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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)

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)

= x² – xz + xy – yz 

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

= (x + y)(x – z).

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, 

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, 

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)

(a²)² – [4 (b – c)²]
= [a² + 4 (b – c)²] [a² – 4 (b – c)²]
= [a² + 4 (b – c)²] [(a + 2b – 2c) (a – 2b + 2c)]

8 (a – b)² [2 (a – b) – 3]
= 8 (a – b)² [2a – 2b – 3]

a (b – 1) – 1 (b – 1)
= (a – 1) (b – 1)

abx² + ab + xa² + xb²
= ax (bx + a) + b (bx + a)
= (ax + b) (bx + a)

a²x² + ax³ + x + a
= x (ax² + 1) + a (ax² + 1)
= (x + a) (ax² + 1)

a² – (b² – 2bc + c²)

= a² – (b – c)²

= (a + b – c) (a – b + c)

(a² + b² + c²) (x + y) [Therefore, taking (x + y) common in each term]

x (256x⁴ – 81)
= x [(16x²)² – 9²]
= x (16x + 9) (16x – 9)

3ab² (25a² – 36b²)
= 3ab² [(5a)² – (6b)²]
= 3ab² (5a + 6b) (5a – 6b)

[(x + y) + (a – b)] [(x + y) – (a – b)]
= (x + y + a – b) (x + y – a + b)

\(2(\frac{25}{x\times x}-\frac{x\times x}{81})\)

= \(2[(\frac{5}{x})^2-(\frac{x}{9})^2]\)

= \(2[(\frac{5}{x}+\frac{x}{9})(\frac{5}{x}-\frac{x}{9})\)

x³ (x² – 16)
= x³ (x² – 4²)
= x³ (x + 4) (x – 4)

\((\frac{1}{4}xy)^2 - (\frac{2}{7}yz)^2\)

 = \((\frac{xy}{4}+\frac{2yz}{7})(\frac{xy}{4}-\frac{2yz}{7})\)

= \(y^2(\frac{x}{4}+\frac{2}{7}z)(\frac{x}{4}-\frac{2}{7}z)\)

(4x)² – (5y)²
= (4x + 5y) (4x – 5y)

Consider 27x² - 12y²,Taking 3 common we get,3 [(3x)² – (2y)²]

As we know a² - b² = (a-b) (a+b)

= 3 (3x + 2y) (3x – 2y)

(12a)² – (17b)²
= (12a + 17b) (12a – 17b)

3 (4m² – 9)
= 3 [(2m)² – 3²]
= 3 (2m + 3) (2m – 3)

5 (25x² – 9y²)
= 5 [(5x)² – (3y)²]
= 5 (5x + 3y) (5x – 3y)

Greatest common factor of the two terms namely 21lmn, - 3lm²n, 4lmn² of expression 21lmn - 3lm²n + 4lmn² is lm

21lmn - 3lm²n + 4lmn² = lm(21 - 3m + 4n)

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

= a(b + n) – m(n + b) 

= (a – m)(b + n).

pq² + q(p – 1) – 1 = pq² + qp – q – 1 

= pq(q + 1) – 1(q + 1) 

= (pq – 1)(q + 1)

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)

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

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

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, 

= 2x 2 + 2x + 3x + 3 

= 2x(x +1) + 3(x + 1) 

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

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).

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)

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).

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).

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

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

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).

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

= x(x – 12)(x + 12) 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)

(x⁴)²–(1)²
= (x⁴ + 1) (x⁴ – 1)

8² – (a + 1)²
= [8 + (a + 1)] [8 – (a + 1)]
= (a + 9) (7 – a)

(6l)² – (m + n)²
= (6l + m + n) (6l – m – n)

(a²)² – (\(\frac{1}{b\times b}\))²
= (a² + \(\frac{1}{b\times b}\)) (a² - \(\frac{1}{b\times b}\))

(5x²y²)² – (1)²
= (5x²y² – 1) (5x²y² + 1)

The numerical coefficients of given numerical are 12, 6, 2

Greatest common factor of 12, 6, 2 is 2.

Common literals appearing in given numerical are a and x

Smallest power of x in three monomials = 2

Smallest power of a in three monomials = 1

Monomials of common literals with smallest power= ax²

Hence, the greatest common factor = 2ax²

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).

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).