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: Identify the root of the divisor

For the divisor \(x + 2\), set \(x+2 = 0\), so \(x=-2\). The coefficients of the dividend \(3x^{3}-16\) (rewritten as \(3x^{3}+0x^{2}+0x - 16\)) are \(3\), \(0\), \(0\), \(-16\).

Step2: Perform synthetic division

Bring down the leading coefficient \(3\).
Multiply \(3\) by \(-2\) to get \(-6\), add to the next coefficient \(0\): \(0 + (-6)=-6\).
Multiply \(-6\) by \(-2\) to get \(12\), add to the next coefficient \(0\): \(0 + 12 = 12\).
Multiply \(12\) by \(-2\) to get \(-24\), add to the last coefficient \(-16\): \(-16+(-24)=-40\).
The quotient polynomial \(q(x)\) has coefficients \(3\), \(-6\), \(12\), so \(q(x)=3x^{2}-6x + 12\), the remainder \(r=-40\), and the divisor \(d(x)=x + 2\).

Answer:

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