As you know, $q(t)$ will be a multiple of $2 t+1$ only, if $2 t+1$ divides $q(t)$ leaving remainder zero. Now, taking $2 t+1=0,$ we have $t=-\frac{1}{2}$

Also, $q\left(-\frac{1}{2}\right)=4\left(-\frac{1}{2}\right)^{3}+4\left(-\frac{1}{2}\right)^{2}-\left(-\frac{1}{2}\right)-1=-\frac{1}{2}+1+\frac{1}{2}-1=0$

So the remainder obtained on dividing $q(t)$ by $2 t+1$ is $0$ .

So, $2 t+1$ is a factor of the given polynomial $q(t),$ that is $q(t)$ is a multiple of $2 t+1$.