Check whether the polynomial $q(t) = 4t^3 + 4t^2 - t - 1$ is a multiple of $2t + 1$.

(A) To check if $q(t)$ is a multiple of $2t + 1$,we need to determine if $2t + 1$ is a factor of $q(t)$.
According to the Factor Theorem,$2t + 1$ is a factor of $q(t)$ if $q(-\frac{1}{2}) = 0$.
Setting $2t + 1 = 0$,we get $t = -\frac{1}{2}$.
Now,substituting $t = -\frac{1}{2}$ into the polynomial $q(t)$:
$q(-\frac{1}{2}) = 4(-\frac{1}{2})^3 + 4(-\frac{1}{2})^2 - (-\frac{1}{2}) - 1$
$q(-\frac{1}{2}) = 4(-\frac{1}{8}) + 4(\frac{1}{4}) + \frac{1}{2} - 1$
$q(-\frac{1}{2}) = -\frac{1}{2} + 1 + \frac{1}{2} - 1 = 0$.
Since the remainder is $0$,$2t + 1$ is a factor of $q(t)$.
Therefore,$q(t)$ is a multiple of $2t + 1$.