# How do you divide (2x^4 -x^3 - 9x^2 - x + 5)/ (5x^2 - 2)  using polynomial long division?

Jul 29, 2018

This really is long division but its layout is different to the conventional approach.

$\frac{2}{5} {x}^{2} - \frac{1}{5} x - \frac{41}{25} + \frac{- 35 x + 44}{25 \left(5 {x}^{2} - 2\right)}$

#### Explanation:

Given: $\frac{\textcolor{b r o w n}{2 {x}^{4} - {x}^{3} - 9 {x}^{2} - x + 5}}{\textcolor{g r e e n}{5 {x}^{2} - 2}}$

Using place keepers of no value. Example: $0 {x}^{3}$

color(white)("ddddddddddddddd")color(brown)(2x^4-color(white)("d")x^3color(white)("d")-9x^2-x+5)
$\textcolor{m a \ge n t a}{+ \frac{2}{5} {x}^{2}} \textcolor{g r e e n}{\left(5 {x}^{2} - 2\right)} \to \underline{2 {x}^{4} + 0 {x}^{3} - \frac{4}{5} {x}^{2} \leftarrow \text{ Subtract}}$
$\textcolor{w h i t e}{\text{ddddddddddddddddd}} 0 - {x}^{3} - \frac{41}{5} {x}^{2} - x + 5$
$\textcolor{m a \ge n t a}{- \frac{1}{5}} \textcolor{g r e e n}{\left(5 {x}^{2} - 2\right)} \to \textcolor{w h i t e}{\text{d.dd")ul( -x^3+color(white)("d.")0x^2+2/5xlarr" Subtract}}$
$\textcolor{w h i t e}{\text{ddddddddddddddddddddd}} 0 - \frac{41}{5} {x}^{2} - \frac{7}{5} x + 5$
$\textcolor{m a \ge n t a}{- \frac{41}{25}} \textcolor{g r e e n}{\left(5 {x}^{2} - 2\right)} \to \textcolor{w h i t e}{\text{dddddddd")ul(-41/5x^2+0x+81/25larr" Sub.}}$
color(white)("d")color(magenta)("Remainder "->color(white)("ddddddddddddd")0-7/5x+44/25)

$\textcolor{m a \ge n t a}{\text{Set remainder as: } \frac{- 35 x + 44}{25}}$ giving:

$\textcolor{m a \ge n t a}{\frac{2}{5} {x}^{2} - \frac{1}{5} x - \frac{41}{25} + \left[\frac{- 35 x + 44}{25} \textcolor{g r e e n}{\div \left(5 {x}^{2} - 2\right)}\right]}$

$\textcolor{m a \ge n t a}{\frac{2}{5} {x}^{2} - \frac{1}{5} x - \frac{41}{25} + \left[\frac{- 35 x + 44}{25 \textcolor{g r e e n}{\left(5 {x}^{2} - 2\right)}}\right]}$

Jul 29, 2018

The remainder is $= \left(- \frac{7}{5} x + \frac{43}{25}\right)$ and the quotient is $= \left(\frac{2}{5} {x}^{2} - \frac{x}{5} - \frac{41}{25}\right)$

#### Explanation:

Perform the polynomial long division

$\textcolor{w h i t e}{a a a a}$$2 {x}^{4} - {x}^{3} - 9 {x}^{2} - x + 5$$\textcolor{w h i t e}{a a a a}$$|$$5 {x}^{2} - 2$

$\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}$$\textcolor{w h i t e}{a a a a}$$|$$\frac{2}{5} {x}^{2} - \frac{x}{5} - \frac{41}{25}$

$\textcolor{w h i t e}{a a a a}$$2 {x}^{4} - 0 {x}^{3} - \frac{4}{5} {x}^{2}$

$\textcolor{w h i t e}{a a a a}$$0 {x}^{4} - 1 {x}^{3} - \frac{41}{5} {x}^{2} - x$

$\textcolor{w h i t e}{a a a a a a a a}$$- 1 {x}^{3} - 00 {x}^{2} + \frac{2}{5} x$

$\textcolor{w h i t e}{a a a a a a a a}$$- 0 {x}^{3} - \frac{41}{5} {x}^{2} - \frac{7}{5} x + 5$

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$- \frac{41}{5} {x}^{2} - \frac{7}{5} x + \frac{82}{25}$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$- 0 {x}^{2} - \frac{7}{5} x + \frac{43}{25}$

Therefore,

$\frac{2 {x}^{4} - {x}^{3} - 9 {x}^{2} - x + 5}{5 {x}^{2} - 2} = \left(\frac{2}{5} {x}^{2} - \frac{x}{5} - \frac{41}{25}\right) + \frac{- \frac{7}{5} x + \frac{43}{25}}{5 {x}^{2} - 2}$

The remainder is $= \left(- \frac{7}{5} x + \frac{43}{25}\right)$ and the quotient is $= \left(\frac{2}{5} {x}^{2} - \frac{x}{5} - \frac{41}{25}\right)$