Question:
Factorise:
(i) 16a2 − 24ab
(ii) 15ab2 − 20a2b
(iii) 12x2y3 − 21x3y2
Solution:
(i) H.C.F. of $16 a^{2}$ and $24 a b$ is $8 a$.
$\therefore 16 a^{2}-24 a b=8 a(2 a-3 b)$
(ii) H.C.F. of $15 a b^{2}$ and $20 a^{2} b$ is $5 a b$.
$\therefore 15 a b^{2}-20 a^{2} b=5 a b(3 b-4 a)$
(iii) H.C.F. of $12 x^{2} y^{3}$ and $21 x^{3} y^{2}$ is $3 x^{2} y^{2}$.
$\therefore 12 x^{2} y^{3}-21 x^{3} y^{2}=3 x^{2} y^{2}(4 y-7 x)$
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