Find the quotient and the remainder when $x^{3}+x^{2}-10x+8$ is divided by $x+3$.

(N/A) To find the quotient and remainder,we perform polynomial long division of $(x^{3}+x^{2}-10x+8)$ by $(x+3)$:
$1$. 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: $(x^3+x^2-10x+8) - (x^3+3x^2) = -2x^2-10x+8$.
$3$. Divide the new first term $(-2x^2)$ by $x$ to get $-2x$.
$4$. Multiply $-2x$ by $(x+3)$ to get $-2x^2-6x$. Subtract this: $(-2x^2-10x+8) - (-2x^2-6x) = -4x+8$.
$5$. Divide the new first term $(-4x)$ by $x$ to get $-4$.
$6$. Multiply $-4$ by $(x+3)$ to get $-4x-12$. Subtract this: $(-4x+8) - (-4x-12) = 20$.
Thus,the Quotient $= x^{2}-2x-4$ and the Remainder $= 20$.