Find the quotient and the remainder when $x^{3}+x^{2}-10x+8$ is divided by $x+4$.

(A) To find the quotient and remainder,we perform polynomial long division of $(x^{3}+x^{2}-10x+8)$ by $(x+4)$:
$1$. 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 $(x+4)$ to get $x^3+4x^2$. Subtract this from the dividend: $(x^3+x^2-10x+8) - (x^3+4x^2) = -3x^2-10x+8$.
$3$. Divide the new first term $(-3x^2)$ by $x$ to get $-3x$.
$4$. Multiply $-3x$ by $(x+4)$ to get $-3x^2-12x$. Subtract this from the current remainder: $(-3x^2-10x+8) - (-3x^2-12x) = 2x+8$.
$5$. Divide the new first term $(2x)$ by $x$ to get $2$.
$6$. Multiply $2$ by $(x+4)$ to get $2x+8$. Subtract this: $(2x+8) - (2x+8) = 0$.
Thus,the Quotient $= x^{2}-3x+2$ and the Remainder $= 0$.