# How do you divide (x^4 - 3x - 5)/(x+2)?

Aug 8, 2016

${x}^{3} - 2 {x}^{2} + 4 x - 11 + \frac{17}{x + 2}$

#### Explanation:

Note that I use place keepers that have no value. This is done to make sure things line up properly. Example$\to 0 {x}^{2}$

Numerator$\text{ "->" } {x}^{4} + 0 {x}^{3} + 0 {x}^{2} - 3 x - 5$
$\textcolor{m a \ge n t a}{{x}^{3}} \left(x + 2\right) \text{ "->" "ul(x^4+2x^3)" "larr" Subtract}$
$\text{ } 0 - 2 {x}^{3} + 0 {x}^{2} - 3 x - 5$
$\textcolor{m a \ge n t a}{- 2 {x}^{2}} \left(x + 2\right) \text{ "->" "ul( -2x^3-4x^2)" "larr" Subtract}$
$\text{ } 0 + 4 {x}^{2} - 3 x - 5$
$\text{ "color(magenta)(4x)(x+2) " "->" "ul(4x^2+8x)" "larr" Subtract}$
$\text{ } 0 - 11 x - 5$
$\textcolor{m a \ge n t a}{- 11} \left(x + 2\right) \text{ "->" "ul(-11x-22 )" "larr" Sub.}$
$\text{ } 0 + 17$

Where the final value of $\textcolor{m a \ge n t a}{\text{17 is the remainder}}$

$\frac{{x}^{4} - 3 x - 5}{x + 2} = \textcolor{m a \ge n t a}{{x}^{3} - 2 {x}^{2} + 4 x - 11 + \frac{17}{x + 2}}$