Divide the polynomial $3x^{4} - 4x^{3} - 3x - 1$ by $x - 1$.

(D) To divide $p(x) = 3x^{4} - 4x^{3} - 3x - 1$ by $x - 1$,we use the long division method:
$1$. Divide the first term of the dividend $(3x^{4})$ by the first term of the divisor $(x)$ to get $3x^{3}$.
$2$. Multiply $3x^{3}$ by $(x - 1)$ to get $3x^{4} - 3x^{3}$. Subtract this from the dividend to get $-x^{3} - 3x - 1$.
$3$. Divide $-x^{3}$ by $x$ to get $-x^{2}$. Multiply $-x^{2}$ by $(x - 1)$ to get $-x^{3} + x^{2}$. Subtract to get $-x^{2} - 3x - 1$.
$4$. Divide $-x^{2}$ by $x$ to get $-x$. Multiply $-x$ by $(x - 1)$ to get $-x^{2} + x$. Subtract to get $-4x - 1$.
$5$. Divide $-4x$ by $x$ to get $-4$. Multiply $-4$ by $(x - 1)$ to get $-4x + 4$. Subtract to get $-5$.
The quotient is $3x^{3} - x^{2} - x - 4$ and the remainder is $-5$.
Alternatively,using the Remainder Theorem,the zero of $x - 1$ is $1$. Substituting $x = 1$ into $p(x)$:
$p(1) = 3(1)^{4} - 4(1)^{3} - 3(1) - 1$
$p(1) = 3 - 4 - 3 - 1 = -5$.
Thus,the remainder is $-5$.