# How do you find the remainder for (2x^2+x-6)div(x+2)?

Jul 28, 2018

$\left(2 {x}^{2} + x - 6\right) = \left(x + 2\right) \left(2 x - 3\right) + \left(0\right)$

$\text{Remainder} = 0$

#### Explanation:

Using synthetic division :

$\diamond \left(2 {x}^{2} + x - 6\right) \div \left(x + 2\right)$

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

We take ,coefficients of $p \left(x\right) \to 2 , 1 , - 6$

$- 2 |$ $2 \textcolor{w h i t e}{\ldots \ldots .} 1 \textcolor{w h i t e}{. .} - 6$
$\underline{\textcolor{w h i t e}{\ldots .}} |$ ul(0color(white)( ...)-4color(white)(.......)6
color(white)(......)2color(white)(...)-3color(white)(.......)color(violet)(ul|0|
We can see that , quotient polynomial :

$q \left(x\right) = 2 x - 3 \mathmr{and} \text{the Remainder} = 0$

Hence ,

$\left(2 {x}^{2} + x - 6\right) = \left(x + 2\right) \left(2 x - 3\right) + \left(0\right)$
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