# How do you divide (3x^3 - 2x^2 - 12x - 2)/(x-7)?

Jun 2, 2017

$3 {x}^{2} + 19 x + 121 + \frac{845}{x - 7}$

#### Explanation:

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

$\text{consider the numerator}$

$\textcolor{red}{3 {x}^{2}} \left(x - 7\right) \textcolor{m a \ge n t a}{+ 21 {x}^{2}} - 2 {x}^{2} - 12 x - 2$

$= \textcolor{red}{3 {x}^{2}} \left(x - 7\right) \textcolor{red}{+ 19 x} \left(x - 7\right) \textcolor{m a \ge n t a}{+ 133 x} - 12 x - 2$

$= \textcolor{red}{3 {x}^{2}} \left(x - 7\right) \textcolor{red}{+ 19 x} \left(x - 7\right) \textcolor{red}{+ 121} \left(x - 7\right) \textcolor{m a \ge n t a}{+ 847} - 2$

$= \textcolor{red}{3 {x}^{2}} \left(x - 7\right) \textcolor{red}{+ 19 x} \left(x - 7\right) \textcolor{red}{+ 121} \left(x - 7\right) + 845$

$\text{quotient "=color(red)(3x^2+19x+121),"remainder } = 845$

$\Rightarrow \frac{3 {x}^{3} - 2 {x}^{2} - 12 x - 2}{x - 7} = 3 {x}^{2} + 19 x + 121 + \frac{845}{x - 7}$