Page 160 - ICSE Math 8

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Example 9: Evaluate (402) – (398) using the identity p – q = (p + q)(p – q).
Solution: Here p = 402, q = 398
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\ (402) – (398) = (402 + 398)(402 – 398) = (800)(4) = 3,200
728 × 728 – 418 × 418
Example 10: Simplify .
(728 + 418)
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(728) – (418) 2 (728 + 418)(728 – 418)
Solution: = = 728 – 418 = 310
(728 + 418) (728 + 418)
EXERCISE 14.3

1. Factorize the following:
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(a) 48x – 243y (b) 36(x + y) – 49(x – 2y) 2 (c) x −
16 169
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(d) 24l – (e) a – 81b (f) 256a – b 8 (g) (x + y) – (x – y) 4
25m 2
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(h) 63x y – 7 (i) 4 – (l – m) 2 (j) ab − bc (k) 16(2p – 1) – 25q 2
36 49
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2. Evaluate using p – q = (p + q)(p – q), where
(a) p = 16, q = 4 (b) p = 8.8, q = 1.2 (c) p = 506, q = 494 (d) p = 52, q = 48
Factorization of a Perfect Square Trinomial
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Any trinomial which can be expressed as x + 2xy + y or x – 2xy + y is known as a perfect square trinomial.
The factorization of algebraic expressions expressible as a perfect square involves the following identities:
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(i) x + 2xy + y = (x + y) = (x + y)(x + y) (ii) x – 2xy + y = (x – y) = (x – y)(x – y)
Example 11: Factorize the following.
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(a) 9a + 24ab + 16b (b) 25a – 4b + 28bc – 49c 2 Maths Info
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Solution: (a) 9a + 24ab + 16b = (3a) + 2(3a)(4b) + (4b) 2 The square of a binomial is called
a perfect square trinomial.
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= (3a + 4b) = (3a + 4b)(3a + 4b)
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(b) 25a – 4b + 28bc – 49c = (5a) – {4b – 28bc + 49c }
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= (5a) – {(2b) – 2(2b)(7c) + (7c) } = (5a) – (2b – 7c) 2
= {5a + 2b – 7c}{5a – (2b – 7c)}
= (5a + 2b – 7c)(5a – 2b + 7c)
Try This
Example 12: Factorize the following.
Factorize:
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(a) (x – 8x y + 16y ) – 289 (b) a – b + a + b 8 (a) a – 2ab + b + b – a
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Solution: (a) (x – 8x y + 16y ) – 289 = {(x ) – 2(x )(4y ) + (4y ) } – (17) 2 (b) x + y – 2 (xy + yz – xz)
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= (x – 4y ) – (17) = (x – 4y + 17)(x – 4y – 17)
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(b) a – b + a + b = {(a ) – (b ) } + (a + b ) {Using x – y = (x – y)(x + y)}
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= (a + b )(a – b ) + (a + b ) {Taking (a + b ) common}
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= (a – b + 1)(a + b )
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