Question

use synthetic division to rewrite the following fraction in the form $q(x) + \frac{r(x)}{d(x)}$, where $d(x)$ is the denominator of the original fraction, $q(x)$ is the quotient, and $r(x)$ is the remainder.\
\\(\frac{3x^5 + 20x^4 + 20x^3 - 25x^2 + 2}{x + 5}\\)

Explanation:

Step1: Identify root of divisor

The divisor is \(x + 5\), so the root is \(x=-5\) (since \(x + 5=0\Rightarrow x=-5\)).

Step2: Set up synthetic division

Write the coefficients of the dividend \(3x^{5}+20x^{4}+20x^{3}-25x^{2}+0x + 2\) (note the missing \(x\) term has coefficient 0) and the root \(-5\):

$$
LATEXBLOCK0
$$

of synthetic division steps: Bring down the first coefficient (3). Multiply by \(-5\) to get \(-15\), add to next coefficient (\(20-15 = 5\)). Multiply 5 by \(-5=-25\), add to next coefficient (\(20-25=-5\)). Multiply \(-5\) by \(-5 = 25\), add to next coefficient (\(-25 + 25=0\)). Multiply 0 by \(-5 = 0\), add to next coefficient (\(0+0 = 0\)). Multiply 0 by \(-5 = 0\), add to last coefficient (\(2+0 = 2\)).

Step3: Form quotient and remainder

The coefficients of the quotient \(q(x)\) are from the results (excluding the last number, which is the remainder \(r(x)\)). The degree of \(q(x)\) is one less than the dividend. So \(q(x)=3x^{4}+5x^{3}-5x^{2}+0x + 0=3x^{4}+5x^{3}-5x^{2}\) and \(r(x)=2\), \(d(x)=x + 5\).
So \(\frac{3x^{5}+20x^{4}+20x^{3}-25x^{2}+2}{x + 5}=3x^{4}+5x^{3}-5x^{2}+\frac{2}{x + 5}\)

Answer:

\(3x^{4}+5x^{3}-5x^{2}+\frac{2}{x + 5}\)