Question

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

Answer 1

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

Explanation:

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`