Answer
$\dfrac{1}{3}$ is not a zero of the function by the factor theorem.
Work Step by Step
The factor theorem states that when $f(a)=0$, then we have $(x-a)$ as a factor of $f(x)$ and when $(x-a)$ is a factor of $f(x)$, then $f(a)=0$.
We are given the function $f(x)=2x^3+3x^2-6x+7$
We simplify the given equation as follows:
$f(\dfrac{1}{3})=2(\dfrac{1}{3})^3+3(\dfrac{1}{3})^2-6(\dfrac{1}{3})+7\\=\dfrac{2}{27}+\dfrac{3}{9}-\dfrac{6}{3}+7 \\=5.417$
This implies that $f(\dfrac{1}{3})=5.417 \ne 0$
So, $\dfrac{1}{3}$ is not a zero of the function by the factor theorem.
