# How do you divide 3x^3 - 3x^2 - 4x + 3 by x + 3?

May 17, 2017

$3 {x}^{2} - 12 x + 32 - \frac{93}{x + 3}$

#### Explanation:

$\text{one way is to use the divisor as a factor in the numerator}$

$\text{consider the numerator}$

$\textcolor{red}{3 {x}^{2}} \left(x + 3\right) \textcolor{m a \ge n t a}{- 9 {x}^{2}} - 3 {x}^{2} - 4 x + 3$

$= \textcolor{red}{3 {x}^{2}} \left(x + 3\right) \textcolor{red}{- 12 x} \left(x + 3\right) \textcolor{m a \ge n t a}{+ 36 x} - 4 x + 3$

$= \textcolor{red}{3 {x}^{2}} \left(x + 3\right) \textcolor{red}{- 12 x} \left(x + 3\right) \textcolor{red}{+ 32} \left(x + 3\right) \textcolor{m a \ge n t a}{- 96} + 3$

$= \textcolor{red}{3 {x}^{2}} \left(x + 3\right) \textcolor{red}{- 12 x} \left(x + 3\right) \textcolor{red}{+ 32} \left(x + 3\right) - 93$

$\text{quotient "=color(red)(3x^2-12x+32)," remainder } = - 93$

$\Rightarrow \frac{3 {x}^{3} - 3 {x}^{2} - 4 x + 3}{x + 3} = 3 {x}^{2} - 12 x + 32 - \frac{93}{x + 3}$

May 17, 2017

The quotient is $= 3 {x}^{2} - 12 x + 32$ and the remainder is $= - 93$

#### Explanation:

We perform a long division

$\textcolor{w h i t e}{a a a a}$$x + 3$$|$$\textcolor{w h i t e}{a a a a}$$3 {x}^{3} - 3 {x}^{2} - 4 x + 3$$\textcolor{w h i t e}{a a a a}$$|$$3 {x}^{2} - 12 x + 32$

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

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a}$$0 - 12 {x}^{2} - 4 x$

$\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}$$- 12 {x}^{2} - 36 x$

$\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}$$+ 0 + 32 x + 3$

$\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 a a}$$+ 32 x + 96$

$\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 a a a a}$$- 0 - 93$

Therefore,

$\frac{3 {x}^{3} - 3 {x}^{2} - 4 x + 3}{x + 3} = 3 {x}^{2} - 12 x + 32 - \frac{93}{x + 3}$