# What are all the possible rational zeros for f(x)=x^4+2x^2-8 and how do you find all zeros?

Sep 9, 2016

$x = \pm \sqrt{2}$ or $x = \pm 2 i$

#### Explanation:

$f \left(x\right) = {x}^{4} + 2 {x}^{2} - 8$

The simplest way to solve for the zeros of this function is to use the transformation $z = {x}^{2}$ the set $f \left(z\right) = 0$

Thus: $f \left(z\right) = {z}^{2} + 2 z - 8 = 0$

Factorise: $\left(z - 2\right) \left(z + 4\right) = 0$

Hence: either $z = 2$ or $z = - 4$

But $z = {x}^{2} \to x = \pm \sqrt{z}$

$\therefore$ either $x = \pm \sqrt{2}$ or $x = \pm \sqrt{- 4} = \pm \sqrt{4} i$

$x = \pm \sqrt{2}$ or $x = \pm 2 i$

Therefore we can see that $f \left(x\right)$ has two irrational zeros and two complex zeros.