# How do you factor x^3 - 2x^2 - x + 2 ?

May 22, 2015

We can factor out $x$ from all the elements that contain it:

$x \left({x}^{2} - 2 x - x\right) - 2$

And then factor the quadratic inside the parenthesis, which, using Bhaskara, will give us the following roots:

$\frac{2 \pm \sqrt{\cancel{\left(4 - 4 \left(1\right) \left(1\right)\right)}}}{2}$
$\frac{2}{2} = 1$

${x}_{1} = {x}_{2} = 1$

Thus, the factors are both $\left(x - 1\right)$

Finally: $x \left(\left(x - 1\right) \left(x - 1\right)\right) + 2$

May 22, 2015

Factor by grouping:

${x}^{3} - 2 {x}^{2} - x + 2$

$= \left({x}^{3} - 2 {x}^{2}\right) + \left(- x + 2\right)$ (Notice the inserted addition before the second pair.)

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

$= \left({x}^{2} - 1\right) \left(x - 2\right)$ Now the first binomial is a difference of squares, so

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

${x}^{3} - 2 {x}^{2} - x + 2 = \left(x + 1\right) \left(x - 1\right) \left(x - 2\right)$