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2. Factorise.
(i) \(4p^2 – 9q^2\)

(ii) \(63a^2 – 112b^2\)

(iii) \(49x^2 – 36\)

(iv) \(16x^5 – 144x^3\)

(v) \((l + m)^2 – (l – m)^2\)

(vi) \(9x^2y^2 – 16\)

(vii) \((x^2 – 2xy + y^2) – z^2\)

(viii) \(25a^2 – 4b^2 + 28bc – 49c^2\)


Answer :

(i)Given: \(4p^2 – 9q^2\)

\(= (2p)^2 – (3q)^2\)

\(= (2p – 3q) (2p + 3q)\quad\)[∵ \(a^2 – b^2 = (a + b)(a – b)\)]

(ii)Given: \(63a^2 – 112b^2\)

\(= 7(9a^2 – 16b^2)\)

\(= 7 [(3a)^2 – (4b)^2]\)

\(= 7(3a – 4b)(3a + 4b)\quad\)[∵ \(a^2 – b^2 = (a + b)(a – b)\)]

(iii) Given:\(49x^2 – 36\)

\(49x^2 – 36 = (7x)^2 – (6)^2\)

\(= (7x – 6) (7x + 6)\quad\)[∵ \(a^2 – b^2 = (a + b)(a – b)\)]

(iv)Given: \(16x^5 – 144x^3 = 16x^3 (x^2 – 9)\)

\(= 16x^3 [(x)^2 – (3)^2]\)

\(= 16x^3(x – 3)(x + 3)\quad \)[∵ \(a^2 – b^2 = (a + b)(a – b)\)]

(v) Given:\((l + m)^2 – (l – m)^2\)

\(= (l + m) – (l – m)] [(l + m) + (l – m)]\quad \)[∵ \(a^2 – b^2 = (a + b)(a – b)\)]

\(= (l + m – l + m)(l + m + l – m)\)

\(= (2m) (2l)\)

\(= 4ml\)

(vi)Given \(9x^2y^2 – 16 = (3xy)^2 – (4)^2\)

\(= (3xy – 4)(3xy + 4)\quad \)[∵ \(a^2 – b^2 = (a + b)(a – b)]\)

(vii)Given: \((x^2 – 2xy + y^2) – z^2\)

\(= (x – y)^2 – z^2\)

\(= (x – y – z) (x – y + z)\quad \)[∵ \(a^2 – b^2 = (a + b)(a – b)]\)

(viii) Given:\(25a^2 – 4b^2 + 28bc – 49c^2\)

\(= 25a^2 – (4b^2 – 28bc + 49c^2)\)

\(= (5a)^2 – (2b – 7c)^2\)

\(= [5a – (2b – 7c)] [5a + (2b – 7c)]\)

\(= (5a – 2b + 7c)(5a + 2b – 7c)\)