Question

1. $(n^5 - 3n^4 + 7n^3 - 14n^2 + 5n + 11) div (n - 2)$

Explanation:

Step1: Set up polynomial long division

Divide $n^5$ by $n$ to get first term.
$\frac{n^5}{n}=n^4$

Step2: Multiply divisor by $n^4$

Subtract result from dividend.
$(n^5-3n^4+7n^3-14n^2+5n+11)-(n^4(n-2))=(n^5-3n^4)-(n^5-2n^4)=-n^4+7n^3-14n^2+5n+11$

Step3: Divide $-n^4$ by $n$

Get next term of quotient.
$\frac{-n^4}{n}=-n^3$

Step4: Multiply divisor by $-n^3$

Subtract from current remainder.
$(-n^4+7n^3)-(-n^3(n-2))=(-n^4+7n^3)-(-n^4+2n^3)=5n^3-14n^2+5n+11$

Step5: Divide $5n^3$ by $n$

Get next quotient term.
$\frac{5n^3}{n}=5n^2$

Step6: Multiply divisor by $5n^2$

Subtract from current remainder.
$(5n^3-14n^2)-(5n^2(n-2))=(5n^3-14n^2)-(5n^3-10n^2)=-4n^2+5n+11$

Step7: Divide $-4n^2$ by $n$

Get next quotient term.
$\frac{-4n^2}{n}=-4n$

Step8: Multiply divisor by $-4n$

Subtract from current remainder.
$(-4n^2+5n)-(-4n(n-2))=(-4n^2+5n)-(-4n^2+8n)=-3n+11$

Step9: Divide $-3n$ by $n$

Get last quotient term.
$\frac{-3n}{n}=-3$

Step10: Multiply divisor by $-3$

Subtract to find final remainder.
$(-3n+11)-(-3(n-2))=(-3n+11)-(-3n+6)=5$

Answer:

Quotient: $n^4 - n^3 + 5n^2 - 4n - 3$, Remainder: $5$
Or written as: $n^4 - n^3 + 5n^2 - 4n - 3 + \frac{5}{n-2}$