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

May 17, 2017

$\frac{x}{2} - 3 + \frac{15 x - 14}{2 {x}^{2} + 4 x - 2}$

Explanation:

$\text{ } {x}^{3} - 4 {x}^{2} + 2 x - 8$
$\textcolor{m a \ge n t a}{+ \frac{x}{2}} \left(2 {x}^{2} + 4 x - 2\right) \to \underline{{x}^{3} + 2 {x}^{2} - x} \leftarrow \text{ Subtract}$
$\text{ } 0 - 6 {x}^{2} + 3 x - 8$
$\textcolor{m a \ge n t a}{- 3} \left(2 {x}^{2} + 4 x - 2\right) \to \text{ " ul(-6x^2-12x+6)larr" Subtract}$
" "0color(magenta)(color(white)(.)+15x-14larr" Remainder")

$\textcolor{m a \ge n t a}{\frac{x}{2} - 3 + \frac{15 x - 14}{2 {x}^{2} + 4 x - 2}}$

May 17, 2017

$\frac{{x}^{3} - 4 {x}^{2} + 2 x - 8}{2 {x}^{2} + 4 x - 2} = \frac{x - 6}{2} + \frac{15 x - 4}{2 \left({x}^{2} + 4 x - 2\right)}$

Explanation:

When working with algebraic fractions, try to factorise first:

(x^3 - 4x^2+2x - 8)/(2x^2+4x-2)" "(larr "try grouping")/(larr"HCF, quadratic trinomial")

$= \frac{{x}^{2} \left(x - 4\right) + 2 \left(x - 4\right)}{2 \left({x}^{2} + 2 x + 1\right)}$

$= \frac{\left(x - 4\right) \left({x}^{2} + 2\right)}{2 \left(x + 1\right) \left(x + 1\right)}$

This cannot be simplified, there are no common factors to cancel

Let's do long division:

(x^3 - 4x^2+2x - 8)/(2x^2+4x-2) = (x^3 - 4x^2+2x - 8)/(2(x^2+2x-1)

Let's divide by $\left({x}^{2} + 2 x - 1\right)$ to start, and divide by $2$ later

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} x - 6$
x^2+2x-1 )bar(x^3 - 4x^2+2x - 8)
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots} \underline{{x}^{3} + 2 {x}^{2} - x} \text{ } \leftarrow$ subtract
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} - 6 {x}^{2} + 3 x - 8$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{- 6 {x}^{2} - 12 x - 6} \text{ } \leftarrow$ subtract
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 15 x - 14 \text{ } \leftarrow$ remainder

We still need to divide by $2$. Therefore we have:

$\frac{{x}^{3} - 4 {x}^{2} + 2 x - 8}{2 {x}^{2} + 4 x - 2} = \frac{x - 6}{2} + \frac{15 x - 4}{2 \left({x}^{2} + 4 x - 2\right)}$