Fill in the missing values below one at a time to find the quotient when
-9x^3 - 12x^2 - 6x - 8 is divided by -3x - 4
To find the quotient when $-9x^3 - 12x^2 - 6x - 8$ is divided by $-3x - 4$, we can use long division.
\[
\begin{array}{c|cccc}
& -3x^2 & x & 2 \\ \cline{2-5}
-3x-4 & -9x^3 & -12x^2 & -6x & -8 \\
\end{array}
\]
First, we divide the term with the highest degree of $-9x^3$ by the term in the divisor, $-3x$, which gives us $3x^2$. We write this above the line.
\[
\begin{array}{c|cccc}
& -3x^2 & x & 2 \\ \cline{2-5}
-3x-4 & -9x^3 & -12x^2 & -6x & -8 \\
& \ \ \underline{-(-9x^3 + 3x^2)} & & & \\
& & -9x^2 & & \\
\end{array}
\]
Next, we multiply $-3x-4$ by $3x^2$ to get $-9x^3 - 12x^2$. We subtract this from $-9x^3 - 12x^2$:
\[
\begin{array}{c|cccc}
& -3x^2 & x & 2 \\ \cline{2-5}
-3x-4 & -9x^3 & -12x^2 & -6x & -8 \\
& \ \ \underline{-(-9x^3 + 3x^2)} & & & \\
& & -9x^2 & & \\
& & \ \ \underline{(-12x^2 + 9x^2)} & & \\
& & & -3x & \\
\end{array}
\]
Now we bring down the next term, $-6x$, and repeat the process. We divide $-3x$ by $-3x$ to get $1$, so we write this above the line.
\[
\begin{array}{c|cccc}
& -3x^2 & x & 2 \\ \cline{2-5}
-3x-4 & -9x^3 & -12x^2 & -6x & -8 \\
& \ \ \underline{-(-9x^3 + 3x^2)} & & & \\
& & -9x^2 & & \\
& & \ \ \underline{(-12x^2 + 9x^2)} & & \\
& & & -3x & \\
& & & \ \ \underline{-(-6x + 3x)} & \\
& & & & -3x \\
\end{array}
\]
Finally, we bring down the last term, $-8$, and divide $-3x$ by $-3x$ again to get $1$. We write this above the line.
\[
\begin{array}{c|cccc}
& -3x^2 & x & 2 \\ \cline{2-5}
-3x-4 & -9x^3 & -12x^2 & -6x & -8 \\
& \ \ \underline{-(-9x^3 + 3x^2)} & & & \\
& & -9x^2 & & \\
& & \ \ \underline{(-12x^2 + 9x^2)} & & \\
& & & -3x & \\
& & & \ \ \underline{-(-6x + 3x)} & \\
& & & & -3x \\
& & & & \ \ \underline{-(-8 + 3x)} \\
& & & & -5 \\
\end{array}
\]
Thus, the quotient when $-9x^3 - 12x^2 - 6x - 8$ is divided by $-3x - 4$ is $3x^2 + x + 2 - \frac{5}{-3x - 4}$.