Divide $p(x) = x^{3} + 7x^{2} + 14x + 1$ by $x + 3$ and find the quotient and the remainder.

(N/A) To divide $p(x) = x^{3} + 7x^{2} + 14x + 1$ by $x + 3$,we use polynomial long division:
$1$. Divide the first term of the dividend $(x^{3})$ by the first term of the divisor $(x)$ to get $x^{2}$.
$2$. Multiply $x^{2}$ by $(x + 3)$ to get $x^{3} + 3x^{2}$. Subtract this from the dividend to get $4x^{2} + 14x + 1$.
$3$. Divide $4x^{2}$ by $x$ to get $4x$. Multiply $4x$ by $(x + 3)$ to get $4x^{2} + 12x$. Subtract this to get $2x + 1$.
$4$. Divide $2x$ by $x$ to get $2$. Multiply $2$ by $(x + 3)$ to get $2x + 6$. Subtract this to get $-5$.
Thus,the quotient is $x^{2} + 4x + 2$ and the remainder is $-5$.