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(i) (3m + 5)2

(ii) (5p – 1)2

(iii) (2n – 1)(2n + 3)

(iv) 4p2 – 25q2

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(i) (3m + 5)2

Comparing (3m + 5)2 with (a + b)2 we have a = 3m and b = 5

(a + b)2 = a2 + 2 ab + b2

(3m + 5)2 = (3m)2 + 2 (3m) (5) + 52

= 32m2 + 30m + 25 

= 9m2 + 30m +25

(ii) (5p – 1)2

Comparing (5p – 1)2 with (a – b)2 we have a = 5p and b = 1

(a – b)2 = a2 – 2ab + b2

(5p – 1)2 = (5p)2 – 2 (5p) (1) + 12

= 52p2 – 10p + 1 = 25p2 – 10p + 1

(iii) (2n – 1)(2n + 3)

Comparing (2n – 1) (2n + 3) with (x + a) (x + b) we have a = -1; b = 3

(x + a) (x + b) = x2 + (a + b)x + ab

(2n +(- 1)) (2n + 3) 

= (2n)2 + (-1 + 3)2n + (-1) (3)

= 22n2 + 2 (2n) – 3 = 4n2 + 4n – 3

(iv) 4p2 – 25q2 = (2p)2 – (5q)2

Comparing (2p)2 – (5q)2 with a2 – b2 we have a = 2p and b = 5q

(a2 – b2) = (a + b)(a – b) 

= (2p + 5q) (2p – 5q)

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