# How do you divide (5x ^ { 3} + 9x ^ { 2} - 26x - 27) \div ( x + 3)?

Jul 27, 2017

7

#### Explanation:

Use the Factor Theorem, $f \left(a\right) = 0$

$x + 3 = 0$
$x = - 3$

$f \left(- 3\right) = 5 {x}^{3} + 9 {x}^{2} - 26 x - 27$
$= 5 {\left(- 3\right)}^{3} + 9 {\left(- 3\right)}^{2} - 26 \left(- 3\right) - 27$
$= 7$

Jul 27, 2017

$5 {x}^{2} - 6 x - 8 - \frac{3}{x + 3}$

#### Explanation:

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

$\text{consider the numerator}$

$\textcolor{red}{5 {x}^{2}} \left(x + 3\right) \textcolor{m a \ge n t a}{- 15 {x}^{2}} + 9 {x}^{2} - 26 x - 27$

$= \textcolor{red}{5 {x}^{2}} \left(x + 3\right) \textcolor{red}{- 6 x} \left(x + 3\right) \textcolor{m a \ge n t a}{+ 18 x} - 26 x - 27$

$= \textcolor{red}{5 {x}^{2}} \left(x + 3\right) \textcolor{red}{- 6 x} \left(x + 3\right) \textcolor{red}{- 8} \left(x + 3\right) \textcolor{m a \ge n t a}{+ 24} - 27$

$= \textcolor{red}{5 {x}^{2}} \left(x + 3\right) \textcolor{red}{- 6 x} \left(x + 3\right) \textcolor{red}{- 8} \left(x + 3\right) - 3$

$\text{quotient "=color(red)(5x^2-6x-8)," remainder } = - 3$

$\Rightarrow \frac{5 {x}^{3} + 9 {x}^{2} - 26 x - 27}{x + 3}$

$= 5 {x}^{2} - 6 x - 8 - \frac{3}{x + 3}$