# How do you long divide  (x^3-13x -12)/ (x-4)?

May 31, 2017

The quotient is $= {x}^{2} + 4 x + 3$ and the remainder is $= 0$

#### Explanation:

Let's perform the long division

$\textcolor{w h i t e}{a a a a}$$x - 4$$|$$\textcolor{w h i t e}{a a a a}$${x}^{3} + 0 {x}^{2} - 13 x - 12$$\textcolor{w h i t e}{a a a a}$$|$${x}^{2} + 4 x + 3$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$${x}^{3} - 4 {x}^{2}$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$0 + 4 {x}^{2} - 13 x$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a}$$+ 4 {x}^{2} - 16 x$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}$$+ 0 + 3 x - 12$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a}$$+ 3 x - 12$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a}$$+ 0 - 0$

Therefore,

$\frac{{x}^{3} - 13 x - 12}{x - 4} = {x}^{2} + 4 x + 3$

May 31, 2017

${x}^{2} + 4 x + 3$

#### Explanation:

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

$\text{consider the numerator}$

$\textcolor{red}{{x}^{2}} \left(x - 4\right) \textcolor{m a \ge n t a}{+ 4 {x}^{2}} - 13 x - 12$

$= \textcolor{red}{{x}^{2}} \left(x - 4\right) \textcolor{red}{+ 4 x} \left(x - 4\right) \textcolor{m a \ge n t a}{+ 16 x} - 13 x - 12$

$= \textcolor{red}{{x}^{2}} \left(x - 4\right) \textcolor{red}{+ 4 x} \left(x - 4\right) \textcolor{red}{+ 3} \left(x - 4\right) \textcolor{m a \ge n t a}{+ 12} - 12$

$= \textcolor{red}{{x}^{2}} \left(x - 4\right) \textcolor{red}{+ 4 x} \left(x - 4\right) \textcolor{red}{+ 3} \left(x - 4\right)$

$\text{quotient "=color(red)(x^2+4x+3)," remainder } = 0$

$\Rightarrow \frac{{x}^{3} - 13 x - 12}{x - 4} = {x}^{2} + 4 x + 3$