Factorise the following.
(A) $4 x^{2} - 20 x + 25$
(B) $a^{3} - 4 a^{2} + 12 - 3 a$

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Solution

(A) $4 x^{2} - 20 x + 25 = (2 x)^{2} - 2 \times 2 x \times 5 + (5)^{2} = (2 x - 5)^{2} (Since a^{2} - 2 a b + b^{2} = (a - b)^{2}) = (2 x - 5) (2 x - 5)$

(B) $a^{3} - 4 a^{2} + 12 - 3 a = a^{3} - 4 a^{2} - 3 a + 12 = a^{2} (a - 4) - 3 (a - 4) = (a - 4) (a^{2} - 3)$

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Factorise the following. (A) 4x2−20x+25 (B) a3−4a2+12−3a

(A) 4x2−20x+25=(2x)2−2×2x×5+(5)2=(2x−5)2(Since a2−2ab+b2=(a−b)2)=(2x−5)(2x−5) (B) a3−4a2+12−3a=a3−4a2−3a+12=a2(a−4)−3(a−4)=(a−4)(a2−3)

Factorise the following.
(A) $4 x^{2} - 20 x + 25$
(B) $a^{3} - 4 a^{2} + 12 - 3 a$

(A) $4 x^{2} - 20 x + 25 = (2 x)^{2} - 2 \times 2 x \times 5 + (5)^{2} = (2 x - 5)^{2} (Since a^{2} - 2 a b + b^{2} = (a - b)^{2}) = (2 x - 5) (2 x - 5)$

(B) $a^{3} - 4 a^{2} + 12 - 3 a = a^{3} - 4 a^{2} - 3 a + 12 = a^{2} (a - 4) - 3 (a - 4) = (a - 4) (a^{2} - 3)$