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(a – 4b)²- (5c)²

= (a – 4b + 5c) (a – 4b – 5c)

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

y² (1 + x²) + x (1 + x²)
= (x – y²) (1 + x²)

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

(a + b)² – c²

= (a + b + c) (a + b – c)

49 – (a² – 8ab + 16b²)

= 49 – (a – 4b)²

We know: a² – b² = (a + b)(a-b)

= (7 + a – 4b) (7 – a + 4b)

= - (a – 4b + 7) (a – 4b – 7)

x (x² – 1)
= x (x + 1) (x – 1)

(x – 3y)² – (5a)²

= (x – 3y + 5a) (x – 3y – 5a)

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

2 [(3ax)² – (4)²]
= 2 (3ax + 4) (3ax – 4)

4x² + 12xy + 9y²
= (2x)² + (3y)² + 2 (2x) (3y)
= (2x + 3y)²

x² + 4 + 4x – 6x – 12 + 9

= x² + 1 – 2x

= (x – 1)²

Here, coefficient of a² = 6, coefficient of a = 17b and constant term = - 3b²

We shall now split up the coefficient of middle term i.e., 17b into two parts whose sum is 17b and product is 6 (-3b²) = - 18b²

Clearly,

18b – b = 17b and

6 (-3b²) = - 36y²

So, we replace middle term 17ab = 18ab – ab

Thus, we have

6a² +17ab – 3b² = 6a² + 18ab - ab – 3b²

= 6a (a + 3b) – b (a + 3b)

= (6a – b) (a + 3b)

Here, coefficient of x² = 7, coefficient of x = -19 and constant term = -6

We shall now split up the coefficient of x i.e., -19 into two parts whose sum is -19 and product is 7 x -6 = -42

Clearly, 2 - 21 = -19 and

2 x (-21) = - 42

So, we write middle term - 19x as 2x - 21x

Thus, we have

7x² - 19x – 6 = 7x² + 2x - 21x – 6

= x (7x + 2) - 3 (7x + 2)

= (7x + 2) (x – 3)

Here, coefficient of x² = 14, coefficient of x = 11y and constant term = - 15y²

We shall now split up the coefficient of middle term i.e., 11y into two parts whose sum is 11y and product is 14 (-15y²) = - 210y²

Clearly,

21y – 10y = 11y and

(21y) (-10y) = - 210y²

So, we replace middle term 11xy = 21xy – 10xy

Thus, we have

14x² + 11xy- 15y² = 14x² + 21xy - 10xy - 15y²

= 2x (7x – 5y) + 3y (7x – 5y)

= (2x + 3y) (7x - 5y)

Here, coefficient of x² = 6, coefficient of x = -13y and constant term = 2y²

We shall now split up the coefficient of middle term i.e., -13y into two parts whose sum is -13y and product is 6 (2y²) = 12y²

Clearly,

-12y – y = -13y and

(-12y) (-y) = 12y²

So, we replace middle term -13xy = -12xy – xy

Thus, we have

6x² -13xy+ 2y² = 6x² - 12xy - xy - 2y²

= (6x – y) (x - 2y)

7x - 6 - 2x² = - 2x² + 7x – 6

Here, coefficient of x² = - 2, coefficient of x = 7 and constant term = -6

We shall now split up the coefficient of x i.e., 7 into two parts whose sum is 7 and product is - 2 x - 6 = 12

Clearly,

4 + 3 = 7 and

4 x 3 = 12

So, we write middle term 7x as 4x + 3x

Thus, we have

- 2x² + 7x – 6 = - 2x² + 4x + 3x – 6

= - 2x (x – 2) + 3 (x – 2)

= (x – 2) (3 – 2x)

f(x)=-2x²+7x-6

=>-f(x)=2x²-7x+6

=>-f(x)=2x²-3x-4x+6

=>-f(x)=x(2x-3)-2(2x-3)

=>-f(x)=(2x-3)(x-2)

=>f(x)=(2x-3)(2-x)

Here, coefficient of x² = 6, coefficient of x = -5y and constant term = - 6y²

We shall now split up the coefficient of middle term i.e., -5y into two parts whose sum is -5y and product is 6 (-6y²) = - 36y²

Clearly,

4y – 9y = -5y and

(4y) (-9y) = - 36y²

So, we replace middle term -5xy = 4xy – 9xy

Thus, we have

6x² -5xy- 6y² = 6x² + 4xy - 9xy - 6y²

= (2x – 3y) (3x + 2y)

(2x²)² + (y²)² + 4x²y² – 4x²y²

= (2x² + y²)² – 4x²y²

= (2x² + y² + 2xy) (2x² + y² – 2xy)

(2x²)² + 1 + 4x² – 4x²
= (2x² + 1)² – 4x²
= (2x² + 2x + 1) (2x² – 2x + 1)

-x² – 4x + 96

= -x² – 12x + 8x + 96

= -x (x + 12) + 8 (x + 12)

= (x + 12) (-x + 8)

(2a – b)² – (4c)²
= (2a – b + 4c) (2a – b – 4c)

3a³ (a² – 16)
= 3a³ (a² – 4²)
= 3a³ (a + 4) (a – 5)

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)

(3z)² – [x² – 2 (x) (2y) + (2y)²]

= (3z)² – (x – 2y)²

= [3z + (x – 2y)] [3z – (x – 2y)]

[(3x + 4y)²]² – (x²)²
= [(3x + 4y)² + x²] [(3x + 4y)² – x²]
= [(3x + 4y)² + x²] [3x + 4y + x] [3x + 4y – x]

(x – 4y)² – (25)²
= (x – 4y + 25) (x – 4y – 25)

(pq)² + (3r)² – 2 (pq) (3r)
= (pq – 3r)²

Here, coefficient of (2a – b)² = 1, coefficient of (2a – b) = 2 and constant term = - 8

We shall now split up the coefficient of middle term i.e., 2 into two parts whose sum is 2 and product is -8 (1) = - 8

Clearly,

4 - 2 = 2 and

4 (-2) = - 8

So, we replace 2 (2a – b) = 4 (2a –b) – 2 (2a – b)

Thus, we have

(2a – b)² + 2 (2a – b) – 8 = (2a – b)² + 4 (2a – b) – 2 (2a – b) - 8

= (2a – b) (2a – b + 4) – 2 (2a – b + 4)

= (2a – b – 2) (2a – b + 4)

7x – 6x² + 20 = - 6x² + 7x + 20

Here, coefficient of x² = -6, coefficient of x = 7 and constant term = 20

We shall now split up the coefficient of x i.e., 7 into two parts whose sum is 7 and product is -6 x 20 = - 120

Clearly,

15 - 8 = 7 and

15 (-8) = - 120

So, we write middle term 7x as 15x - 8x

Thus, we have

-6x² + 7x + 20 = -6x² + 15x - 8x + 20

= -3x (2x – 5) - 4 (2x – 5)

= - (3x + 4) (2x - 5)

x² + 4y² – 4xy – 5x + 10y + 6

Here, coefficient of (x – 2y)² = 1, coefficient of (x – 2y ) = - 5 and constant = 6

We shall now split up the coefficient of middle term i.e., - 5 into two parts whose sum is -5 and product is 6 (1) = 6

Clearly,

- 2 - 3 = - 5 and

- 2 (-3) = 6

So, we replace - 5 (x – 3y) = - 2 (x – 2y) – 3 (x – 2y)

Thus, we have

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

= (x – 2y - 2) (x - 2y - 3)

Here, coefficient of x² = 15, coefficient of x = -16yz and constant term = - 15y²z²

We shall now split up the coefficient of middle term i.e., -16yz into two parts whose sum is -16yz and product is 15 (-15y²z²) = - 225y²z²

Clearly,

-25yz + 9yz = -16yz and

(-25yz) (9yz) = - 225y²z²

So, we replace middle term -16xyz = -25yz – 9yz

Thus, we have

15x² -16xyz- 15y²z² = 15x² - 25yz + 9yz - 15y²z²

= 5x (3x – 5yz) + 3yz (3x – 5yz)

= (5x + 3yz) (3x - 5yz)

3 + 23y – 8y² = - 8y² + 23y + 3

Here, coefficient of y² = -8, coefficient of y = 23 and constant term = 3

We shall now split up the coefficient of x i.e., 23 into two parts whose sum is 23 and product is -8 (3) = - 24

Clearly,

24 - 1 = 23 and

24 (-1) = - 24

So, we write middle term 23y as 24y - y

Thus, we have 

- 8y² + 23y + 3 = - 8² + 24y - y + 3

= - 8y (y – 3) - 1 (y – 3)

= - (8y + 1) (y – 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)

(x + 2y)² – [2 (2x – y)]²
= [(x + 2y) + 2 (2x – y)] [x + 2y – 2 (2x – y)]
= (x + 4x + 2y – 2y) (x – 4x + 2y + 2y)
= (5x) (4y – 3x)

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

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

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)

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

9 (4a² + 4a + 1)
= 9 [(2a)² + 2 (2a) + 1¹]
= 9 (2a + 1)²

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)

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

p + q = -11, pq = -42
Clearly,
3 – 14 = -11, 3 (-14) = -42
Therefore, split (-11x) as 3x – 14x
Therefore,
x² - 11x – 42 = x² + 3x – 14x – 42

= x (x + 3) – 14 (x + 3)

= (x – 14) (x + 3)

11x² – 54x + 63

Here, coefficient of x² = 11, coefficient of x = - 54 and constant term = 63

We shall now split up the coefficient of x i.e., -54 into two parts whose sum is - 54 and product is 11 x 63 = 693

Clearly,

-33x - 21x = - 54x and

(-33) x (-21) = 693

So, we write middle term - 54x as - 33x - 21x

Thus, we have

11x² – 54x + 63 = 11x² - 33x - 21x – 6

= 11x (x – 3) - 21 (x – 3)

= (11x – 21) (x – 3)

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)

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)

(2x + 1)² – (3x²)²
= (2x + 1 + 3x²) (2x + 1 – 3x²)
= (3x² + 2x + 1) (-3x² + 2x + 1)

Greatest common factor of the two terms namely 9x²y and 3axy of expression 9x²y + 3axy is 3xy 9x²y + 3axy = 3xy(3x² +a)

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

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 – 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)