# How do you divide (2x^3 - 6x^2 + 8) / (x^2 - 4)?

Feb 2, 2018

#### Explanation:

$\frac{2 {x}^{3} - 6 {x}^{2} + 8}{{x}^{2} - 4}$

$\textcolor{w h i t e}{\ldots \ldots . .} \textcolor{w h i t e}{.} 2 x - 6$
${x}^{2} - 4 | \overline{2 {x}^{3} - 6 {x}^{2} \textcolor{w h i t e}{\ldots \ldots} + 8}$
color(white)(............)ul(2x^3 color(white)(.........)-8x
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots .} - 6 {x}^{2} + 8 x + 8$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots} \underline{- 6 {x}^{2} \textcolor{w h i t e}{\ldots \ldots .} + 24}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 8 x - 16$

$\frac{2 {x}^{3} - 6 {x}^{2} + 8}{{x}^{2} - 4} = 2 x - 6$ and remainder of $\frac{8 x - 16}{{x}^{2} - 4}$