Question

Factorize:
$x^3-3x+2$

Answer 1

Given: $x^3-3x+2$

Now if we put $x=1$ then $\to (1{)}^3-3(1)+2=0$

It's coming zero, it means $(x-1)$ is a factor of $x^3-3x+2$

Now remaining roots we will find by dividing.

$x-1)\overline{x^3-3x+2}(x^2+x-2x^3-x^2\_\_\_\_\_\_\_\_x^2-3x+2x^2-x\_\_\_\_\_\_\_\_-2x+2-2x+2\_\_\_\_\_\_\_0$

So, $x^3-3x+2=(x-1)(x^2+x-2)$

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

$=(x-1)\left(x\right(x+2)-1(x+2\left)\right)$

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

So, $x^3-3x+2=(x-1{)}^2(x+2)$