# How do you divide (3x^3 - 12x^2 - 11x - 20)/(x+5)?

Dec 24, 2017

(3x^3−12x^2−11x−20)/(x+5)=3x^2-27x+124-640/(x+5)

#### Explanation:

(3x^3−12x^2−11x−20)-:(x+5)=3x^2-27x+124
(3x^3+15x^2)/
$\textcolor{w h i t e}{. .} 0 - 27 {x}^{2} - 11 x - 20$
color(white)(....)(-27x^2-135x)/
$\textcolor{w h i t e}{\ldots \ldots \ldots . .} 0 + 124 x - 20$
color(white)(..............)(+124x+620)/
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots .} 0 - 640$

(3x^3−12x^2−11x−20)/(x+5)=3x^2-27x+124+(-640)/(x+5)

Dec 24, 2017

$3 {x}^{2} - 27 x + 124 - \frac{640}{x + 5}$

#### Explanation:

$\text{one way is to use the divisor as a factor in the numerator}$

$\text{consider the numerator}$

$\textcolor{red}{3 {x}^{2}} \left(x + 5\right) \textcolor{m a \ge n t a}{- 15 {x}^{2}} - 12 {x}^{2} - 11 x - 20$

$= \textcolor{red}{3 {x}^{2}} \left(x + 5\right) \textcolor{red}{- 27 x} \left(x + 5\right) \textcolor{m a \ge n t a}{+ 135 x} - 11 x - 20$

$= \textcolor{red}{3 {x}^{2}} \left(x + 5\right) \textcolor{red}{- 27 x} \left(x + 5\right) \textcolor{red}{+ 124} \left(x + 5\right) \textcolor{m a \ge n t a}{- 620} - 20$

$= \textcolor{red}{3 {x}^{2}} \left(x + 5\right) \textcolor{red}{- 27 x} \left(x + 5\right) \textcolor{red}{+ 124} \left(x + 5\right) - 640$

$\text{quotient "=color(red)(3x^2-27x+124)," remainder } = - 640$

$\Rightarrow \frac{3 {x}^{3} - 12 {x}^{2} - 11 x - 20}{x + 5}$

$= 3 {x}^{2} - 27 x + 124 - \frac{640}{x + 5}$