Answer
$\left\{-\sqrt2, -\sqrt3, \sqrt2, \sqrt3\right\}$
Work Step by Step
The zeros of a function can be found by setting $f(x)=0$ then solving the resulting equation.
Set $f(x)=0$ then solve the equation to obtain:
\begin{align*}
0&=x^4-5x^2+6\\
0&=(x^2)^2-5x^2+6
\end{align*}
Let $u=x^2$.
The equation above becomes:
\begin{align*}
0&=u^2-5u+6\\
0&=(u-3)(u-2)
\end{align*}
Use the Zero-Product Property by equating each factor to zero then solve each equation to obtain:
\begin{align*}
u-3&=0 &\text{or}& &u-2=0\\
u&=3 &\text{or}& &u=2
\end{align*}
Replace $u$ with $x^2$ to obtain:
\begin{align*}
x^2&=3 &\text{or}& &x^2=2\\
x&=\pm\sqrt3 &\text{or}& &x=\pm\sqrt2
\end{align*}
Thus, the zeros of the given function are: $\left\{-\sqrt2, -\sqrt3, \sqrt2, \sqrt3\right\}$
