Divide $x^{6}+5 x^{3}+7 x+3$ by $x^{2}+2$.

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