How do you factor completely x^3-2x^2-9x+18?

May 8, 2017

$\left(x - 3\right) \left(x + 3\right) \left(x - 2\right)$

Explanation:

Factor by grouping:

$\left(\textcolor{b l u e}{{x}^{3} - 2 {x}^{2}}\right)$ $+$ $\left(\textcolor{red}{- 9 x + 18}\right)$

Starting on the left we can factor out an ${x}^{2}$

$\textcolor{b l u e}{{x}^{2} \left(x - 2\right)}$

On the right we can then factor out a $- 9$

$\textcolor{red}{- 9 \left(x - 2\right)}$

Observe:

$\textcolor{b l u e}{{x}^{2} \left(x - 2\right)}$ $+$ $\textcolor{red}{- 9 \left(x - 2\right)}$

*Notice how we have two $x - 2$. We can then simply rewrite the expression as follows.

$\left({x}^{2} - 9\right) \left(x - 2\right)$

*Note: all we did was combine $\textcolor{b l u e}{{x}^{2}}$ and $\textcolor{red}{- 9}$ and wrote $\left(x - 2\right)$ as one term instead of two.

We're not done just yet. We can still factor $\left({x}^{2} - 9\right)$ into $\left(x - 3\right) \left(x + 3\right)$

So a completely factored expression is then:

$\left(x - 3\right) \left(x + 3\right) \left(x - 2\right)$