Answer
$\dfrac{3}{5}$ 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 $f(x)=2x^6-5x^4+x^3-x+1$
We simplify the given equation as follows:
$f(\dfrac{3}{5})=2(\dfrac{3}{5})^6-5(\dfrac{3}{5})^4+(\dfrac{3}{5})^3-(\dfrac{3}{5})+1\\=0.061312$
This implies that $f(\dfrac{3}{5}) \ne 0$
So, $\dfrac{3}{5}$ is not a zero of the function by the factor theorem.
