One of the factors of  $4x^3+4x^2-x-1$ is

$4x^3+4x^2-x-1=f\left(x\right)$

If $x=-1$ then $f(-1)=4{(-1)}^3+4{(-1)}^2-(-1)-1$

$=-4+4+1-1=0$

$\therefore (x+1)$ is a factor of $f\left(x\right)$

Put $x=-\frac{1}{2}$

$=4{\left(-\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)}^3+4{\left(-\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)}^2-\left(-\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)-1$

$=-\frac{4}{8}+\frac{4}{4}+\frac{1}{2}-1$

$=\frac{-4+8+4-8}{8}=\frac{0}{8}=0$

$\therefore 2x+1$ is a factor of $f\left(x\right)$