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

(N/A) The dividend polynomial is $p(x) = x^{4} + 0x^{3} - 3x^{2} + 4x + 5$ and the divisor polynomial is $s(x) = x^{2} - x + 1$.
Performing polynomial long division:
$1$. Divide the first term of the dividend $(x^{4})$ by the first term of the divisor $(x^{2})$ to get $x^{2}$.
$2$. Multiply $x^{2}$ by $(x^{2} - x + 1)$ to get $x^{4} - x^{3} + x^{2}$. Subtract this from the dividend to get $x^{3} - 4x^{2} + 4x + 5$.
$3$. Divide the first term of the new polynomial $(x^{3})$ by $x^{2}$ to get $x$. Multiply $x$ by $(x^{2} - x + 1)$ to get $x^{3} - x^{2} + x$. Subtract to get $-3x^{2} + 3x + 5$.
$4$. Divide the first term $(-3x^{2})$ by $x^{2}$ to get $-3$. Multiply $-3$ by $(x^{2} - x + 1)$ to get $-3x^{2} + 3x - 3$. Subtract to get the remainder $8$.
Thus,the quotient is $q(x) = x^{2} + x - 3$ and the remainder is $r(x) = 8$.