Answer
zeros of $ f(x)$ are $2,\dfrac{1}{2}, -3 $
Work Step by Step
When a polynomial $ f(x)$ is divided by $(x-a)$, then the remainder is $ r=f(a)$
We have the dividend $ f(x)= 2x^3+x^2-13x+6$ and it is divided by $(x-2)$.
So, $ f(x)=(x-2) (2x^2+5x-3)$
or, $ f(x)=(x-2)(2x-1)(x+3)$
We need to solve for $ x $, so set each factor equal to $0$.
$ x=2,\dfrac{1}{2}, -3 $
Thus, the zeros of $ f(x)$ are $2,\dfrac{1}{2}, -3 $
