Answer
Remainder = $0$
$(x-\frac{1}{2})$ is a factor of $f(x)$.
Work Step by Step
The Remainder Theorem states that when a function $f(x)$ is divided by $(x-R)$ , then the remainder will be: $f(R)$.
We have: $f(x)=2x^4-x^3+2x-1$
$f\left(\frac{1}{2}\right)=(2)\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^3+(2)\left(\frac{1}{2}\right)-1\\=(2)\left(\frac{1}{16})-(\frac{1}{8}\right)+1-1\\
=\frac{1}{8}-\frac{1}{8}\\
=0$
The Factor Theorem states that if $f(a)=0$, then $(x-a)$ is a factor of $f(x)$ and vice versa.
Therefore, by the Factor Theorem $(x-\frac{1}{2})$ is a factor of $f(x)$.
