Question

Factorise:
(i) 14x³ + 21x⁴y − 28x²y²
(ii) −5 − 10t + 20t²

Answer 1

(i) H.C.F. of $14x^3$, $21x^{4}y$ and $28x^2{y}^2$ is $7x^2$.

∴ $14x^3+21x^{4}y-28x^2{y}^2=7x^2\left(2x+3x^{2}y-4y^2\right)$

(ii) H.C.F. of $-5$, $-10t$ and $20t^2$ is 5.

∴ $-5-10t+20t^2=5\left(-1-2t+4t^2\right)$