Factorisation

(i) 12x + 15

Taking 3 as common from the whole, we get,

12x + 15 = 3(4x + 5).

(ii) 14m – 21,

Taking 7 as common from the whole, we get,

14m – 21 = 7(2m – 3)

(iii) 9n – 12n²,

Taking 3n as common from the whole, we get,

9n – 12n² = 3n (3 – 4n).

(i) Let’s take HCF of 16a² – 24ab

Taking 8a as common from the whole, we get,

16a² – 24ab = 8a(2a – 3b).

(ii) 15ab² – 20a²b,

Taking 5ab as common from the whole, we get,

15ab² – 20a²b = 5ab(3b – 4a)

(iii) 12x²y³ – 21x³y²,

Taking 3x²y² as common from the whole, we get,

12x²y³ – 21x³y² = 3x²y²(4y – 7x)

(i) 24x³ – 36x²y,

Taking 12x² as common from the whole, we get,

24x³ – 36x²y = 12x²(2x – 3y)

(ii) 10x³ – 15x²

Taking 5x² as common from the whole, we get,

10x³ – 15x² = 5x²(2x – 3)

(iii) 36x³y – 60x²y³z

Taking 12x²y as common from the whole, we get,

36x³y – 60x²y³z = 12x²y(3x – 5y²z)

(i) Let’s find out the HCF of 9x³ , 6x² , 12x

3x is the highest common factor which divides 9x³, 6x² and 12x.

So,

9x³ – 6x² + 12x = 3x(3x² - 2x + 4)

(ii) Let’s find out the HCF of 8x³, 72xy and 12x

4x is the highest common factor which divides 8x³, 72xy and 12x.

So,

8x³ – 72xy + 12x = 4x(2x² – 18y + 3)

(iii) Let’s find out the HCF of 18a³b³, 27a²b³, 36a³b²

9a²b² is the highest common factor which divides 18a³b³, 27a²b³, 36a³b².

So,

18a³b³ – 27a²b³ + 36a³b² = 9a²b² (2ab - 3b + 4a)

(i) Let’s find out the HCF of 14x³, 21x⁴y and 28x²y²

7x² is the highest common factor of 14x³, 21x⁴y, 28x²y²

So,

14x³ + 21x⁴y – 28x²y² = 7x²(2x + 3x²y – 4y²)

(ii) Let’s find out the HCF of 5, 10t and 20t²,

5 is the highest common factor of 5, 10t and 20t².

So,

- 5 – 10t + 20t² = - 5(1 + 2t – 4t²)

(Note: As we have learned in the previous chapter when we multiplied – sign with – sign it become +)

(i) x(x + 3) + 5(x + 3)

Taking x + 3 as common from the whole, we get,

(x + 3)(x + 5).

(ii) 5x(x – 4) -7(x – 4)

Taking x – 4 as common from the whole, we get,

5x(x – 4) -7(x – 4) = (x – 4)(5x – 7).

(iii) 2m(1 – n) +3(1 – n)

Taking 1 – n as common from the whole, we get,

2m(1 – n) +3(1 – n) = (1 – n)(2m + 3).

6a(a – 2b) + 5b(a – 2b)

Taking a – 2b as common from the whole, we get,

= (a – 2b)(6a + 5b).

x³(2a – b) + x²(2a – b)

Taking 2a – b as common from the whole, we get,

= (2a – b)(x³ + x²).

9a(3a – 5b) – 12a²(3a – 5b)

Taking 3a – 5b as common from the whole, we get,

= (3a – 5b)(9a – 12a²).

(x + 5)² – 4(x + 5)

Taking (x + 5) as common from the whole, we get,

= (x + 5){(x + 5) – 4}

= (x + 5)(x + 5 – 4)

= (x + 5)(x + 1)

So,

The factors of (x + 5)² – 4(x + 5) are: (x + 5) and (x + 1)

3(a – 2b)² -5(a – 2b)

= (a – 2b) {3(a – 2b) – 5}

= (a – 2b){(3a – 6b) – 5}

= (a – 2b)(3a – 6b – 5)

So,

We get,

3(a – 2b)² -5(a – 2b) = (a – 2b)(3a – 6b – 5)

2a + 6b – 3(a + 3b)²


= 2(a + 3b) - 3(a + 3b)²

= (a + 3b){2 - 3(a + 3b)}

16(2p – 3q)² – 4(2p – 3q)

= (2p – 3q){16(2p – 3q) – 4}

= (2p – 3q){(32p – 48q) – 4}

= (2p – 3q)(32p – 48q – 4)

= 4(2p – 3q)(8p – 12q -1)

So,

We get,

16(2p – 3q)² – 4(2p – 3q) = 4(2p – 3q)(8p – 12q -1)

x(a – 3) + y(3 – a)

= x(a – 3) – y(a – 3)

= (a – 3)(x – y)

12(2x – 3y)² – 16(3y – 2x)

= 12(2x – 3y)² + 16(2x -3y)

= (2x – 3y) {12(2x – 3y) + 16}

= (2x – 3y)(24x – 36y + 16)

= 4(2x - 3y)(6x – 9y + 4)

So,

We get,

12(2x – 3y)² – 16(3y – 2x) = 4(2x - 3y)(6x – 9y + 4)

(x + y)(2x + 5) – (x + y)(x + 3)

= (x + y){(2x + 5) – (x + 3)}

= (x + y)(2x + 5 – x – 3)

= (x + y)(2x – x + 5 – 3)

= (x + y)(x + 2)

So,

We get,

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

ar + br + at + bt

First group the terms together;

= (ar + br) + (at + bt)

= r(a + b) + t(a + b)

= (a + b)(r + t)

So,

We get,

ar + br + at + bt = (a + b)(r + t)

x² – ax – bx + ab

Let’s arrange the terms in a suitable form;

x² – ax – bx + ab

= x² – bx – ax + ab

= (x² – bx) – (ax – ab)

= x(x – b) – a(x – b)

= (x – b)(x –a)

So we get,

x² – ax – bx + ab = (x – b)(x –a)

ab² – bc² – ab + c²

Let’s first arrange the terms in a suitable form;

ab² – bc² – ab + c²

= ab² – ab – bc² + c²

= (ab² – ab) – (bc² - c²)

= ab(b – 1) – c²(b - 1)

= (b – 1)(ab – c²)

So we get,

ab² – bc² – ab + c² = (b – 1)(ab – c²)

Let’s first arrange the terms in a suitable form;

x² – xz + xy – yz

= x² + xy – xz – yz

= (x² + xy) – (xz + yz)

= x(x + y) – z(x + y)

= (x + y)(x – z)

So we get,

x² – xz + xy – yz = (x + y)(x – z)

6ab – b² + 12ac – 2bc

= 6ab + 12ac – b² – 2bc

= (6ab + 12ac) – (b² + 2bc)

= 6a(b + 2c) – b(b + 2c)

= (b + 2c)(6a – b)

So we get,

6ab – b² + 12ac – 2bc = (b + 2c)(6a – b)

(x – 2y)² + 4x – 8y

= (x – 2y)² + 4(x – 2y)

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

= (x -2y){(x -2y) +4}

= (x – 2y)(x – 2y + 4)

So we get,

(x – 2y)² + 4x – 8y = = (x – 2y)(x – 2y + 4)

y² – xy(1 – x) – x³

= y² – xy + x²y – x³

= (y² – xy) + (x²y – x³)

= y(y – x) + x²(y – x)

= (y –x)(y + x²)

So we get,

y² – xy(1 – x) – x³ = (y –x)(y + x²)

(ax + by)² + (bx – ay)²

By using the formulas;

(a + b)² = a² + b² + 2ab and

(a – b)² = a² + b² – 2ab

= (a²x² + b²y² + 2axby) + (b²x² + a²y² – 2bxay)

= a²x² + a²y² + b²y² + b²x² + 2axby – 2bxay

= a²(x² + y²) + b²x² + b²y² + 2axby – 2axby

= a²(x² + y²) + b²(x² + y²)

= (x² + y²)(a² + b²)

So we get,

(ax + by)² + (bx – ay)² = (x² + y²)(a² + b²)

ab² + (a – 1)b -1

= ab² + ba – b – 1

= (ab² + ba) – (b + 1)

= ab (b + 1) - 1(b + 1)

= (b + 1)(ab – 1)

So we get,

ab² + (a – 1)b -1 = (b + 1)(ab – 1)

x³ – 3x² + x – 3

= (x³ – 3x²) + (x -3)

= x²(x – 3) + 1(x – 3)

= (x – 3)(x² + 1)

So we get,

x³ – 3x² + x – 3 = (x – 3)(x² + 1)

ab(x² + y²) – xy(a² + b²)

= abx² + aby² – a²xy – b²xy

= abx² – a²xy + aby² – b²xy

= ax(bx – ay) + by(ay – bx)

= ax(bx – ay) – by(bx – ay)

= (bx – ay)(ax – by)

So we get,

ab(x² + y²) – xy(a² + b²) = (bx – ay)(ax – by)

x² – x(a + 2b) + 2ab

= x² – ax – 2bx + 2ab

= x² – 2bx – ax + 2ab

= (x² – 2bx) – (ax – 2ab)

= x(x – 2b) – a(x – 2b)

= (x – 2b)(x – a)

So we get,

x² – x(a + 2b) + 2ab = (x – 2b)(x – a)

We have,

x² – 36

Which is,

= (x)² – (6)²

By using the formula a² – b² = (a + b)(a – b)

We get,

x² – 36 = (x)² – (6)²

= (x + 6)(x – 6)

We have,

4a² – 9

= (2a)² – (3)²

By using the formula a² – b² = (a + b)(a – b)

We get,

4a² – 9 = (2a)² – (3)²

= (2a + 3)(2a – 3)

We have,

81 – 49x²

= (9)² – (7x)²

By using the formula a² – b² = (a + b)(a – b)

We get,

81 – 49x² = (9)² – (7x)²

= (9 + 7x)(9 – 7x)

We have,

4x² – 9y²

= (2x)² – (3y)²

By using the formula a² – b² = (a + b)(a – b)

We get,

4x² – 9y² = (2x)² – (3y)²

= (2x + 3y)(2x – 3y)

We have,

16a² – 225b²

= (4a)² – (15b)²

By using the formula a² – b² = (a + b)(a – b)

We get,

16a² – 225b² = (4a)² – (15b)²

= (4a + 15b)(4a – 15b)

We have,

9a²b² – 25

= (3ab)² – (5)²

By using the formula a² – b² = (a + b)(a – b)

We get,

9a²b² – 25 = (3ab)² – (5)²

= (3ab + 5)(3ab – 5)

We have,

16a² – 144

= (4a)² – (12)²

By using the formula a² – b² = (a + b)(a – b)

We get,

16a² – 144 = (4a)² – (12)²

= (4a + 12)(4a – 12)

= 4(a + 3) 4(a – 3)

= 16(a + 3)(a – 3)

We have,

63a² – 112b²

= 7(9a² – 16b²)

By using the formula a² – b² = (a + b)(a – b)

We get,

63a² – 112b² = 7(9a² – 16b²)

= 7{(3a)² – (4b)²}

= 7(3a + 4b)(3a – 4b)

We have,

20a² – 45b²

= 5(4a² – 9b²)

By using the formula a² – b² = (a + b)(a – b)

We get,

20a² – 45b² = 5(4a² – 9b²)

= 5{(2a)² – (3b)²}

= 5(2a + 3b)(2a – 3b)

We have,

12x² – 27

= 3(4x² – 9)

By using the formula a² – b² = (a + b)(a – b)

We get,

12x² – 27 = 3(4x² – 9)

= 3{(2x)² – (3)²}

= 3(2x + 3)(2x – 3)

We have,

x³ – 64x

= x(x² – 64)

By using the formula a² – b² = (a + b)(a – b)

We get,

x³ – 64x = x(x² – 64)

= x{(x)² – (8)²}

= x(x + 8)(x – 8)

We have,

16x⁵ – 144x³

= 3x³(x² – 9)

By using the formula a² – b² = (a + b)(a – b)

We get,

16x⁵ – 144x³ = 3x³(x² – 9)

= 16x³{(x)² – (3)²}

= 16x³(x + 3)(x – 3)

We have,

3x⁵ – 48x³

= 3x³(x² – 16)

By using the formula a² – b² = (a + b)(a – b)

We get,

3x⁵ – 48x³ = 3x³(x² – 16)

= 3x³{(x)² – (4)²}

= 3x³(x + 4)(x – 4)

We have,

16p³ – 4p

= 4p(4p² – 1)

By using the formula a² – b² = (a + b)(a – b)

We get,

16p³ – 4p = 4p(4p² – 1)

= 4p{(2p)² – (1)²}

= 4p(2p + 1)(2p – 1)

We have,

63a²b² – 7

= 7(9a²b² – 1)

By using the formula a² – b² = (a + b)(a – b)

We get,

63a²b² – 7 = 7(9a²b² – 1)

= 7{(3ab)² – (1)²}

= 7(3ab + 1)(3ab – 1)

We have,

1 – (b – c)²

= (1)² – (b –c)²

By using the formula a² – b² = (a + b)(a – b)

We get,

1 – (b – c)² = (1)² – (b –c)²

= {1 + (b – c)}{1 – (b – c)}

= (1 + b – c)(1 – b + c)

Given,

(2a + 3b)² – 16c²

= (2a + 3b)² – (4c)²

By using the formula a² – b² = (a + b)(a – b)

We get,

(2a + 3b)² – 16c² = (2a + 3b)² – (4c)²

= {(2a + 3b) + 4c}{(2a + 3b) – 4c}

= (2a + 3b + 4c)(2a + 3b – 4c)

Given,

(2x + 5y)² – (1)²

= (2x + 5y)² – (1)²

By using the formula a² – b² = (a + b)(a – b)

We get,

(2x + 5y)² – (1)² = (2x + 5y)² – (1)²

= {(2x + 5y) + 1}{(2x + 5y) – 1}

= (2x + 5y + 1)(2x + 5y – 1)

Given,

36c² – (5a + b)²

= (6c)² – (5a + b)²

By using the formula a² – b² = (a + b)(a – b)

We get,

36c² – (5a + b)² = (6c)² – (5a + b)²

= {(6c) + (5a + b)}{(6c) – (5a + b)}

= (6c + 5a + b)(6c – 5a – b)

Given,

(3x – 4y)² – 25z²

= (3x – 4y)² – (5z)²

By using the formula a² – b² = (a + b)(a – b)

We get,

(3x – 4y)² – 25z² = (3x – 4y)² – (5z)²

={(3x – 4y) + 5z}{(3x – 4y) – 5z}

= (3x – 4y + 5z)(3x – 4y – 5z)

Given,

x² – y² – 2y – 1

= x² – (y² + 2y + 1)

By using the formula a² – b² = (a + b)(a – b)

We get,

x² – y² – 2y – 1 = x² – (y² + 2y + 1)

= (x)² – (y + 1)²

={x + (y +1)}{x – (y + 1)}

= (x + y + 1)(x – y – 1)

Given,

25 – a² – b² – 2ab

= 25 – (a² + b² + 2ab)

By using the formula a² – b² = (a + b)(a – b)

We get,

25 – a² – b² – 2ab = 25 – (a² + b² + 2ab)

= 25 – (a + b)²

=(5)² – (a + b)²

= {5 + (a + b)}{5 – (a + b)}

= (5 + a + b)(5 – a – b)

Given,

25a² – 4b² + 28bc – 49c²

= 25a² – (4b² – 28bc + 49c²)

By using the formula a² – b² = (a + b)(a – b)

We get,

25a² – 4b² + 28bc – 49c² = 25a² – (4b² – 28bc + 49c²)

= (5a)² – (2b – 7c)²

= {5a + (2b – 7c)}{5a – (2b – 7c)}

= (5a + 2b – 7c)(5a – 2b + 7c)

Given,

9a² – b² + 4b – 4

= 9a² – (b² – 4b + 4)

By using the formula a² – b² = (a + b)(a – b)

We get,

9a² – b² + 4b – 4 = 9a² – (b² – 4b + 4)

= (3a)² – (b – 2)²

= {3a + (b – 2)}{3a – (b – 2)}

= (3a + b - 2)(3a – b + 2)

Given,

100 – (x – 5)²

By using the formula a² – b² = (a + b)(a – b)

We get,

100 – (x – 5)² = (10)² – (x – 5)²

= {10 + (x – 5)}{10 – (x – 5)}

= (10 + x - 5)(10 – x + 5)

= (5 + x)(15 – x)

Given,

{(405)² – (395)²}

By using the formula a² – b² = (a + b)(a – b)

We get,

{(405)² – (395)²} = (405 + 395)(405 – 395)

= (800 × 10)

= 8000

We have,

{(7.8)² – (2.2)²}

By using the formula a² – b² = (a + b)(a – b)

We get,

{(7.8)² – (2.2)²} = (7.8 + 2.2)(7.8 – 2.2)

= (10 × 5.6)

= 56

So,

{(7.8)² – (2.2)²} = 56

Given,

x² + 8x + 16

By using the formula (a + b)² = a² + 2ab + b²

We get,

= x² + 2 × (x) × 4 + (4)²

= (x + 4)²

Given;

x² + 14x + 49

By using the formula (a + b)² = a² + 2ab + b²

We get,

= x² + 2 × (x) × 7 + (7)²

= (x + 7)²

Given,

1 + 2x + x² = x² + 2x + 1

By using the formula (a + b)² = a² + 2ab + b²

We get,

= x² + 2 × (x) × 1 + (1)²

= (x + 1)²

Given,

9 + 6z + z² = z² + 6z + 9

By using the formula (a + b)² = a² + 2ab + b²

We get,

= z² + 2 × z × 3 + (3)²

= (3 + z)²

Given;

x² + 6ax + 9a²

By using the formula (a + b)² = a² + 2ab + b²

We get,

= x² + 2 × (x) × 3a + (3a)²

= (x + 3a)²

Given;

4y² + 20y + 25

By using the formula (a + b)² = a² + 2ab + b²

We get,

= (2y)² + 2 × 2y × 5 + (5)²

= (2y + 5)²

Given,

36a² + 36a + 9

By using the formula (a + b)² = a² + 2ab + b²

We get,

= (6a)² + 2×6a×3 + (3)²

= (6a + 3)²

Given,

9m² + 24m + 16

By using the formula (a + b)² = a² + 2ab + b²

We get,

= (3m)² + 2×3m×4 + (4)²

= (3m + 4)²

Given,

By using the formula (a + b)² = a² + 2ab + b²

We get,

Given,

49a² + 84ab + 36b²

By using the formula (a + b)² = a² + 2ab + b²

We get,

= (7a)² + 2 × 7a × 6b + (6b)²

= (7a + 6b)²

Given,

P² – 10p + 25

By using the formula (a - b)² = a² – 2ab + b²

We get,

= p² – 2 × p × 5 + (5)²

= (p – 5)²

Given,

121a² – 88ab + 16b²

By using the formula (a – b)² = a² – 2ab + b²

We get,

= (11a)² – 2 × 11a × 4b + (4b)²

= (11a – 4b)²

Given,

1 – 6x + 9x² = 9x² – 6x + 1

By using the formula (a – b)² = a² – 2ab + b²

We get,

= (3x)² – 2 × (3x) × 1 + (1)²

= (3x – 1)²

Given,

9y² – 12y + 4

By using the formula (a – b)² = a² – 2ab + b²

We get,

= (3y)² – 2 × 3y × 2 + (2)²

= (3y – z)²

Given,

16x² – 24x + 9

By using the formula (a - b)² = a² - 2ab + b²

= (4x)² – 2 × (4x) × 3 + (3)²

= (4x – 3)²

Given,

m² – 4mn + 4n²

By using the formula (a - b)² = a² - 2ab + b²

= m² - 2×m×2n + (2n)²

= (m – 2n)²

Given,

a²b² – 6ab + 9c²

By using the formula (a - b)² = a² + b² - 2ab

We get,

= (ab)² - 2×ab×3c + (3c)²

= (ab – 3c)²

Given,

m⁴ + 2m²n² + n⁴

By using the formula (a + b)² = a² + b² + 2ab

We get,

= (m²)² + 2×m²×n² + (n²)²

= (m² + n²)

Given,

(l + m)² – 4lm

By using the formula (a + b)² = a² + b² + 2ab

We get,

(l + m)² – 4lm = (l² + m² + 2lm) – 4lm

= l² + m² + 2lm – 4lm

= l² + m² – 2lm

= (l)² + (m)² - 2×l×m

= (l – m)²

Given,

x² + 5x + 6

Now first find the numbers whose-

Sum = 5 and

Product = 6

Required numbers are 2 and 3,

So we get;

x² + 5x + 6

= x² + 2x + 3x + 6

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

= (x + 2)(x + 3)

Given,

y² + 10y + 24

Now first find the numbers whose-

Sum = 10 and

Product = 24

Required numbers are 6 and 4,

So we get;

y² + 10y + 24 = y² + 6y + 4y + 24

= y(y + 6) + 4(y + 6)

= (y + 6)(y + 4)

z² + 12z + 27

Now first find the numbers whose-

Sum = 12 and

Product = 27

Required numbers are 9 and 3,

So we get;

z² + 12z + 27

= z² + 9z + 3z + 27

= z(z + 9) + 3(z + 9)

= (z + 9)(z + 3)

Given,

p² + 6p + 8

Now first find the numbers whose-

Sum = 6 and

Product = 8

Required numbers are 4 and 2,

So we get;

p² + 6p + 8

= p² + 4p + 2p + 8

= p(p + 4) + 2(p + 4)

= (p + 4)(p + 2)

Given,

x² + 15x + 56

Now first find the numbers whose-

Sum = 15 and

Product = 56

Required numbers are 7 and 8,

So we get;

x² + 15x + 56

= x² + 7x + 8x + 56

= x(x + 7) + 8(x + 7)

= (x + 7)(x + 8)

y² + 19y + 60

Now first find the numbers whose-

Sum = 19 and

Product = 60

Required numbers are 15 and 4,

So we get;

y² + 19y + 60

= y² + 15y + 4y + 60

= y(y + 15) + 4(y + 15)

= (y + 15)(y + 4)

Given,

x² + 13x + 40

Now first find the numbers whose-

Sum = 13 and

Product = 40

Required numbers are 8 and 5,

So we get;

x² + 13x + 40

= x² + 8x + 5x + 40

= x(x + 8) + 5(x + 8)

= (x + 8)(x + 5)

Given,

q² – 10q + 21

Now first find the numbers whose-

Sum = - 10 and

Product = 21

Required numbers are 7 and 3,

So we get;

q² – 10q + 21

= q² – 7q – 3q + 21

= q(q – 7) – 3(q – 7)

= (q – 7)(q – 3)

Given,

p² + 6p – 16

Now first find the numbers whose-

Sum = 6 and

Product = - 16

Required numbers are 8 and 2,

So we get;

p² + 6p – 16

= p² + 8p – 2p – 16

= p(p + 8) – 2(p + 8)

= (p + 8)(p – 2)

Given,

x² – 10x + 24

Now first find the numbers whose-

Sum = - 10 and

Product = 24

Required numbers are 6 and 4,

So we get;

x² – 10x + 24

= x² – 6x – 4x + 24

= x(x – 6) – 4(x – 6)

= (x – 6)(x – 4)

Given,

x² – 23x + 42

Now, first we have to find out the numbers whose-

Sum = - 23 and

Product = 42

The numbers are 21 and 2,

So,

x² – 23x + 42 = x² – 21x – 2x + 42

= x(x – 21) – 2(x – 21)

= (x – 21)(x – 2)

Given,

x² – 17x + 16

Now, first we have to find out the numbers whose-

Sum = - 17 and

Product = 16

The numbers are 16 and 1,

So,

x² – 17x + 16 = x² – 16x – 1x + 16

= x(x – 16) – 1(x – 16)

= (x – 16)(x – 1)

Given,

y² – 21y + 90

Now, first we have to find out the numbers whose-

Sum = - 21 and

Product = 90

The numbers are 15 and 6,

So,

y² – 21y + 90 = y² – 15y – 6y + 90

= y(y – 15) – 6(y – 15)

= (y – 15)(y – 6)

Given,

x² – 22x + 117

Now, first we have to find out the numbers whose-

Sum = - 22 and

Product = 117

The numbers are 13 and 9,

So,

x² – 22x + 117 = x² – 13x – 9x + 117

= x(x – 13) – 9(x – 13)

= (x – 13)(x – 9)

x² – 9x + 20

Now, first we have to find out the numbers whose-

Sum = - 9 and

Product = 20

The numbers are 5 and 4,

So,

x² – 9x + 20 = x² – 5x – 4x + 20

= x(x – 5) – 4(x – 5)

= (x – 5)(x – 4)

x² + x – 132

Now, first we have to find out the numbers whose-

Sum = 1 and

Product = - 132

The numbers are 12 and 11,

So,

x² + x – 132 = x² + 12x – 11x – 132

= x(x + 12) – 11(x + 12)

= (x + 12)(x – 11)

x² + 5x – 104

Now, first we have to find out the numbers whose-

Sum = 5 and

Product = - 104

The numbers are 13 and 8,

So,

x² + 5x – 104 = x² + 13x – 8x – 104

= x(x + 13) – 8(x + 13)

= (x + 13)(x – 8)

y² + 7y – 144

Now, first we have to find out the numbers whose-

Sum = 7 and

Product = - 144

The numbers are 16 and - 9,

So,

y² + 7y – 144

= y² + 16y – 9y – 144

= y(y + 16) – 9(y + 16)

= (y + 16)(y – 9)

Given,

z² + 19z – 150

Now, first we have to find out the numbers whose-

Sum = 19 and

Product = - 150

The numbers are 25 and 6,

So,

z² + 19z – 150

= z² + 25z – 6z – 150

= z(z + 25) – 6(z + 25)

= (z + 25)(z – 6)

Given,

y² + y – 72

Now, first we have to find out the numbers whose-

Sum = 1 and

Product = - 72

The numbers are 9 and 8,

So,

y² + y – 72

= y² + 9y – 9y – 72

= y(y + 9) – 9(y + 9)

= (y + 9)(y – 9)

a² + 6a – 91

Now, first we have to find out the numbers whose-

Sum = 6 and

Product = - 91

The numbers are 13 and 7,

So,

a² + 6a – 91

= a² + 13a – 7a – 91

= a(a + 13) – 7(a + 13)

= (a + 13) (a – 7)

p² – 4p – 77

Now, first we have to find out the numbers whose-

Sum = - 4 and

Product = - 77

The numbers are 11 and 7,

So,

p² – 4p – 77

= p² – 11p + 7p – 77

= p(p – 11) + 7(p – 11)

= (p – 11)(p + 7)

x² – 7x – 30

Now, first we have to find out the numbers whose-

Sum = - 7 and

Product = - 30

The numbers are 10 and 3,

So,

x² – 7x – 30

= x² – 10x + 3x – 30

= x(x – 10) + 3(x – 10)

= (x – 10)(x + 3)

x² – 11x – 42

Now, first we have to find out the numbers whose-

Sum = - 11 and

Product = - 42

The numbers are 14 and 3,

So,

x² – 11x – 42

= x² – 14x + 3x – 42

= x(x – 14) + 3(x + 14)

= (x – 14)(x + 3)

x² – 5x – 24

Now, first we have to find out the numbers whose-

Sum = - 5 and

Product = - 24

The numbers are - 8 and 3,

So,

x² – 5x – 24

= x² – 8x + 3x – 24

= x(x – 8) + 3(x – 8)

= (x – 8)(x + 3)

Given;

y² – 6y – 135

Now first find the numbers whose-

Sum = - 6 and

Product = - 135

Required numbers are 15 and 9,

So we get;

y² – 6y – 135

= y² – 15y + 9y – 135

= y(y – 15) + 9(y – 15)

= (y – 15)(y + 9)

Given

z² – 12z – 45

Now first find the numbers whose-

Sum = - 12 and

Product = - 45

Required numbers are 15 and 3,

So we get;

z² – 12z – 45

= z² – 15z + 3z - 45

= z(z – 15) + 3(z – 15)

= (z – 15)(z + 3)

Given,

x² – 4x – 12

Now first find the numbers whose-

Sum = - 4 and

Product = - 12

Required numbers are 6 and 2,

So we get;

x² – 4x – 12

= x² – 6x + 2x – 12

= x(x – 6) + 2(x – 6)

= (x – 6)(x + 2)

Given,

3x² + 10x + 8

Now first find the numbers whose-

Sum = 10 and

Product = 3 × 8 = 24

Required numbers are 6 and 4,

So we get;

3x² + 10x + 8

= 3x² + 6x + 4x + 8

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

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

Given,

3y² + 14y + 8

Now first find the numbers whose-

Sum = 14 and

Product = 3 × 8 = 24

Required numbers are 12 and 2,

So we get;

3y² + 14y + 8 = 3y² + 12y + 2y + 8

= 3y(y + 4) + 2(y + 4)

= (y + 4)(3y + 2)

Given,

3z² – 10z + 8

Now, first we have to find out the numbers whose-

Sum = - 10 and

Product = 3 × 8 = 24

The numbers are 6 and 4,

So,

3z² – 10z + 8

= 3z² – 6z – 4z + 8

= 3z(z – 2) – 4(z – 2)

= (z – 2)(3z – 4)

Given,

2x² + x – 45

Now first find the numbers whose-

Sum = 1 and

Product = - 45 × 2 = - 90

Required numbers are 10 and 9,

So we get;

2x² + x – 45

= 2x² + 10x – 9x – 45

= 2x(x + 5) – 9(x + 5)

= (x + 5)(2x – 9)

Given,

6p² + 11p – 10

Now first find the numbers whose-

Sum = 11 and

Product = - 10 × 6 = - 60

Required numbers are 15 and 4,

So we get;

= 6p² + 15p – 4p – 10

= 3p(2p + 5) – 2(2p + 5)

= (2p + 5)(3p – 2)

Given,

2x² – 17x – 30

Now first find the numbers whose-

Sum = - 17 and

Product = - 30 × 2 = - 60

Required numbers are 20 and 3,

So we get;

2x² – 17x – 30

= 2x² – 20x + 3x – 30

= 2x(x – 10) + 3(x – 10)

= (x – 10)(2x + 3)

Given,

7y² – 19y – 6

Now first find the numbers whose-

Sum = - 19 and

Product = - 6 × 7 = - 42

Required numbers are 21 and 2,

So we get;

7y² – 19y – 6

= 7y² – 21y + 2y – 6

= 7y(y – 3) + 2(y – 3)

= (y – 3)(7y + 2)

Given,

28 – 31x – 5x²

Now first find the numbers whose-

Sum = - 31 and

Product = - 5 × 28 = 140

Required numbers are 35 and 4,

So we get;

28 – 31x – 5x²

= 28 + 4x – 35x – 5x²

= 4(7 + x) – 5x(7 + x)

= (7 + x)(4 – 5x)

Given,

3 + 23z – 8z²

Now first find the numbers whose-

Sum = 23 and

Product = - 8 × 3 = 24

Required numbers are 24 and 1,

So we get;

3 + 23z – 8z²

= 3 + 24z – z – 8z²

= 3(1 + 8z) – z(1 + 8z)

= (1 + 8z)(3 – z)

Given,

6x² – 5x – 6

Now first find the numbers whose-

Sum = - 5 and

Product = - 6 × 6 = - 36

Required numbers are 9 and 4,

So we get;

= 6x² – 9x + 4x – 6

= 3x(2x – 3) + 2(2x – 3)

= (2x – 3)(3x + 2)

Given,

3m² + 24m + 36

Now first find the numbers whose-

Sum = 24 and

Product = 36 × 3 = 108

Required numbers are 18 and 6,

So we get;

3m² + 24m + 36

= 3m² + 18m + 6m + 36

= 3m(m + 6) + 6(m + 6)

= (m + 6)(3m + 6)

Given,

4n² – 8n + 3

Now first find the numbers whose-

Sum = - 8 and

Product = 4 × 3 = 12

Required numbers are 6 and 2,

So we get;

4n² – 8n + 3

= 4n² – 2n – 6n + 3

= 2n(2n – 1) – 3(2n – 3)

= (2n – 1)(2n – 3)

Given,

6x² – 17x – 3

Now, first we have to find out the numbers whose-

Sum = - 17 and

Product = 6 × - 3 = - 18

The numbers are 18 and 1,

So,

6x² – 17x – 3

= 6x² – 18x + 1x – 3

= 6x(x – 3) + 1(x – 3)

= (x – 3)(6x + 1)

Given,

7x² – 19x – 6

Now, first we have to find out the numbers whose-

Sum = - 19 and

Product = 7 × - 6 = - 42

The numbers are 21 and 2,

So,

7x² – 19x – 6

= 7x² – 21x + 2x – 6

= 7x(x – 3) + 2(x – 3)

= (x – 3)(7x + 2)

(7a² – 63b²) = 7(a² – 9b²) (taking 7 as common from whole)

= 7(a – 3b)(a + 3b) a² – b² = (a – b)(a + b)

(2x – 32x³) = 2x(1 – 16x²) (taking 2x as common from whole)

= 2x(1 – 4x)(1 + 4x) a² – b² = (a – b)(a + b)

X³ – 144x = x(x² – 144) (taking x as common from whole)

= x(x – 12)(x + 12) a² – b² = (a – b)(a + b)

2 – 50x²= 2(1 – 25x²) (taking 2 as common from whole)

= 2(1 – 5x)(1 + 5x) a² – b² = (a – b)(a + b)

a²+bc+ab+ac = a²+ab + bc + ac

Rearranging the terms and taking a and c as common respectively.

= a(a + b) + c(a + b)

= (a + c)(a + b).

pq² + q(p – 1) – 1 = pq² + qp – q – 1

= pq(q + 1) – 1(q + 1)

= (pq – 1)(q + 1)

= ab – mn + an – bm = ab + an – mn – bm

= a(b + n) – m(n + b)

= (a – m)(b + n).

ab – a – b + 1

= a(b – 1) – 1(b – 1) (taking a and – 1 as common )

= (a – 1)(b – 1).

= x² – xz + xy – yz

= x(x – z) +y(x – z) (taking x and y as common resp.)

= (x + y)(x – z).

12m² – 27 = 3(4m² – 9) (taking 3 as common from whole)

= 3(2m – 3)(2m + 3) a² – b² = (a – b)(a + b)

x³ – x = x(x² – 1) (taking x as common from whole)

= x(x – 1)(x + 1) a² – b² = (a – b)(a + b)

1 – 2ab – (a² + b²) = 1 – 2ab – a² – b²

= 1 – (2ab + a² + b²)

= 1 – (a + b)²

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

x² + 6x + 8

Factorizing the equation and taking x and 2 as common,

= x² + 4x + 2x + 8

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

= (x + 2)(x + 4).

x² + 4x – 21

Factorizing the equation and taking x and – 3 as common,

= x² + 7x – 3x – 21

= x(x + 7) – 3(x + 7)

= (x – 3)(x + 7).

y² + 2y – 3

Factorizing the equation and taking y and – 1 as common,

= y² + 3y – y – 3

= y(y + 3) – 1(y + 3)

= (y + 3)(y – 1).

40 + 3x – x²

Factorizing the equation and taking 8 and – x as common,

= 40 + 8x – 3x – x²

= 8(5 + x) – x(5 + x)

= (8 – x)(5 + x).

2x² + 5x + 3

Factorizing the equation and taking 2x and 3 as common,

= 2x² + 2x + 3x + 3

= 2x(x +1) + 3(x + 1)

= (2x + 3)(x + 1).

6a² – 13a + 6

Factorizing the equation and taking 3a and – 2 as common,

= 6a² – 9a – 4a+ 6

= 3a(2a – 3) – 2(2a – 3)

= (3a – 2)(2a – 3).

4z² – 8z + 3

Factorizing the equation and taking 2z and – 1 as common,

= 4z² – 6z – 2z + 3

= 2z(2z – 3) – 1(2z – 3)

= (2z – 1)(2z – 3).

3 + 23y – 8y²

Factorizing the equation and taking 3 and – y as common,

= 3 + 24y – y – 8y²

= 3(1 + 8y) – y(1 + 8y)

= (3 – y)(1 + 8y).