Factorise-i -2-2-2 -2 ii -2-10-21

(i) (x^2 2xy + y^2) − z^2 nbsp; nbsp; = (x y)^2 z^2 nbsp; nbsp; [∵ (a − b)^2 = a^2 2ab + b^2] = (x y z)(x y+z) [∵ a^2 b^2 = (a + ...

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Standard VIII

Mathematics

Factors of Algebraic Expressions and Factorisation

Factorise:i �...

Question

Factorise:
(i) $(x^{2} - 2 x y + y^{2}) - z^{2}$
(ii) $q^{2} - 10 q + 21$

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Solution

$(i) (x^{2} - 2 x y + y^{2}) - z^{2} = (x - y)^{2} - z^{2} [∵ (a - b)^{2} = a^{2} - 2 a b + b^{2}] = (x - y - z) (x - y + z) [∵ a^{2} - b^{2} = (a + b) (a - b)]$

$(ii) q^{2} - 10 q + 21 = q^{2} + (- 3 - 7) q + (- 3) (- 7) = (q - 3) (q - 7) [∵ (x + a) (x + b) = x^{2} + (a + b) x + a b]$

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Factorise: (i) (x2−2xy+y2)−z2 (ii) q2−10q+21

(i) (x2−2xy+y2)−z2=(x−y)2−z2 [∵(a−b)2=a2−2ab+b2]=(x−y−z)(x−y+z) [∵a2−b2=(a+b)(a−b)] (ii) q2−10q+21=q2+(−3−7)q+(−3)(−7)=(q−3)(q−7) [∵(x+a)(x+b)=x2+(a+b)x+ab]

Factorise:
(i) $(x^{2} - 2 x y + y^{2}) - z^{2}$
(ii) $q^{2} - 10 q + 21$

$(i) (x^{2} - 2 x y + y^{2}) - z^{2} = (x - y)^{2} - z^{2} [∵ (a - b)^{2} = a^{2} - 2 a b + b^{2}] = (x - y - z) (x - y + z) [∵ a^{2} - b^{2} = (a + b) (a - b)]$

$(ii) q^{2} - 10 q + 21 = q^{2} + (- 3 - 7) q + (- 3) (- 7) = (q - 3) (q - 7) [∵ (x + a) (x + b) = x^{2} + (a + b) x + a b]$