Page 129 - Start Up Mathematics_8 (Non CCE)

Factorization of Algebraic Expressions Expressible as a Perfect Square
The factorization of algebraic expressions expressible as a perfect square involves the following identities:
2
2
2
2
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2
(i) p + 2pq + q = (p + q) = (p + q)(p + q) (ii) p – 2pq + q = (p – q) = (p – q)(p – q)
Example 10: Factorize the following:
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2
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(a) 9a + 24ab + 16b (b) (l + m) – 4lm (NCERT)
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4
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(c) 25a – 4b + 28bc – 49c 2 (NCERT) (d) a – (a – b) 4
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Solution: (a) 9a + 24ab + 16b = (3a) + 2(3a)(4b) + (4b) 2

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= (3a + 4b) 2 {Using p + 2pq + q = (p + q) }
= (3a + 4b)(3a + 4b)
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2
2
2
2

2
(b) (l + m) – 4lm = l + m + 2lm – 4lm {Using p + 2pq + q = (p + q) }
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= l + m – 2lm
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= l – 2lm + m 2
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2

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= (l – m) = (l – m)(l – m) {Using p – 2pq + q = (p – q) }
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(c) 25a – 4b + 28bc – 49c = (5a) – {4b – 28bc + 49c }
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= (5a) – {(2b) – 2(2b)(7c) + (7c) }
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2

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= (5a) – (2b – 7c) 2 {Using p – 2pq + q = (p – q) }
= {5a + 2b – 7c}{5a – (2b – 7c)}
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{Using p – q = (p + q)(p – q)}
= (5a + 2b – 7c)(5a – 2b + 7c)
2 2
2 2
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(d) a – (a – b) = (a ) – {(a – b) }
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= {a + (a – b) }{a – (a – b) } {Using p – q = (p + q)(p – q)}
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= (a + a – 2ab + b )(a + a – b){a – (a – b)}
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{Using (p – q) = p – 2pq + q and p – q = (p + q)(p – q)}
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= (2a – 2ab + b )(2a – b)(a – a + b)
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= (2a – 2ab + b )(2a – b)b
Example 11: Factorize the following:
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2 2
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4
(a) (x – 8x y + 16y ) – 289 (b) a – 2ab + b + b – a
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16
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(c) x + y – 2 (xy + yz – xz) (d) a – b + a + b 8
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2 2
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2 2
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2 2
Solution: (a) (x – 8x y + 16y ) – 289 = {(x ) – 2(x )(4y ) + (4y ) } – (17) 2
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2 2
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= (x – 4y ) – (17) 2 {Using p – 2pq + q = (p – q) }
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= (x – 4y + 17)(x – 4y – 17)
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{Using p – q = (p – q)(p + q)}
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(b) a – 2ab + b + b – a = (a – 2ab + b ) + (b – a) {Using p – 2pq + q = (p – q) }
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= (a – b) – 1(a – b) {Taking (–1) common from the last term}
= (a – b – 1)(a – b)
{Taking (a – b) common from both the terms}
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