# How do you find all the zeros of F(x) = x^4 - 13x^2 + 36?

Jul 16, 2016

Zeros: $\pm 3$, $\pm 2$

#### Explanation:

Since there are no terms of odd degree, we can treat this as a quadratic in ${x}^{2}$ to simplify it, then factorize each of the resulting quadratic factors.

First note that $36 = 9 \times 4$ and $9 + 4 = 13$, so we find:

${x}^{4} - 13 {x}^{2} + 36$

$= {\left({x}^{2}\right)}^{2} - 13 \left({x}^{2}\right) + 36$

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

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

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

Hence zeros:

$\pm 3$, $\pm 2$