Math worksheet on polynomial long division with a fun title.

To solve the problem, we need to perform long division of polynomials for each given expression. Let's go through each one step by step.

---

Problem 1:

\[
(2x^3 - 9x + 5) \div (x - 7)
\]

#### Step 1: Set up the division.
Divide \(2x^3\) by \(x\) to get \(2x^2\). Multiply \(2x^2\) by \(x - 7\) and subtract:
\[
\begin{array}{r|rrrr}
x - 7 & 2x^3 & 0x^2 & -9x & +5 \\
& 2x^2 & -14x^2 & & \\
\hline
& & 14x^2 & -9x & +5 \\
& & 14x^2 & -98x & \\
\hline
& & & 89x & +5 \\
& & & 89x & -623 \\
\hline
& & & & 628 \\
\end{array}
\]
The quotient is \(2x^2 + 14x + 89\) and the remainder is \(628\).

So, the result is:
\[
2x^2 + 14x + 89 + \frac{628}{x - 7}
\]

---

Problem 2:

\[
(2x^3 + 8x^2 + 15) \div (x + 3)
\]

#### Step 1: Set up the division.
Divide \(2x^3\) by \(x\) to get \(2x^2\). Multiply \(2x^2\) by \(x + 3\) and subtract:
\[
\begin{array}{r|rrrr}
x + 3 & 2x^3 & +8x^2 & 0x & +15 \\
& 2x^2 & +6x^2 & & \\
\hline
& & 2x^2 & 0x & +15 \\
& & 2x^2 & +6x & \\
\hline
& & & -6x & +15 \\
& & & -6x & -18 \\
\hline
& & & & 33 \\
\end{array}
\]
The quotient is \(2x^2 + 2x - 6\) and the remainder is \(33\).

So, the result is:
\[
2x^2 + 2x - 6 + \frac{33}{x + 3}
\]

---

Problem 3:

\[
(3x^3 + 2x - 1) \div (x + 2)
\]

#### Step 1: Set up the division.
Divide \(3x^3\) by \(x\) to get \(3x^2\). Multiply \(3x^2\) by \(x + 2\) and subtract:
\[
\begin{array}{r|rrrr}
x + 2 & 3x^3 & 0x^2 & +2x & -1 \\
& 3x^2 & +6x^2 & & \\
\hline
& & -6x^2 & +2x & -1 \\
& & -6x^2 & -12x & \\
\hline
& & & 14x & -1 \\
& & & 14x & +28 \\
\hline
& & & & -29 \\
\end{array}
\]
The quotient is \(3x^2 - 6x + 14\) and the remainder is \(-29\).

So, the result is:
\[
3x^2 - 6x + 14 - \frac{29}{x + 2}
\]

---

Problem 4:

\[
(5x^3 - 12x^2 + 11x + 12) \div (3x + 3)
\]

#### Step 1: Set up the division.
Divide \(5x^3\) by \(3x\) to get \(\frac{5}{3}x^2\). Multiply \(\frac{5}{3}x^2\) by \(3x + 3\) and subtract:
\[
\begin{array}{r|rrrr}
3x + 3 & 5x^3 & -12x^2 & +11x & +12 \\
& \frac{5}{3}x^2 & +5x^2 & & \\
\hline
& & -\frac{41}{3}x^2 & +11x & +12 \\
& & -\frac{41}{3}x^2 & -\frac{41}{3}x & \\
\hline
& & & \frac{74}{3}x & +12 \\
& & & \frac{74}{3}x & +74 \\
\hline
& & & & -62 \\
\end{array}
\]
The quotient is \(\frac{5}{3}x^2 - \frac{41}{9}x + \frac{74}{9}\) and the remainder is \(-62\).

So, the result is:
\[
\frac{5}{3}x^2 - \frac{41}{9}x + \frac{74}{9} - \frac{62}{3x + 3}
\]

---

Problem 5:

\[
(x^3 + 729) \div (x + 8)
\]

#### Step 1: Set up the division.
Divide \(x^3\) by \(x\) to get \(x^2\). Multiply \(x^2\) by \(x + 8\) and subtract:
\[
\begin{array}{r|rrrr}
x + 8 & x^3 & 0x^2 & 0x & +729 \\
& x^2 & +8x^2 & & \\
\hline
& & -8x^2 & 0x & +729 \\
& & -8x^2 & -64x & \\
\hline
& & & 64x & +729 \\
& & & 64x & +512 \\
\hline
& & & & 217 \\
\end{array}
\]
The quotient is \(x^2 - 8x + 64\) and the remainder is \(217\).

So, the result is:
\[
x^2 - 8x + 64 + \frac{217}{x + 8}
\]

---

Problem 6:

\[
(3x^4 + 15x^3 + 15x^2 - 3x - 6) \div (3x + 6)
\]

#### Step 1: Set up the division.
Divide \(3x^4\) by \(3x\) to get \(x^3\). Multiply \(x^3\) by \(3x + 6\) and subtract:
\[
\begin{array}{r|rrrrr}
3x + 6 & 3x^4 & +15x^3 & +15x^2 & -3x & -6 \\
& x^3 & +2x^3 & & & \\
\hline
& & 13x^3 & +15x^2 & -3x & -6 \\
& & 13x^3 & +26x^2 & & \\
\hline
& & & -11x^2 & -3x & -6 \\
& & & -11x^2 & -22x & \\
\hline
& & & & 19x & -6 \\
& & & & 19x & +38 \\
\hline
& & & & & -44 \\
\end{array}
\]
The quotient is \(x^3 + 3x^2 - \frac{11}{3}x + \frac{19}{3}\) and the remainder is \(-44\).

So, the result is:
\[
x^3 + 3x^2 - \frac{11}{3}x + \frac{19}{3} - \frac{44}{3x + 6}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad 2x^2 + 14x + 89 + \frac{628}{x - 7} \\
2. & \quad 2x^2 + 2x - 6 + \frac{33}{x + 3} \\
3. & \quad 3x^2 - 6x + 14 - \frac{29}{x + 2} \\
4. & \quad \frac{5}{3}x^2 - \frac{41}{9}x + \frac{74}{9} - \frac{62}{3x + 3} \\
5. & \quad x^2 - 8x + 64 + \frac{217}{x + 8} \\
6. & \quad x^3 + 3x^2 - \frac{11}{3}x + \frac{19}{3} - \frac{44}{3x + 6}
\end{aligned}
}
\]