Answer
Zeros:$2,-2$
$x$-intercepts: $2,-2$
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 $F(x)=0$:
$$2x^4-5x^2-12=0$$
Let $u=x^2$, the original equation becomes
$$2u^2-5u-12=0$$
By factoring
$$(2u+3)(u-4) = 0$$
Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain::
\begin{align*}
2u+3 &=0 &\text{ or }& &u-4=0\\
2u &= -3 &\text{ or }& &u=4\\
u &= -\dfrac{3}{2} &\text{ or }& &u=4\\
\end{align*}
To solve for $x$, we use $u=x^2$
$$\because u = x^2$$
$$\therefore x = \pm \sqrt{u}$$
For $u=-\dfrac{3}{2}$
$$x=\pm \sqrt{\left(-\dfrac{3}{2}\right)} \hspace{20pt} \text{ No real solution} $$
For $u=4$
$$x= \pm \sqrt{4}$$
$$x= \pm 2$$
$\therefore x = 2,-2$
