Answer:
[tex]\large{\text{Yes.}\ f(x)=x-\dfrac{1}{2} \text{is a factor of}\ f(x)=2x^4+x^3+x-\dfrac{3}{4}.}[/tex]
Step-by-step explanation:
[tex]\text{If}\ (x-a)\ \text{is a factor of}\ p(x),\ \text{then}\ p(a)=0.[/tex]
[tex]\text{Thereofre if}\ h(x)=x-\dfrac{1}{2}\ \text{is a factor of}\ f(x)=2x^4+x^3+x-\dfrac{3}{4},\ \text{then}\ f\left(\dfrac{1}{2}\right)=0.[/tex]
[tex]\text{Substitute:}\\\\f\left(\dfrac{1}{2}\right)=2\left(\dfrac{1}{2}\right)^4+\left(\dfrac{1}{2}\right)^3+\dfrac{1}{2}-\dfrac{3}{4}=2\left(\dfrac{1}{16}\right)+\dfrac{1}{8}+\dfrac{1}{2}-\dfrac{3}{4}\\\\=\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1\cdot2}{2\cdot2}-\dfrac{3}{4}=\dfrac{1+1}{8}+\dfrac{2}{4}-\dfrac{3}{4}=\dfrac{2}{8}+\dfrac{2}{4}-\dfrac{3}{4}\\\\=\dfrac{1}{4}+\dfrac{2}{4}-\dfrac{3}{4}=\dfrac{1+2-3}{4}=\dfrac{0}{4}=0[/tex]
