Question

1. $(3n^4 + 6n^3 - 15n^2 + 32n - 25) div (n + 4)$

Explanation:

Step1: Use Polynomial Long Division

Divide the leading term of the dividend \(3n^4\) by the leading term of the divisor \(n\), we get \(3n^3\). Multiply the divisor \((n + 4)\) by \(3n^3\): \(3n^3(n + 4)=3n^4+12n^3\). Subtract this from the dividend:
\[

$$\begin{align*} &(3n^4 + 6n^3 - 15n^2 + 32n - 25)-(3n^4+12n^3)\\ =&3n^4 + 6n^3 - 15n^2 + 32n - 25 - 3n^4 - 12n^3\\ =& - 6n^3 - 15n^2 + 32n - 25 \end{align*}$$

\]

Step2: Divide the new leading term

Divide the leading term of \(-6n^3\) by \(n\) to get \(-6n^2\). Multiply the divisor by \(-6n^2\): \(-6n^2(n + 4)=-6n^3-24n^2\). Subtract this from the result of Step1:
\[

$$\begin{align*} &(-6n^3 - 15n^2 + 32n - 25)-(-6n^3-24n^2)\\ =&-6n^3 - 15n^2 + 32n - 25 + 6n^3 + 24n^2\\ =&9n^2 + 32n - 25 \end{align*}$$

\]

Step3: Divide the new leading term

Divide the leading term of \(9n^2\) by \(n\) to get \(9n\). Multiply the divisor by \(9n\): \(9n(n + 4)=9n^2+36n\). Subtract this from the result of Step2:
\[

$$\begin{align*} &(9n^2 + 32n - 25)-(9n^2+36n)\\ =&9n^2 + 32n - 25 - 9n^2 - 36n\\ =& - 4n - 25 \end{align*}$$

\]

Step4: Divide the new leading term

Divide the leading term of \(-4n\) by \(n\) to get \(-4\). Multiply the divisor by \(-4\): \(-4(n + 4)=-4n-16\). Subtract this from the result of Step3:
\[

$$\begin{align*} &(-4n - 25)-(-4n-16)\\ =&-4n - 25 + 4n + 16\\ =& - 9 \end{align*}$$

\]

Step5: Combine the results

The quotient is \(3n^3-6n^2 + 9n-4\) and the remainder is \(-9\). So we can write the division as:
\[
\frac{3n^4 + 6n^3 - 15n^2 + 32n - 25}{n + 4}=3n^3-6n^2 + 9n-4-\frac{9}{n + 4}
\]

Answer:

\(3n^3 - 6n^2 + 9n - 4-\frac{9}{n + 4}\)