Question:
Factorise:
(i) 14x3 + 21x4y − 28x2y2
(ii) −5 − 10t + 20t2
Solution:
(i) H.C.F. of $14 x^{3}, 21 x^{4} y$ and $28 x^{2} y^{2}$ is $7 x^{2}$.
$\therefore 14 x^{3}+21 x^{4} y-28 x^{2} y^{2}=7 x^{2}\left(2 x+3 x^{2} y-4 y^{2}\right)$
(ii) H.C.F. of $-5,-10 t$ and $20 t^{2}$ is 5 .
$\therefore-5-10 t+20 t^{2}=5\left(-1-2 t+4 t^{2}\right)$
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