# How do you factor x^4-9x^2+18?

Jun 4, 2016

$\left(t - 6\right) \left(t - 3\right)$

#### Explanation:

Say:
${x}^{2} = t$

Then:
${t}^{2} - 9 t + 18$

After this you can factorise for t:

So factorisation for t would be: $\left(t - 6\right) \left(t - 3\right)$.

Now, we have to find x; in order to do that, we have to find the roots of t's. Since this is a quadratic equation it has to be equal zero, so:
$\left(t - 6\right) \left(t - 3\right) = 0$
Accordingly: $t - 6 = 0$ and $t - 3 = 0$. From here we get: $t 1 = 6$ and $t 2 = 3$.

Let's find x1 and x2:
$t 1 = {x}^{2}$
$6 = {x}^{2}$
$\pm \sqrt{6} = x 1 , x 2$ (two roots: $x 1 = \sqrt{6}$, $x 2 = - \sqrt{6}$ )

and x
$t 2 = {x}^{2}$
$3 = {x}^{2}$
$\pm \sqrt{3} = x 3 , x 4$ (two roots: $x 3 = \sqrt{3}$, $x 4 = - \sqrt{4}$ )

Why $\pm$? Because any number square-rooted can be negative or positive. RECALL: ${\left(- 8\right)}^{2} = 64$ as well as ${8}^{2} = 64$ but not $- {8}^{2}$ is not equal $64$.

To get the correct answers we have to find which of the four roots fits into the equation to satisfy it to be zero.

First, let's try with the first one:
${\left(\sqrt{6}\right)}^{4} - 9 {\left(\sqrt{6}\right)}^{2} + 18 = 0 \implies 36 - 9 \cdot 6 + 18 = 0 \implies 36 - 54 + 18 = 0 \implies 0 = 0$

Secondly,
${\left(- \sqrt{6}\right)}^{4} - 9 {\left(- \sqrt{6}\right)}^{2} + 18 = 0 \implies 36 - 9 \cdot 6 + 18 \implies 0 = 0$

Thirdly,
${\left(\sqrt{3}\right)}^{4} - 9 {\left(\sqrt{3}\right)}^{2} + 18 = 0 \implies 9 - 9 \cdot 3 + 18 = 0 \implies 9 - 27 + 18 = 0 \implies 0 = 0$

Lastly,
${\left(- \sqrt{3}\right)}^{4} - 9 {\left(- \sqrt{3}\right)}^{2} + 18 = 0 \implies 9 - 9 \cdot 3 + 18 = 0 \implies 0 = 0$

So the solutions are: $- \sqrt{3} , \sqrt{3} , - \sqrt{6} , \sqrt{6}$