Question

use synthetic division to find $(3x^3 - 16) \div (x + 2)$. write your answer in the form $q(x) + \frac{r}{d(x)}$, where $q(x)$ is a polynomial, $r$ is an integer, and $d(x)$ is a linear polynomial. simplify any fractions.

Explanation:

Step1: Rewrite the polynomial

First, rewrite \(3x^{3}-16\) as \(3x^{3}+0x^{2}+0x - 16\) to include all the missing terms with coefficient \(0\). The divisor is \(x + 2\), so we use \(c=-2\) for synthetic division (since \(x+2=x-(-2)\)).

Step2: Set up synthetic division

Set up the synthetic division with \(c = - 2\) and the coefficients of the dividend \(3,0,0,-16\):

\[

$$\begin{array}{r|rrrr} -2 & 3 & 0 & 0 & -16\\ & & -6 & 12 & -24\\ \hline & 3 & -6 & 12 & -40\\ \end{array}$$

\]

Step3: Interpret the results

The numbers at the bottom are the coefficients of the quotient polynomial and the remainder. The quotient polynomial \(q(x)\) has degree one less than the dividend. So \(q(x)=3x^{2}-6x + 12\) and the remainder \(r=-40\), and the divisor \(d(x)=x + 2\).

Answer:

\(3x^{2}-6x + 12+\frac{-40}{x + 2}\) (or \(3x^{2}-6x + 12-\frac{40}{x + 2}\))