polynomials, the common factors are x(x – 7), extra common factors are (x – 2) (x + 7).

Therefore, required G.C.F is (x(x – 7), L.C.M is x(x – 7) (x – 2) (x + 7).

(a) x⁶ + 9x⁴y + 27x²y² + 27y³, x² + 3y, and x³ + x²y + 3xy + 3y²

First polynomial = x⁶ + 9x⁴y + 27x²y² + 27y³

According to the (a + b)³ = a³ + b³ + 3a²b + 3ab²

= (x²)³ + (3y)³ + 3 (x²)² (3y) + 3 x² (3y)²

= (x² + 3y)³

Second Polynomial = x²+3y

Third Polynomial = x³ + x²y + 3xy + 3y²

= x²(x + y) + 3y(x + y) = (x + y) (x² + 3y)

In three polynomials, the common factor is (x² + 3y), extra common factors are (x + y), (x² + 3y)².

Therefore, required, greatest common factor is (x² + 3y), lowest common multiple is (x² + 3y)³ (x + y).

(b) 2x² – 14x + 20, 6x² – 2x – 20, 3x² – 7x – 20

First polynomial = 2x² – 14x + 20

= 2x² – 4x – 10x + 20 = 2x(x – 2) – 10(x – 2)

= (2x – 10) (x – 2) = 2(x – 5) (x – 2)

Second Polynomial = 6x² – 2x – 20

= 6x² – 12x + 10x – 20 = 6x(x – 2) + 10(x – 2)

= (6x + 10) (x – 2) = 2(3x + 5) (x – 2)

Third Polynomial = 3x² – 7x – 20

= 3x² -12x + 5x – 20 = 3x(x – 4) +5 (x – 4)

= (3x + 5) (x – 4)

In three polynomials, the common factors between first and second polynomial is 2(x – 2), second and third polynomial is (3x + 5), extra common factors are (x – 4) (x – 5).

Therefore, required H.C.F is 1, required L.C.M is 2(x – 2) (3x + 5) (x – 4) (x – 5).

(c) a² + 3a – 4, a² + 5a + 4, and a² – 1

First polynomial = a² + 3a – 4

= a² + 4a – a – 4 = a(a + 4) -1(a + 4)

= (a + 4) (a – 1)

Second Polynomial = a² + 5a + 4

= a² + 4a + a + 4 = a(a + 4) +1(a + 4)

= (a + 1) (a + 4)

Third Polynomial = a² – 1

= a² – 1² = (a + 1) (a – 1)

In three polynomials, the common factors between first polynomial and second polynomial is (a + 4), the second and third polynomial is (a + 1), first and the third polynomial is (a + 1).

Therefore required H.C.F = 1, L.C.M = (a + 4) (a + 1) (a – 1)

(a) pq – np, pq – mq, q² – 3nq + 2n², pq – 2np – mq + 2mn, and pq – np – mq + mn

First polynomial = pq – np

= p(q – n)

Second Polynomial = pq – mq

= q(p – m)

Third Polynomial = q² – 3nq + 2n²

= q² – 2nq – nq + 2n² = q(q – 2n) -n (q – 2n)

= (q – 2n) (q – n)

Fourth Polynomial = pq – 2np – mq + 2mn

= p(q – 2n) – m(q – 2n)

= (p – m) (q – 2n)

Fifth Polynomial = pq – np – mq + mn

= p(q – n) – m(q – n) = (p-m) (q-n)

In all polynomials, the common factors are zero.

Therefore, required lowest common multiple is pq(p – m) (q – n) (q – 2n), G.C.F is 1.

(b) x³ – 7x + 6, x³ + 2x² – 7x + 4

First Polynomial = x³ – 7x + 6

= (x – 2) (x + 3) (x – 1)

Second polynomial = x³ + 2x² – 7x + 4

= (x + 4) (x² – 2x + 1)

= (x + 4) (x – 1) (x – 1)

In both polynomials, the common factor is (x – 1), the extra common factors are (x – 1) (x + 4) ( x + 3) (x – 2).

Therefore, required L.C.M is (x – 1)² (x + 4) ( x + 3) (x – 2), H.C.F is (x – 1).

Given that,

f(x) = (a² – 1) = (a² – 1²)

= (a + 1) (a – 1)

g(x) = (a – 1).

L.C.M of f(x), g(x) = (a – 1) (a + 1)

H.C.F of f(x), g(x) = (a – 1)

f(x) x g(x) = (a + 1) (a – 1) (a – 1)

LCM x GCD = (a – 1) (a + 1) (a – 1)

∴ Hence proved.

Let f(a) = a²+ 36a +315

f(a) x g(a) = a⁴ +53a³ + 957a² + 6435a +9450

G.C.F of f(a), g(a) is (a + 15)

Factorize f(a)

= a²+ 36a +315 = a²+ 15a + 21a + 315

= a(a + 15) + 21(a + 15) = (a + 15) (a + 21)

f(a) x g(a) = (a + 15) (a + 21) (a + 15) * x = a⁴ +53a³ + 957a² + 6435a +9450

Factors of a⁴ +53a³ + 957a² + 6435a +9450 = (a + 15) (a + 21) (a + 15) (a + 2)

Therefore , g(a) = (a + 15) (a + 2)= a² + 17a + 30

L.C.M of polynomials = (a + 15) (a + 2) (a + 21).

Given that,

L.C.M of polynomials = x³ – 4x² – 19x – 14

H.C.F of polynomials = (x – 7)

Factorize the least common multiple,

= x³ – 4x² – 19x – 14 = (x – 7) (x + 1) (x + 2)

p(x) = (x – 7) (x + 1)

= x² – 7x + x – 7 = x² – 6x – 7

q(x) = (x – 7) (x + 2)

= x² – 7x + 2x – 14 = x² – 5x – 14

Given two polynomials are

f = 2x² + 18x + 3xy + 27y, g = 6x² + 14x + 9xy + 21y

The greatest common divisor of polynomials is (2x + 3y)

The relation between H.C.F and L.C.M of two polynomials says that the product of polynomials is equal to the product of their L.C.M and H.C.F

(2x² + 18x + 3xy + 27y) (6x² + 14x + 9xy + 21y) = (2x + 3y) * L.C.M

L.C.M = [(2x² + 18x + 3xy + 27y) (6x² + 14x + 9xy + 21y)] / (2x + 3y)

= [(2x + 3y) (x + 9y) (2x + 3y) (3x + 7y)] / (2x + 3y)

= (2x + 3y) (x + 9y) (3x + 7y) = 6x³ + 77x²y + 228xy² + 189y³.

(a) x² – 1, x² + 2x + 1

First polynomial = x² – 1

= x² – 1² = (x – 1) ( x + 1)

Second polynomial = x² + 2x + 1

= x² + x + x + 1, by splitting the middle term

= x(x + 1) + 1(x + 1) = (x + 1) (x + 1)

In both the polynomials, the common factors are (x+1), extra common factors are (x + 1) (x – 1).

Therefore, the required L.C.M. = (x + 1) (x + 1) (x – 1) =

(b) x+8, x+2

First polynomial = x + 8

Second polynomial = x + 2

There are no factors for the given polynomials.

Therefore, required lowest common multiple = (x + 8) (x + 2).

(c) y² – 2², (y + 2)²

First polynomial = y² – 2²

= (y – 2) (y + 2)

Second polynomial = (y + 2)²

= (y + 2) (y + 2)

In both the polynomials, the common factors are (y + 2), the extra common factor of the first polynomial (y – 2), the second polynomial is (y + 2).

Therefore, required the least common multiple = (y + 2) (y + 2) (y – 2)

(d) p², p + 3

First polynomial = p² = p * p

Second polynomial = p + 3

There are no factors for the given polynomials.

Therefore, required L.C.M = p * p (p + 3)

(a) x³ – y³, x² – xy + y², x² – 2xy + y²

First Polynomial = x³ – y³

= (x – y)(x² + xy + y²)

Second Polynomial = x² – xy + y²

Third Polynomial = x² – 2xy + y²

= (x – y)²

There are no common factors in the polynomials.

Therefore, the Lowest common multiple of given polynomials is its product.

Hence, required L.C.M = (x – y)²(x² + xy + y²)

(b) m(m – 1), m², (m – 1)²

First Polynomial = m(m – 1)

Second Polynomial = m * m

Third Polynomial = (m – 1) (m – 1)

Least common multiple of m(m – 1), m², (m – 1)² = m² (m – 1)².

(c) x² – 1, x² + 2x + 1, (x -1) (x + 2)

First polynomial = x² – 1

= x² – 1² = (x + 1) (x – 1)

Second polynomial = x² + 2x + 1

= x² + x + x + 1 = x(x + 1) + 1(x + 1)

= (x+1) (x+1)

Third Polynomial = (x -1) (x + 2)

Therefore, required L.C.M = (x + 1) (x – 1) (x+1) (x + 2) = (x – 1) (x+1)² (x + 2)

(d) x² + 8x + 12, x² + 2x – 24, and x² + 15x + 54

First Polynomial = x² + 8x + 12

= x² + 6x + 2x + 12 = x(x + 6) + 2(x + 6)

= (x+2) (x+6)

Second Polynomial = x² + 2x – 24

= x² + 6x – 4x – 24 = x(x + 6) – 4(x + 6)

= (x-4)(x+6)

Third polynomial = x² + 15x + 54

= x² + 9x + 6x + 54 = x(x + 9) + 6(x + 9)

= (x+9) (x+6)

The common factor is (x+6), extra common factors are (x+2), (x-4), (x+9).

Therefore, the required least common factor = (x+6) (x+2) (x-4) (x+9).

(a) x² + 15x + 56, x² + 5x – 24, and x² + 8x

First Polynomial = x² + 15x + 56

= x² + 8x + 7x + 56 = x(x + 8) + 7(x + 8)

= (x + 7) (x + 8)

Second Polynomial = x² + 5x – 24

= x² + 8x – 3x – 24 = x(x + 8) – 3(x + 8)

= (x – 3) (x + 8)

Third Polynomial = x² + 8x

= x(x + 8)

The common factor is (x +8), extra common factors are x, (x – 3), (x + 7)

Therefore, the required L.C.M is x(x + 8) (x – 3) (x + 7).

(b) x³ + 3x² + 4x + 12, x⁴ + x³ + 4x² + 4x

First polynomial = x³ + 3x² + 4x + 12

= x²(x + 3) + 4(x + 3)

= (x + 3) (x² + 4)

Second Polynomial = x⁴ + x³ + 4x² + 4x

= x(x³ + x² + 4x + 4)

= x(x²(x + 1) + 4(x + 1)) = x(x² + 4) (x + 1)

The common factor is (x² + 4), extra common factors are x(x +1), (x + 3).

Therefore, required L.C.M is x(x² + 4) (x + 1) (x + 3).

(c) 4(m² – 36), 6(m² + 4m – 12) and 12(m² – 12m + 36)

First Polynomial = 4(m² – 36)

= 4(m² – 6²) = 4(m + 6) (m – 6)

Second polynomial = 6(m² + 4m – 12)

= 6(m² + 6m – 2m -12) = 6(m(m + 6) – 2(m + 6))

= 6(m – 2)(m + 6)

Third Polynomial = 12(m² – 12m + 36)

= 12(m² – 6m – 6m + 36) = 12(m(m – 6) – 6(m – 6))

= 12(m – 6) (m – 6)

Therefore, the required L.C.M is 24 (m – 6)² (m + 6) (m – 2).

(a) 9u³(u⁴ – 2v³ + 1)², 6u⁴v⁵(u⁴ – 2v³ + 1)³

Factorize the given two polynomials,

3u³(u⁴ – 2v³ + 1)² = 3x3xuxuxu (u⁴ – 2v³ + 1)(u⁴ – 2v³ + 1)

6u⁴v⁵(u⁴ – 2v³ + 1)³ = 3x2xuxuxuxuxvxvxvxvxv (u⁴ – 2v³ + 1)(u⁴ – 2v³ + 1)(u⁴ – 2v³ + 1)

Therefore, in the two polynomials, the common factors are 3 x uxuxu (u⁴ – 2v³ + 1)(u⁴ – 2v³ + 1)

So, the H.C.F = 3u³(u⁴ – 2v³ + 1)²

(b) 15x³yz²(y + z), 9xy⁴z³(2x + y + z)

Factorize the given two polynomials,

15x³yz²(y + z) =3*5*x*x*x*y*z*z (y + z)

9xy⁴z³(2x + y + z) = 3*3*y*y*y*y*z*z*z (2x + y + z)

Therefore, in two polynomials, the common factors are 3*x*y*z*z = 3xyz²

So, the required greatest common factor = 3xyz²

(c) x⁴ – 27a³x, (x – 3a)²

Factorize the given two polynomials,

x⁴ – 27a³x = x(x³ – 27a³) = x(x³ – (3a)³)

= x(x – 3a) (x²+3ax+9a²) i.e (a³ – b³) = (a – b) (a² + ab + b²)

(x – 3a)² = (x – 3a)(x – 3a)

So, in two polynomials the common factor is (x-3a).

Therefore, the required highest common factor = (x – 3a).

(d) 2x² + 9x + 4, 2x² + 11x + 5

By factorizing the two polynomials,

2x² + 9x + 4 = 2x² + 8x + x + 4

= 2x(x + 4) +1(x + 4) = (2x+1)(x+4)

2x² + 11x + 5 = 2x² + 10x + x + 5

= 2x(x+5) +1(x+5) = (2x+1)(x+5)

So, in both polynomials, the common factor is (2x+1).

Therefore, required G.C.F = (2x+1).

(a) (s²-25t²)⁹, (s+5t)⁸, s²+10st+25t²

Factorize all three polynomials,

(s²-25t²)⁹ = (s² – (5t)²)⁹ = [(s+5t) (s-5t)]⁹

= (s+5t)⁹ (s-5t)⁹

(s+5t)⁸ = (s+5t) (s+5t) (s+5t) (s+5t) (s+5t) (s+5t) (s+5t) (s+5t)

s²+10st+25t² = s²+5st+5st+25t²

= s(s+5t) +5t(s+5t) = (s+5t) (s+5t)

So, in all polynomials, the common factor is (s+5t) (s+5t) = (s+5t)²

Therefore, required G.C.F = (s+5t)²

(b) 84a⁴b⁶(a⁴-b⁴), 92ab⁷(a²+b²)², 66a⁸b²

84a⁴b⁶(a⁴-b⁴) = 2x2x3x7xaxaxaxaxbxbxbxbxbxbx((a²)²-(b²)²)

= 2x2x3x7xaxaxaxaxbxbxbxbxbxbx(a²-b²)(a²+b²)

= 2x2x3x7xaxaxaxaxbxbxbxbxbxbx(a+b)(a-b)(a²+b²)

92ab⁷(a²+b²)² =2x2x23xaxbxbxbxbxbxbxbx(a²+b²)(a²+b²)

66a⁸b² = 2x3x11xaxaxaxaxaxaxaxaxbxb

So, the common factors in all polynomials is 2xaxbxb = 2ab²

Therefore, the required G.C.F is 2ab².

(c) 3x⁴ + 8x³ + 4x², 3x⁵ + 11x⁴ + 6x³ and 3x⁴ – 16x³ – 12x³

Factorizing all the polynomials,

3x⁴ + 8x³ + 4x² = x²(3x² + 8x + 4)

= x²(3x² + 6x + 2x + 4) = x²(3x(x + 2) +2(x + 2)

= x²(3x + 2) (x + 2)

3x⁵ + 11x⁴ + 6x³ = x³(3x² + 11x + 6)

= x³(3x² + 9x + 2x + 6)

= x³(3x(x + 3) + 2(x +3)

= x³(3x + 2) (x + 3)

3x⁴ – 16x³ – 12x³ = x²(3x² – 16x – 12)

= x²(3x² – 18x + 2x – 12) = x²(3x(x – 6) + 2(x – 6))

= x²(3x + 2) (x – 6)

So, the common factors are x²(3x + 2)

Therefore, the required highest common factor is x²(3x + 2).

(d) 4y²-y-18, (4y-9)⁴, 4y-9

Factorize all polynomials,

4y²-y-18 = 4y²-9y+8y-18

= y(4y-9)+2(4y-9)

= (y+2) (4y+9)

(4y-9)⁴ =  (4y+9) (4y+9) (4y+9) (4y+9)

4y-9 = (4y+9)

So, the common factor is (4y-9).

Therefore, the required greatest common factor is (4y+9).

(a) 5x³-10x²-65x-50, 5x³-25x²-5x+25

Given polynomials are f(x) = 5x³-10x²-65x-50, g(x) = 5x³-25x²-5x+25

Take 5 common from both polynomials, f(x) = 5(x³-2x²-13x-10), g(x) = 5(x³-5x²-x+5)

By using the division method,

Thus, H.C.F of x³-2x²-13x-10, x³-5x²-x+5 = (x-2)

Therefore, G.C.F of 5x³-10x²-65x-50, 5x³-25x²-5x+25 = 5 (x – 2)

(b) x³-x²-2x+2, x³-1

Given polynomials are f(x) = x³-x²-2x+2, g(x) = x³-1

Arranging the polynomials in the descending order g(x) = x³+0x²+0x-1

By using the division method,

Thus, H.C.F of x³-x²-2x+2, x³-1 is (x – 1).

(c) x²+7x+12, 13x⁵+39x⁴

Given polynomials are f(x) = x²+7x+12, g(x) = 13x⁵+39x⁴

Taking x⁴ common in g(x) i.e g(x) = x⁴(13x + 39)

By using the division method,

(a) 18x⁴-36x³+18x², 45x⁶-45x³

Factorize the polynomials,

18x⁴-36x³+18x² = 18x² (x² – 2x + 1)

= 18x² (x² – x – x + 1)

= 18x² (x(x – 1) -1 (x – 1)) = 18x² (x – 1) (x – 1)

45x⁶-45x³ = 45x³(x³ – 1)

= 45x³(x – 1) (x² + x + 1)

Thus, H.C.F of 18x⁴-36x³+18x², 45x⁶-45x³ = 9x²(x – 1).

(b) k² – x², k² – kx and k²x – kx²

Factorize the polynomials,

k² – x² = (k – x) (k + x)

k² – kx = k(k – x)

k²x – kx² = kx(k – x)

Thus, G.C.F of k² – x², k² – kx and k²x – kx² = (k – x).

(a) 15x²y, 3x³y

Numerical coefficients = 15, 3

Since, 15=3×5, 3=3×1.

L.C.M of 15, 3 = 3×5 = 15

G.C.F of 15, 3 = 3

Literal coefficients = x²y, x³y

Since in, x²y, x³y

The highest power and lowest powers of x is x³ & x².

The highest power and lowest powers of y is y & y.

Thus, L.C.M of x²y, x³y = x³y

H.C.F of x²y, x³y = x²y

Therefore, L.C.M of 15x²y, 3x³y = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 15 x x³y = 15x³y.

G.C.F of 15x²y, 3x³y = H.C.F of numerical coefficients x H.C.F. of literal coefficients

= 3 x x²y = 3x²y.

(b) 45pq³r⁴, 18p³q²r

Numerical coefficients = 45, 18

Since, 45=9*5, 18=9*2

The least common multiple of 45, 18 = 9*5*2 = 90

The G.C.F of 45, 18 = 9

Literal coefficients = pq³r⁴, p³q²r

Since in, pq³r⁴, p³q²r

The highest power and lowest powers of p is p³ & p.

The highest power and lowest powers of q is q³ & q².

The highest power and lowest powers of r is r⁴ & r.

Thus, L.C.M of pq³r⁴, p³q²r = p³q³r⁴

G.C.F of pq³r⁴, p³q²r = pq²r

Therefore, L.C.M of 45pq³r⁴, 18p³q²r = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 90 x p³q³r⁴ = 90p³q³r⁴

The H.C.F of 45pq³r⁴, 18p³q²r = H.C.F of numerical coefficients x H.C.F. of literal coefficients

= 9 x pq²r = 9pq²r.

(c) 9u²v, 90uv³

Numerical coefficients = 9, 90

Since, 9=9*1, 90=9*10

The L.C.M of 9, 90 = 9*10 = 90

The H.C.F of 9, 90 = 9

Literal coefficients = u²v, uv³

Since in, u²v, uv³

The highest power and lowest powers of u is u² & u.

The highest power and lowest powers of v is v³ & v.

Thus, L.C.M of u²v, uv³ = u²v³

H.C.F of u²v, uv³ = uv

Therefore, the lowest common multiple of 9u²v, 90uv³ = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 90 x u²v³ = 90u²v³

H.C.F of 9u²v, 90uv³ = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 9 x uv = 9uv.

(a) 14c²d³e, 38fe²g, 28b²c³d

Numerical coefficients = 14, 38, 28

Since, 14=2×7, 38=2×19, 28=2x2x7

Thus, greatest common factor of 14, 38, 28 = 2

The least common multiple of 14, 38, 28 = 2x7x19x2 = 532

Literal coefficients = c²d³e, fe²g, b²c³d

since in, c²d³e, fe²g, b²c³d

The highest power and lowest powers of b is b² & b².

The highest power and lowest powers of c is c³ & c².

The highest power and lowest powers of d is d³ & d.

The highest power and lowest powers of e is e² & e.

The highest power and lowest powers of f is f & f.

The highest power and lowest powers of g is g & g.

Thus, L.C.M of c²d³e, fe²g, b²c³d = b²c³d³e²fg

H.C.F of c²d³e, fe²g, b²c³d = c²

Therefore, Lowest common multiple of 14c²d³e, 38fe²g, 28b²c³d = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 532 x b²c³d³e²fg = 532b²c³d³e²fg

H.C.F of 14c²d³e, 38fe²g, 28b²c³d = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 2 x c² = 2c².

(b) 32x⁵yz², 40x⁴yz², 24x²yz³

Numerical coefficients = 32, 40, and 24

Since, 32=2×8, 40=8×5, 24=3×8

The G.C.F of 32, 40, and 24 = 8

The L.C.M of 32, 40, and 24 = 8x3x2x5 = 240

Literal coefficients = x⁵yz², x⁴yz², x²yz³

The highest power and lowest powers of x is x⁵ & x².

The highest power and lowest powers of y is y & y.

The highest power and lowest powers of z is z³ & z².

Thus, L.C.M of x⁵yz², x⁴yz², x²yz³ = x⁵yz³

H.C.F of x⁵yz², x⁴yz², x²yz³ = x²yz²

L.C.M of 32x⁵yz², 40x⁴yz², 24x²yz³ = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 240 x x⁵yz³ = 240x⁵yz³

H.C.f of 32x⁵yz², 40x⁴yz², 24x²yz³ = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 8 x x²yz² = 8x²yz².

(c) 3pq³, 2p²r and 5p³q²r²

Numerical coefficients = 3, 2, 5

H.C.F of 3, 2, 5 = 1

Least common multiple of 3, 2, 5 = 30

Literal coefficients = pq³, p²r and p³q²r²

Since in, pq³, p²r and p³q²r²

The highest power and lowest powers of p is p³ & p.

The highest power and lowest powers of q is q³ & q².

The highest power and lowest powers of r is r² & r.

Thus, L.C.M of pq³, p²r and p³q²r² = p³q³r²

H.C.F of pq³, p²r and p³q²r² = pr

Therfore, L.C.M of 3pq³, 2p²r and 5p³q²r² = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 30 x p³q³r² = 30p³q³r²

H.C.f of 3pq³, 2p²r and 5p³q²r² = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 1 x pq²r = pr.

(a) 30x³y², 24x²y, 6x²y³

Numerical coefficients = 30, 24, 6

Since 30=2x5x3, 24=2x2x2x3, 6=2×3

H.C.F of 30, 24, 6 = 2×3 = 6

Least common multiple of 30, 24, 6 = 2x3x5x2x2 = 120

Literal coefficients = x³y², x²y, x²y³

Since in, x³y², x²y, x²y³

The highest power and lowest powers of x is x³ & x².

The highest power and lowest powers of y is y³ & y.

Thus, L.C.M of x³y², x²y, x²y³ = x³y³

H.C.F of x³y², x²y, x²y³ = x²y

Therfore, L.C.M of 30x³y², 24x²y, 6x²y³ = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 120 x x³y³ = 120x³y³

H.C.F of 30x³y², 24x²y, 6x²y³ = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 6 x x²y = 6x²y.

(b) 75a², 24a, 30a⁴b

Numerical coefficients = 75, 24, 30

Since, 75=5x5x3, 24=4x3x2, 30=5x3x2

The highest common factor of 75, 24, 30 = 3

The lowest common multiple of 75, 24, 30 = 5x5x4x2x5x2 = 360

Literal coefficients = a², a, a⁴b

Since in, a², a, a⁴b

The highest power and lowest powers of a is a⁴ & a.

The highest power and lowest powers of b is b & b.

Thus, H.C.F of a², a, a⁴b = a.

L.C.M of a², a, a⁴b = a⁴b

Therfore, L.C.M of 75a², 24a, 30a⁴b = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 360 x a⁴b = 360a⁴b

H.C.F of 75a², 24a, 30a⁴b = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 3 x a = 3a.

(c) 4mn, 10n

Numerical coefficients = 4, 10

Since, 4=2×2, =5×2

The greatest common factor of 4, 10 is 2.

The least common multiple of 4, 10 = 2x5x2 = 20

Literal coefficients = mn, n

Since in, mn, n

The highest power and lowest powers of m is m & m.

The highest power and lowest powers of n is n & n.

Thus, H.C.F of mn, n = n.

L.C.M of mn, n = mn

Therefore, H.CF of 4mn, 10n = H.C.F of numerical coefficients x H.C.F of literal coefficients

= 2 x n = 2n.

L.C.M of 4mn, 10n = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 20 x mn = 20mn.

(a) 7x⁴y³, 28xy², 7x²y⁴, 28x³y²

Numerical Coefficients = 7, 28, 7, and 28

Since, 7=1×7, 28=7×4, 7=1×7, and 28=7×4

The H.C.F of 7, 18, 7, and 28 = 7

The L.C.M of 7, 18, 7, and 28 = 7×4 = 28

Literal Coefficients = x⁴y³, xy², x²y⁴, x³y²

Since in, x⁴y³, xy², x²y⁴, x³y²

The lowest and highest power of x is x & x⁴.

The lowest and highest power of y is y² & y⁴.

Thus, least common multiple of x⁴y³, xy², x²y⁴, x³y² = x⁴y⁴

Highest common factor of x⁴y³, xy², x²y⁴, x³y² = xy²

Therefore, G.C.F of 7x⁴y³, 28xy², 7x²y⁴, 28x³y² = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 7 x xy² = 7xy²

L.C.M of 7x⁴y³, 28xy², 7x²y⁴, 28x³y² = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 28 x x⁴y⁴ = 28x⁴y⁴.

(b) 14p²q³, 22pq², 9p⁴q, 5p³qr²

Numerical Coefficients = 14, 22, 9, 5

Since, 14=2×7, 22=2×11, 9=3×3, 5=1×5

Thus, L.C.M of 14, 22, 9, 5 = 2x7x11x9x5 = 6930

H.C.F of G.C.F of 14, 22, 9, 5 = 1

Literal Coefficients = p²q³, pq², p⁴q, p³qr²

Since in, p²q³, pq², p⁴q, p³qr²

The lowest and highest power of p is p & p⁴.

The lowest and highest power of q is q & q³.

The lowest and highest power of r is r² & r².

The L.C.M of p²q³, pq², p⁴q, p³qr² = p⁴q³r²

H.C.F of p²q³, pq², p⁴q, p³qr²=pq

Therfore, G.C.F of 14p²q³, 22pq², 9p⁴q, 5p³qr² = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 1 x pq = pq

L.C.M of 14p²q³, 22pq², 9p⁴q, 5p³qr² = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 6930 x p⁴q³r² = 6930p⁴q³r².

(c) 12a⁴b³c, 24a²b⁵c, 16a⁴bc², 14ab³c, 50a²b²c²

Numerical Coefficients = 12, 25, 16, 14, 50

Since, 12=2x2x3, 24=2x2x2x3, 16=2x2x2x2, 14=2×7, 50=5x5x2

L.C.M of 12, 24, 16, 14, 50 = 4x6x14x25 = 8400

H.C.F of 12, 24, 16, 14, 50 = 2

Literal Coefficients = a⁴b³c, a²b⁵c, a⁴bc², ab³c, a²b²c²

Since in, a⁴b³c, a²b⁵c, a⁴bc², ab³c, a²b²c²

The lowest and highest power of a is a & a⁴.

The lowest and highest power of b is b & b⁵.

The lowest and highest power of c is c & c².

H.C.F of a⁴b³c, a²b⁵c, a⁴bc², ab³c, a²b²c² = abc

L.C.M of a⁴b³c, a²b⁵c, a⁴bc², ab³c, a²b²c² = a⁴b⁵c²

Therfore, H.C.F of 12a⁴b³c, 24a²b⁵c, 16a⁴bc², 14ab³c, 50a²b²c² = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 2 x abc = 2abc

L.C.M of 12a⁴b³c, 24a²b⁵c, 16a⁴bc², 14ab³c, 50a²b²c² = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 8400 x a⁴b⁵c² = 8400a⁴b⁵c².