Answer
$\left( -3,0 \right),\left( -2,0 \right)\left( 2,0 \right)$ and $\left( 3,0 \right)$
Work Step by Step
$f\left( x \right)={{x}^{4}}-13{{x}^{2}}+36$
Assume $z={{x}^{2}}\,\text{and}\,{{z}^{2}}={{x}^{4}}$
Then simplify the expression.
$\begin{align}
& f\left( x \right)={{x}^{4}}-13{{x}^{2}}+36 \\
& f\left( x \right)={{z}^{2}}-13z+36
\end{align}$
$\begin{align}
& {{z}^{2}}-13z+36=0 \\
& \left( z-4 \right)\left( z-9 \right)=0
\end{align}$
Apply zero product rule,
If $\left( z-4 \right)=0$ then,
$\begin{align}
& \left( z-4 \right)=0 \\
& z=4
\end{align}$
If $\left( z-9 \right)=0$ then,
$\begin{align}
& \left( z-9 \right)=0 \\
& z=9
\end{align}$
Now replace the z with ${{x}^{2}}$ and solve these equations.
$\begin{align}
& {{x}^{2}}=4 \\
& x=\sqrt{4} \\
& x=\pm 2
\end{align}$
Or
$\begin{align}
& {{x}^{2}}=9 \\
& x=\sqrt{9} \\
& x=\pm 3
\end{align}$
Thus the $x$ $intercepts$ are $\left( -3,0 \right)\left( -2,0 \right)\left( 2,0 \right)$ and $\left( 3,0 \right)$
