How do you find all the zeros of f(x)=x^4-5x^2-36?

Aug 14, 2016

$f \left(x\right)$ has zeros $\pm 3$ and $\pm 2 i$

Explanation:

Note that $9 \cdot 4 = 36$ and $9 - 4 = 5$.

So we find:

${x}^{4} - 5 {x}^{2} - 36$

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

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

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

Hence zeros: $\pm 3$ and $\pm 2 i$