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

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Answer 1 · 2018-08-02T11:27:36

$-3x^3-x^2+3x-18=(2x+2)(-3/2x^2+x+1/2)-19$

Let ,
$F(x)=(-3x^3-x^2+3x-18)/(2x+2)=(-3x^3-x^2+3x-18)/(2(x+1))$

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

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

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

So,

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

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

$:.F(x)=(-3x^2+2x+1)/2-19/(2(x+1))$
Hence ,
$"quotient polynomial : "q(x)=-3/2x^2+x+1/2$

And remainder =-19

OR
$-3x^3-x^2+3x-18=(2x+2)(-3/2x^2+x+1/2)-19$

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