# How do you divide (-3x^3 - x^2 + 3x -18)/(2x+2)?

Aug 2, 2018

$- 3 {x}^{3} - {x}^{2} + 3 x - 18 = \left(2 x + 2\right) \left(- \frac{3}{2} {x}^{2} + x + \frac{1}{2}\right) - 19$

#### Explanation:

Let ,
$F \left(x\right) = \frac{- 3 {x}^{3} - {x}^{2} + 3 x - 18}{2 x + 2} = \frac{- 3 {x}^{3} - {x}^{2} + 3 x - 18}{2 \left(x + 1\right)}$

We try to obtain factor $\left(x + 1\right)$ from the numerator-where it is

possible and left the free term , adjusting with $\left(- 18\right) .$
"Numerator"=-3x^3color(red)(-x^2)+color(blue)(3x)color(brown)(-18

=-3x^3color(red)(-3x^2+2x^2)color(blue)(+2x+x)+color(brown)(1-19
=-3x^2(x+1)+2x(x+1)+1(x+1)color(brown)(-19
=(x+1)[-3x^2+2x+1]color(brown)(-19

So,

F(x)=((x+1)[-3x^2+2x+1]color(brown)(-19))/(2(x+1)

$\therefore F \left(x\right) = \frac{\cancel{\left(x + 1\right)} \left[- 3 {x}^{2} + 2 x + 1\right]}{2 \cancel{\left(x + 1\right)}} - \frac{\textcolor{b r o w n}{19}}{2 \left(x + 1\right)}$

$\therefore F \left(x\right) = \frac{- 3 {x}^{2} + 2 x + 1}{2} - \frac{19}{2 \left(x + 1\right)}$
Hence ,
$\text{quotient polynomial : } q \left(x\right) = - \frac{3}{2} {x}^{2} + x + \frac{1}{2}$

And remainder =-19

OR
$- 3 {x}^{3} - {x}^{2} + 3 x - 18 = \left(2 x + 2\right) \left(- \frac{3}{2} {x}^{2} + x + \frac{1}{2}\right) - 19$