Answer
$P\left( \dfrac{1}{3} \right)=-\dfrac{187}{81}$
Work Step by Step
Substituting $x$ with $
\dfrac{1}{3}
$ in $
P(x)=2x^4-3x^2-2
$, then
\begin{array}{l}
P\left( \dfrac{1}{3} \right)=2\left( \dfrac{1}{3} \right)^4-3\left( \dfrac{1}{3} \right)^2-2
\\\\
P\left( \dfrac{1}{3} \right)=2\left( \dfrac{1}{81} \right)-3\left( \dfrac{1}{9} \right)-2
\\\\
P\left( \dfrac{1}{3} \right)=\dfrac{2}{81}-\dfrac{1}{3}-2
\\\\
P\left( \dfrac{1}{3} \right)=\dfrac{2}{81}-\dfrac{27}{81}-\dfrac{162}{81}
\\\\
P\left( \dfrac{1}{3} \right)=-\dfrac{187}{81}
.\end{array}
