Answer
True
Work Step by Step
$g\left( x \right)={{x}^{2}}-9$
To find the zeros of the polynomial, set the polynomial equal to 0 and solve for the variable value.
So, $g\left( x \right)=0$
And
${{x}^{2}}-9=0$
Apply the identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$,
$\begin{align}
& {{x}^{2}}-9=0 \\
& {{x}^{2}}-{{3}^{2}}=0 \\
& \left( x+3 \right)\left( x-3 \right)=0
\end{align}$
Thus:
$\begin{align}
\left( x+3 \right)\left( x-3 \right)=0 & \\
\left( x+3 \right)=0,\left( x-3 \right)=0 & \\
x=-3,x=3 & \\
\end{align}$
Thus, the zeros of $g\left( x \right)={{x}^{2}}-9$ are $-3$ and $3$.
