# How do you factor x^5-5x^3-36x completely?

Feb 5, 2015

We need to start out by factoring out any common factors. We notice that we can factor out x.

$x \left({x}^{4} - 5 {x}^{2} - 36\right)$

Note that ${x}^{4} = {\left({x}^{2}\right)}^{2}$

$x \left({\left({x}^{2}\right)}^{2} - 5 {x}^{2} - 36\right)$

Let $r = {x}^{2}$

$x \left({r}^{2} - 5 r - 36\right)$

Now factor $\left({r}^{2} - 5 r - 36\right)$

$x \left(r - 9\right) \left(r + 4\right)$

Replace $r$ with ${x}^{2}$

$x \left({x}^{2} - 9\right) \left({x}^{2} + 4\right)$

We should recognize $\left({x}^{2} - 9\right)$ as a difference of squares.

Leaving me with

$x \left(x - 3\right) \left(x + 3\right) \left({x}^{2} + 4\right)$