Question

find the quotient and remainder using synthetic division: $\frac{x^{4}+9x^{3}-113x - 32}{x + 6}$
the quotient is
the remainder is
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Explanation:

Step1: Identify root and coefficients

Divisor: $x+6$, so root $c=-6$.
Dividend coefficients (include $0x^2$): $1, 9, 0, -113, -32$

Step2: Set up synthetic division

Write $-6$ left, coefficients top:

$$\begin{array}{r|rrrrr} -6 & 1 & 9 & 0 & -113 & -32 \\ \hline & & & & & \\ \end{array}$$

Step3: Bring down leading coefficient

Bring down $1$ to bottom row:

$$\begin{array}{r|rrrrr} -6 & 1 & 9 & 0 & -113 & -32 \\ \hline & & & & & \\ & 1 & & & & \\ \end{array}$$

Step4: Multiply and add (column 2)

Multiply $1\times(-6)=-6$, add to $9$: $9+(-6)=3$

$$\begin{array}{r|rrrrr} -6 & 1 & 9 & 0 & -113 & -32 \\ \hline & & -6 & & & \\ & 1 & 3 & & & \\ \end{array}$$

Step5: Multiply and add (column 3)

Multiply $3\times(-6)=-18$, add to $0$: $0+(-18)=-18$

$$\begin{array}{r|rrrrr} -6 & 1 & 9 & 0 & -113 & -32 \\ \hline & & -6 & -18 & & \\ & 1 & 3 & -18 & & \\ \end{array}$$

Step6: Multiply and add (column 4)

Multiply $-18\times(-6)=108$, add to $-113$: $-113+108=-5$

$$\begin{array}{r|rrrrr} -6 & 1 & 9 & 0 & -113 & -32 \\ \hline & & -6 & -18 & 108 & \\ & 1 & 3 & -18 & -5 & \\ \end{array}$$

Step7: Multiply and add (column 5)

Multiply $-5\times(-6)=30$, add to $-32$: $-32+30=-2$

$$\begin{array}{r|rrrrr} -6 & 1 & 9 & 0 & -113 & -32 \\ \hline & & -6 & -18 & 108 & 30 \\ & 1 & 3 & -18 & -5 & -2 \\ \end{array}$$

Step8: Form quotient polynomial

Bottom row (excluding last term) are coefficients of quotient (degree 3): $1, 3, -18, -5$
Quotient: $x^3 + 3x^2 - 18x - 5$
Last term is remainder: $-2$

Answer:

The quotient is $x^3 + 3x^2 - 18x - 5$
The remainder is $-2$