# How do you divide ( -3x^3 - 2x^2+13x+4 )/(3-x )?

May 3, 2017

Using the Long Division method ->

#### Explanation:

First step I would do is remove the negative sign from the fraction.
$- \frac{3 {x}^{3} + 2 {x}^{2} - 13 x - 4}{-} \left(x - 3\right)$ $= \frac{3 {x}^{3} + 2 {x}^{2} - 13 x - 4}{x - 3}$
Proceed with Long Division.

..........................$\left(+ 3 {x}^{2} + 11 x + 20\right)$
$\textcolor{w h i t e}{\frac{x - 3}{\textcolor{b l a c k}{x - 3}} \frac{3 {x}^{3} + 2 {x}^{2} - {x}^{2} - 7 x - 7}{\textcolor{b l a c k}{\text{)} \overline{3 {x}^{3} + 2 {x}^{2} - 13 x - 4}}}}$
.............$- \left(3 {x}^{3} - 9 {x}^{2}\right)$
..................------------------------------------
.................$\left(0 {x}^{3} + 11 {x}^{2} - 13 x\right)$
..........................$- \left(11 {x}^{2} - 33 x\right)$
..................------------------------------------
..........................$\left(- 0 {X}^{2} + 20 x - 4\right)$
...........................................$- \left(20 x - 60\right)$
.................--------------------------------------
................................................$\left(0 x + 56\right)$

Note: Ignore the full stops, I'm new at this :/

You may want to copy it on paper to make it clearer.

So, to simplify the fraction from the above:

$- \frac{3 {x}^{3} + 2 {x}^{2} - 13 x - 4}{-} \left(x - 3\right)$ = $3 {x}^{2} + 11 x + 20 + \left(\frac{56}{x - 3}\right)$

May 3, 2017

This really is long division. It just looks different!

$3 {x}^{2} + 11 x + 20 + \frac{56}{x - 3}$

#### Explanation:

For convenience I have written $3 - x$ as $- x + 3$

$\text{ } - 3 {x}^{3} - 2 {x}^{2} + 13 x + 4$
$\textcolor{m a \ge n t a}{3 {x}^{2}} \left(- x + 3\right) \to \underline{- 3 {x}^{3} + 9 {x}^{2}} \leftarrow \text{ Subtract}$
$\text{ } 0 - 11 {x}^{2} + 13 x + 4$
$\textcolor{m a \ge n t a}{11 x} \left(- x + 3\right) \to \text{ "ul(-11x^2+33x )larr" Subtract}$
$\text{ "0" } - 20 x + 4$
$\textcolor{m a \ge n t a}{20} \left(- x + 3\right) \to \text{ "ul(-20x+60) larr" Subtract}$
" "color(magenta)(0-56 larr "Remainder")

$\frac{- 3 {x}^{3} - 2 {x}^{2} + 13 x + 4}{3 - x} = \textcolor{m a \ge n t a}{3 {x}^{2} + 11 x + 20 - \frac{56}{3 - x}}$

To match Ching's answer consider $- \frac{56}{3 - x}$

Not that $3 - x$ is the same as $- \left(+ x - 3\right)$

So we have $\text{ } - \left(\frac{56}{- \left(x - 3\right)}\right) \to + \frac{56}{x - 3}$

Giving: $\text{ } 3 {x}^{2} + 11 x + 20 + \frac{56}{x - 3}$