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

Mar 10, 2016

$x = \left\{- 3 , - 2 , 2 , 3\right\}$

#### Explanation:

let $u = {x}^{2}$ then $f \left(x\right) = {x}^{4} - 5 {x}^{2} - 36 \to f \left(u\right) = {u}^{2} - 5 u - 36$
${u}^{2} - 5 u - 36 = 0$
$\left(u - 9\right) \left(u + 4\right) = 0$
$u - 9 = 0 , u + 4 = 0$
$u = 9 \mathmr{and} u = - 4$
${x}^{2} = 9 , {x}^{2} = 4$
$x = \pm 3 , x = \pm 2$
$x = \left\{- 3 , - 2 , 2 , 3\right\}$