Question

Factorise : [4 MARKS]
$1)x^3-2x^2-x+2$
$2)2y^3+y^2-2y-1$

Answer 1

Concept: 1 Mark
Factorisation: 1.5 Marks each

1)Let $p\left(x\right)=x^3-2x^2-x+2$

$p\left(1\right)=1^3-2{\left(1\right)}^2-1+2$
$=1-2-1+2=0$ [Using hit and trial method]

By Factor theorem, we can say that $(x-1)$ is the factor of $p\left(x\right)$

We can also find other factors of p(x) by doing long division but we will just split the terms to take (x − 1) common from p(x).

$p\left(x\right)=x^3-2x^2-x+2$

$=x^3-x^2-x^2+x-2x+2$

$=x^2(x-1)-x(x-1)-2(x-1)$

$=(x-1)(x^2-x-2)$

$=(x-1)(x-2)(x+1)$

$2\left)p\right(y)=2y^3+y^2-2y-1$

$p\left(1\right)=2{\left(1\right)}^3+{\left(1\right)}^2-2\left(1\right)-1=2+1-2-1=0$[Using hit and trial method]

Therefore, by factor theorem, we can say that $(y-1)$ is a factor of p(y).

Now, we will split the terms accordingly to take (y−1) common from p(y). We can also do long division to find other factors but it may take more time.

$p\left(y\right)=2y^3+y^2-2y-1$

$=2y^3+2y^2-y^2-y-y-1$

$=2y^2(y+1)-y(y+1)-1(y+1)$

$=(y+1)(2y^2-y-1)$

$=(y+1)(y-1)(2y+1)$