Divide $3 x^{4}+5 x^{3}-7 x^{2}+2 x+2$ by $x^{2}+3 x+1$.

(N/A) To divide $3 x^{4}+5 x^{3}-7 x^{2}+2 x+2$ by $x^{2}+3 x+1$,we use long division:
$1$. Divide the first term of the dividend $(3 x^{4})$ by the first term of the divisor $(x^{2})$ to get $3 x^{2}$.
$2$. Multiply $3 x^{2}$ by $(x^{2}+3 x+1)$ to get $3 x^{4}+9 x^{3}+3 x^{2}$.
$3$. Subtract this from the dividend: $(3 x^{4}+5 x^{3}-7 x^{2}+2 x+2) - (3 x^{4}+9 x^{3}+3 x^{2}) = -4 x^{3}-10 x^{2}+2 x+2$.
$4$. Divide the first term of the new polynomial $(-4 x^{3})$ by $x^{2}$ to get $-4 x$.
$5$. Multiply $-4 x$ by $(x^{2}+3 x+1)$ to get $-4 x^{3}-12 x^{2}-4 x$.
$6$. Subtract this: $(-4 x^{3}-10 x^{2}+2 x+2) - (-4 x^{3}-12 x^{2}-4 x) = 2 x^{2}+6 x+2$.
$7$. Divide $2 x^{2}$ by $x^{2}$ to get $2$.
$8$. Multiply $2$ by $(x^{2}+3 x+1)$ to get $2 x^{2}+6 x+2$.
$9$. Subtracting this results in a remainder of $0$.
Thus,the quotient is $3 x^{2}-4 x+2$ and the remainder is $0$.