# How do you divide (b^3+2b^2-15b+49)div(b+6) using synthetic division?

Jul 29, 2018

$\left({b}^{3} + 2 {b}^{2} - 15 b + 49\right) = \left(b + 6\right) \left({b}^{2} - 4 b + 9\right) + \left(- 5\right)$

#### Explanation:

$\left({b}^{3} + 2 {b}^{2} - 15 b + 49\right) \div \left(b + 6\right)$

Using synthetic division :

We have , $p \left(b\right) = {b}^{3} + 2 {b}^{2} - 15 b + 49 \mathmr{and} \text{divisor : } b = - 6$

We take ,coefficients of $p \left(b\right) \to 1 , 2 , - 15 , 49$

$- 6 |$ $1 \textcolor{w h i t e}{\ldots \ldots . .} 2 \textcolor{w h i t e}{\ldots \ldots .} - 15 \textcolor{w h i t e}{\ldots \ldots .} 49$
$\underline{\textcolor{w h i t e}{\ldots .}} |$ ul(0color(white)( .....)-6color(white)(..........)24color(white)(...)-54
color(white)(......)1color(white)(......)-4color(white)(........)color(white)(..)9color(white)(.....)color(violet)(ul|-5|
We can see that , quotient polynomial :

$q \left(b\right) = {b}^{2} - 4 b + 9 \mathmr{and} \text{the Remainder} = - 5$

Hence ,

$\left({b}^{3} + 2 {b}^{2} - 15 b + 49\right) = \left(b + 6\right) \left({b}^{2} - 4 b + 9\right) + \left(- 5\right)$