Question

one factor of this polynomial is (x + 8).
$x^3 + 5x^2 - 11x + 104$
use synthetic division to find the other factor of the polynomial.
$\bigcirc$ $x^3 + 3x^2 + 13x$
$\bigcirc$ $x^3 - 3x^2 + 13x$

Explanation:

Step1: Identify root for synthetic division

Since the factor is \((x + 8)\), the root \(r=-8\). The coefficients of the polynomial \(x^{3}+5x^{2}-11x + 104\) are \(1,5,-11,104\).

Step2: Perform synthetic division

Set up synthetic division with root \(-8\):
\[

$$\begin{array}{r|rrrr} -8 & 1 & 5 & -11 & 104 \\ & & -8 & 24 & -104 \\ \hline & 1 & -3 & 13 & 0 \\ \end{array}$$

\]
The quotient polynomial has coefficients \(1,-3,13\) and since the original polynomial is cubic and we divided by a linear factor, the quotient is quadratic: \(x^{2}-3x + 13\)? Wait, no, wait. Wait, the original polynomial is \(x^{3}+5x^{2}-11x + 104\), when we divide by \((x + 8)\), the result should be a quadratic polynomial. Wait, but the options given are cubic? Wait, maybe there is a mistake in the options or in my understanding. Wait, no, wait, the original polynomial is \(x^{3}+5x^{2}-11x + 104\), and when we do synthetic division with root \(-8\), we get:

First term: \(1\) (carry down).

Multiply \(1\) by \(-8\) to get \(-8\), add to \(5\): \(5+(-8)=-3\).

Multiply \(-3\) by \(-8\) to get \(24\), add to \(-11\): \(-11 + 24 = 13\).

Multiply \(13\) by \(-8\) to get \(-104\), add to \(104\): \(104+(-104)=0\).

So the quotient is \(x^{2}-3x + 13\)? But the options are cubic. Wait, maybe the original polynomial is being factored as \((x + 8)\times(\text{quadratic})\), but the options given are cubic. Wait, maybe there is a typo in the options. Wait, no, the options provided are:

Option 1: \(x^{3}+3x^{2}+13x\)

Option 2: \(x^{3}-3x^{2}+13x\)

Wait, maybe the question is wrong, but according to the synthetic division, the quotient is \(x^{2}-3x + 13\), but if we multiply \((x + 8)\) with \(x^{2}-3x + 13\), we get \(x^{3}-3x^{2}+13x+8x^{2}-24x + 104=x^{3}+5x^{2}-11x + 104\), which matches the original polynomial. But the options are cubic. Wait, maybe the options are missing the constant term? Wait, no, the options are \(x^{3}+3x^{2}+13x\) and \(x^{3}-3x^{2}+13x\). Wait, if we take the quotient as \(x^{2}-3x + 13\) and multiply by \(x\), we get \(x^{3}-3x^{2}+13x\), which is option 2. Wait, maybe the question is asking for the product of \((x + 8)\) and the quotient? No, the question says "find the other factor". Wait, the original polynomial is \((x + 8)\times(\text{quadratic})\), but the options are cubic. Wait, maybe there is a mistake in the problem. But according to the synthetic division, the quotient (the other factor, ignoring the linear factor) is \(x^{2}-3x + 13\), but if we consider that maybe the original polynomial was miswritten, and the options are cubic, then if we multiply \((x + 8)\) with \(x^{2}-3x + 13\), we get \(x^{3}+5x^{2}-11x + 104\), but the options are \(x^{3}-3x^{2}+13x\) (option 2) which is \(x(x^{2}-3x + 13)\), so maybe the question has a mistake, but based on the synthetic division result, the quadratic factor is \(x^{2}-3x + 13\), and if we multiply by \(x\) we get \(x^{3}-3x^{2}+13x\), which is option 2.

Answer:

B. \(x^{3}-3x^{2}+13x\)