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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, 

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, 

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, 

m² – 4mn + 4n² 

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

= m² - 2×m×2n + (2n)² 

= (m – 2n)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

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 – 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, 

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, 

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)

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)

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, 

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, 

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, 

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, 

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, 

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, 

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)

x (a² + b²) – y (a² + b²)

= (x – y) (a² + b²)

(x – y) (x – y + 1) [Therefore, taking (x – y) common)

a (x – y) – 2b (x – y) + c (x – y)² [Therefore, (y – x) = - (x – y)]

= (x – y) [a – 2b + c (x – y)]

= (x – y) (a – 2b + cx – cy)

[6 – 4 (a + 2b)] (a + 2b) [Therefore, taking (a + 2b) common]

= (6 – 4a – 8b) (a + 2b)

49 – (x² + y² – 2xy)

= 7² – (x – y)²

= [7 + (x – y)] [7 – x + y]

(a + b)² – c²

= (a + b + c) (a + b – c)

x² (a – 2b) (x + 1) [Therefore, taking x² (a – 2b) as common]

a² – 2ab – ac + 2bc
= a (a – c) – 2b (a – c)
= (a – 2b) (a – c)

a² + ab + ac – bc
= a (a – c) + b (a – c)
= (a + b) (a – c)

x (x – 1) – 11y (x – 1)
= (x – 11y) (x – 1)

a² – 14a – 51

Coefficient of a² = 1

Therefore, we have

a² – 14a – 51 = a² – 14a + 7² – 7² – 51 (Therefore, adding and subtracting 7²)

= (a – 7)² – 10² (Completing the square)

= (a – 7 + 10) (9 – 7 – 10)

= (a + 3) (a – 17)

Greatest common factor of the two terms namely 28a², 14a²b², - 21a⁴ of expression 28a² + 14a²b² - 21a⁴ is 7a²

28a² + 14a²b²-21a⁴ = 7a²(4 + 2b² - 3a²)

In order to factorize the given expression, we find to find two numbers p and q such that:

p + q = 14, pq = 48

Clearly,

8 + 6 = 14, 8 (6) = 48

Therefore, split (14a) as 8a + 6a

Therefore,

a² + 14a + 48 = a² + 8a + 6a + 48

= a (a + 8) + 6 (a + 8) 

= (a + 6) (a + 8)

a² – 3a – 54

In order to factorize the given expression, we find to find two numbers p and q such that:

p + q = -3, pq = -54

Clearly,

6 – 9 = - 3, 6 (-9) = -54

Therefore,

split – 3a as 6a – 9a

Therefore,

a² – 3a – 54 = a² + 6a – 9a – 54

= (a - 9) (a + 6)

Therefore,

(a + 7) (a – 10) + 16 = (a – 9) (a + 6)

In order to factorize the given expression, we find to find two numbers p and q such that:

p + q = 3, pq = -88
Therefore, split 3a as 11a – 8a
Therefore,
a² + 3a – 88 = a² + 11a – 8a – 88

= a (a + 11) – 8 (a + 11)

= (x – 8) (a + 11)

In order to factorize the given expression, we find to find two numbers p and q such that:

p + q = 14, pq = 45

Clearly,

5 + 9 = 14, 5 (9) = 45

Therefore, split 14x as 5x + 9x

Therefore,

x² + 14x + 45 = x² + 5x + 9x + 45

= x (x + 5) – 9 (x + 5)

= (x + 9) (x + 5)

In order to factorize the given expression, we find to find two numbers p and q such that:

p + q = -14, pq = -51
Clearly,
3 – 17 = -14, 3 (-17) = -51
Therefore, split 14a as 3a – 17a
Therefore,
a² – 14a – 51 = a² + 3a – 17a – 51

= a (a + 3) – 17 (a + 3)

= (a – 17) (a + 3)

(7a + 3b) (2x – 3) [Therefore, taking (2x – 3) common]

(x – 2y) [5 (x – 2y) + 3] [Therefore, taking (x – 2y) common]

= (x – 2y) (5x – 10y + 3)

(9a – 12a²) (6a – 5b) [Therefore, taking (6a – 5b) common]

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)

12(2x – 3y)² – 16(3y – 2x) 

= 12(2x – 3y)² + 16(2x -3y) 

[Taking (2x - 3y) common from the expression] 

= (2x – 3y) {12(2x – 3y) + 16} 

= (2x – 3y)(24x – 36y + 16) 

[Taking 4 common from the expression] 

= 4(2x - 3y)(6x – 9y + 4) 

So, 

We get, 

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

6a(a – 2b) + 5b(a – 2b)

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

= (a – 2b)(6a + 5b).

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)

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)

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)

(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)

(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)

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)

(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)

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)

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)