# How do you find the quotient of (x^3 + 4x -7) by x-3?

Nov 15, 2017

${x}^{2} + 3 x + 13$

#### Explanation:

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

$\text{consider the numerator}$

$\textcolor{red}{{x}^{2}} \left(x - 3\right) \textcolor{m a \ge n t a}{+ 3 {x}^{2}} + 4 x - 7$

$= \textcolor{red}{{x}^{2}} \left(x - 3\right) \textcolor{red}{+ 3 x} \left(x - 3\right) \textcolor{m a \ge n t a}{+ 9 x} + 4 x - 7$

$= \textcolor{red}{{x}^{2}} \left(x - 3\right) \textcolor{red}{+ 3 x} \left(x - 3\right) \textcolor{red}{+ 13} \left(x - 3\right) \textcolor{m a \ge n t a}{+ 39} - 7$

$= \textcolor{red}{{x}^{2}} \left(x - 3\right) \textcolor{red}{+ 3 x} \left(x - 3\right) \textcolor{red}{+ 13} \left(x - 3\right) + 32$

$\text{quotient "=color(red)(x^2+3x+13)," remainder } = 32$

Nov 15, 2017

Really this is the same as Jim's solution. It just looks different.

${x}^{2} + 3 x + 13 + \frac{32}{x - 3}$

#### Explanation:

Note that I am using a place keepers $0 {x}^{2}$. It has no value.

$\textcolor{w h i t e}{\text{dddddddd.ddd.ddd}} {x}^{3} + 0 {x}^{2} + 4 x - 7$
$\textcolor{m a \ge n t a}{+ {x}^{2}} \textcolor{g r e e n}{\left(x - 3\right)} \to \textcolor{w h i t e}{\text{ddd") ul(x^3-3x^2 larr" Subtract}}$
$\textcolor{w h i t e}{\text{ddddddddddddddd")0color(white)("d}} + 3 {x}^{2} + 4 x - 7$
$\textcolor{m a \ge n t a}{+ 3 x} \textcolor{g r e e n}{\left(x - 3\right)} \to \textcolor{w h i t e}{\text{ddddddd") ul(3x^2-9x larr" Subtract}}$
$\textcolor{w h i t e}{\text{ddddddddddddddddddd") 0color(white)("d}} + 13 x - 7$
$\textcolor{m a \ge n t a}{+ 13} \textcolor{g r e e n}{\left(x - 3\right)} \to \textcolor{w h i t e}{\text{ddddddddddd")ul(13x-39 larr" Subtract}}$
$\textcolor{w h i t e}{\text{dddddddddddddddddddddddd")0color(white)("d")color(magenta)(+32 larr" Remainder}}$

color(magenta)(x^2+3x+13+32/color(green)((x-3))