Answer
Zeros: $1,-1$
$x$-intercepts: $1,-1$
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 $G(x)=0$:
$$3x^4-2x^2-1=0$$
Let $u=x^2$, the original equation becomes
$$3u^2-2u-1=0$$
By factoring
$$(3u+1)(u-1) = 0$$
Use the Zero-Product Property by equating each factor to zero, then solve each equation ton obtain:
\begin{align*}
3u+1 &=0 &\text{ or }& &u-1=0\\
3u &= -1 &\text{ or }& &u=1\\
u &= -\dfrac{1}{3} &\text{ or }& &u=1\\
\end{align*}
To solve for $x$, we use $u=x^2$
$$\because u = x^2$$
$$\therefore x = \pm \sqrt{u}$$
For $u=-\dfrac{1}{3}$
$$x=\pm \sqrt{\left(-\dfrac{1}{3}\right)} \hspace{20pt} \text{ No real solution} $$
For $u=1$
$$x= \pm \sqrt{1}$$
$\therefore x = 1,-1$
