(A) To find the quotient and remainder,we perform polynomial long division of $(x^{3}+x^{2}-10x+8)$ by $(x-2)$:
$1$. 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 $(x-2)$ to get $x^{3}-2x^{2}$.
$3$. Subtract $(x^{3}-2x^{2})$ from $(x^{3}+x^{2}-10x+8)$ to get $3x^{2}-10x+8$.
$4$. Divide $3x^{2}$ by $x$ to get $3x$. Multiply $3x$ by $(x-2)$ to get $3x^{2}-6x$.
$5$. Subtract $(3x^{2}-6x)$ from $(3x^{2}-10x+8)$ to get $-4x+8$.
$6$. Divide $-4x$ by $x$ to get $-4$. Multiply $-4$ by $(x-2)$ to get $-4x+8$.
$7$. Subtract $(-4x+8)$ from $(-4x+8)$ to get $0$.
Thus,the Quotient $= x^{2}+3x-4$ and the Remainder $= 0$.