# How do find the quotient of (10b^2 + b - 1)/(2b + 3)?

Aug 6, 2018

$\frac{10 {b}^{2} + b - 1}{2 b + 3} = \frac{10 {b}^{2} + 15 b - 14 b - 21 + 20}{2 b + 3}$
$= \frac{5 b \left(2 b + 3\right) - 7 \left(2 b + 3\right) + 20}{2 b + 3} = \left(5 b - 7\right) + \frac{20}{2 b + 3}$
$\therefore \text{quotient polynomial} = q \left(b\right) = 5 b - 7$

#### Explanation:

Using synthetic division :

$\diamond \left(10 {b}^{2} + b - 1\right) \div \left(2 b + 3\right)$

We have , $p \left(b\right) = 10 {b}^{2} + b - 1 \mathmr{and} \text{divisor :} b = - \frac{3}{2}$

We take ,coefficients of $p \left(b\right) \to 10 , 1 , - 1$

. $- \frac{3}{2} |$ $10 \textcolor{w h i t e}{\ldots \ldots . .} 1 \textcolor{w h i t e}{\ldots \ldots} - 1$
$\underline{\textcolor{w h i t e}{\ldots \ldots .}} |$ ul(0color(white)( .......)-15color(white)(......)21
color(white)(........)10color(white)(......)-14color(white)(......)color(violet)(ul|20|
Hence ,

$\left(10 {b}^{2} + b - 1\right) = \left(b + \frac{3}{2}\right) \left(10 b - 14\right) + \left(20\right)$

$\left(10 {b}^{2} + b - 1\right) = \frac{1}{2} \left(2 b + 3\right) \left(10 b - 14\right) + \left(20\right)$

$\left(10 {b}^{2} + b - 1\right) = \left(2 b + 3\right) \left(5 b - 7\right) + \left(20\right)$

We can see that , quotient polynomial :

$q \left(b\right) = \frac{1}{2} \left(10 b - 14\right) = 5 b - 7 \mathmr{and} \text{the Remainder} = 20$