Answer
Zeros: $2\sqrt{2}, -2 \sqrt{2}$
$x$-intercepts: $2\sqrt{2}, -2 \sqrt{2}$
Work Step by Step
To find the zeros of a function $f$, solve the equation $f(x)=0$
The zeros of the function are also the $x-$intercepts.
Let $P(x)=0$:
$$x^4-6x^2-16=0$$
Let $u=x^2$, the original equation becomes
$$u^2-6u-16=0$$
By factoring
$$(u+2)(u-8) = 0$$
Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain:
\begin{align*}
u+2 &=0 &\text{ or }& &u-8=0\\
u &= -2 &\text{ or }& &u=8\\
\end{align*}
To solve for $x$, we use $u=x^2$
$$\because u = x^2$$
$$\therefore x = \pm \sqrt{u}$$
For $u=-2$
$$x=\pm \sqrt{-2} \hspace{20pt} \text{ No real solution}$$
For $u=8$
$$x= \pm \sqrt{8}$$
$$x= \pm 2\sqrt{2}$$
$\therefore x = 2\sqrt{2}, -2 \sqrt{2}$
