# How do you solve x^5-5x^3+4x=0?

Oct 23, 2015

$x = 0$, $x = 1$, $x = 2$, $x = - 1$, and $x = - 2$

#### Explanation:

Start off by factoring out an $x$ as follows

$x \left({x}^{4} - 5 {x}^{2} + 4\right) = 0$

We can factor ${x}^{4} - 5 {x}^{2} + 4$

Doing so we have

$x \left({x}^{2} - 4\right) \left({x}^{2} - 1\right) = 0$

So in order for this to be equal to zero

$x = 0$

OR

${x}^{2} - 4 = 0$

OR

${x}^{2} - 1 = 0$

Well $x = 0$

${x}^{2} = 4$

$x = \pm \sqrt{4} = \pm 2$

${x}^{2} = 1$

$x = \pm \sqrt{1} = \pm 1$