# How do you factor  x^4-8x^2-9?

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

#### Explanation:

first lets substitute ${x}^{2} = t$
we get ${t}^{2} - 8 \cdot t - 9$
now split the middle term ${t}^{2} + t - 9 \cdot t - 9$
factorize now $t \left(t + 1\right) - 9 \left(t + 1\right)$
now we get $\left(t + 1\right) \left(t - 9\right)$
now substitute back $t = {x}^{2}$
we get $\left({x}^{2} + 1\right) \left({x}^{2} - 9\right)$
we can still factorize the ${x}^{2} - 9$ term
so we get $\left({x}^{2} + 1\right) \left(x + 3\right) \left(x - 3\right)$