# How do you divide (3x ^ { 3} - 35x ^ { 2} + 128x - 140) \div ( x - 5)?

Jul 1, 2017

$3 {x}^{2} - 20 x + 28$

#### Explanation:

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

$\text{consider the numerator}$

$\textcolor{red}{3 {x}^{2}} \left(x - 5\right) \textcolor{m a \ge n t a}{+ 15 {x}^{2}} - 35 {x}^{2} + 128 x - 140$

$= \textcolor{red}{3 {x}^{2}} \left(x - 5\right) \textcolor{red}{- 20 x} \left(x - 5\right) \textcolor{m a \ge n t a}{- 100 x} + 128 x - 140$

$= \textcolor{red}{3 {x}^{2}} \left(x - 5\right) \textcolor{red}{- 20 x} \left(x - 5\right) \textcolor{red}{+ 28} \left(x - 5\right) \textcolor{m a \ge n t a}{+ 140} - 140$

$= \textcolor{red}{3 {x}^{2}} \left(x - 5\right) \textcolor{red}{- 20 x} \left(x - 5\right) \textcolor{red}{+ 28} \left(x - 5\right) + 0$

$\text{quotient "=color(red)(3x^2-20x+28)," remainder } = 0$

$\Rightarrow \frac{3 {x}^{3} - 35 {x}^{2} + 128 x - 140}{x - 5}$

$= \frac{\cancel{\left(x - 5\right)} \left(3 {x}^{2} - 20 x + 28\right)}{\cancel{\left(x - 5\right)}}$

$= 3 {x}^{2} - 20 x + 28$