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

May 31, 2015

First notice that substituting $x = 1$ results in ${x}^{3} - 3 x + 2 = 0$

So $\left(x - 1\right)$ is a factor.

Use synthetic division to find the remaining factor...

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

Notice that $x = 1$ is also a root of ${x}^{2} + x - 2$

So we have another $\left(x - 1\right)$ factor...

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

So the complete factorization is:

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

May 31, 2015

$f \left(x\right) = {x}^{3} - 3 x + 2 = 0$

Since $\left(a + b + c + d = 0\right)$, then one factor is $\left(x - 1\right)$

Factored form: $f \left(x\right) = \left(x - 1\right) \left({x}^{2} + x - 2\right) = \left(x - 1\right) \left(x - 1\right) \left(x + 2\right) =$

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