How do you divide ( -3x^3 - 2x^2+13x+4 )/(3-x )?

2 Answers
May 3, 2017

Using the Long Division method ->

Explanation:

First step I would do is remove the negative sign from the fraction.
-(3x^3+2x^2-13x-4)/-(x-3) =(3x^3+2x^2-13x-4)/(x-3)
Proceed with Long Division.

..........................(+3x^2+11x+20)
color(white)((x-3)/color(black)(x-3)(3x^3+2x^2-x^2-7x-7)/color(black)(")"bar(3x^3+2x^2-13x-4)))
.............-(3x^3-9x^2)
..................------------------------------------
.................(0x^3+11x^2-13x)
..........................-(11x^2-33x)
..................------------------------------------
..........................(-0X^2+20x-4)
...........................................-(20x-60)
.................--------------------------------------
................................................(0x+56)

Note: Ignore the full stops, I'm new at this :/

You may want to copy it on paper to make it clearer.

So, to simplify the fraction from the above:

-(3x^3+2x^2-13x-4)/-(x-3) = 3x^2+11x+20+(56/(x-3))

May 3, 2017

This really is long division. It just looks different!

3x^2+11x+20+56/(x-3)

Explanation:

For convenience I have written 3-x as -x+3

" "-3x^3-2x^2+13x+4
color(magenta)(3x^2)(-x+3)->ul( -3x^3+9x^2) larr" Subtract"
" "0-11x^2+13x+4
color(magenta)(11x)(-x+3)->" "ul(-11x^2+33x )larr" Subtract"
" "0" " -20x+4
color(magenta)(20)(-x+3)->" "ul(-20x+60) larr" Subtract"
" "color(magenta)(0-56 larr "Remainder")

(-3x^3-2x^2+13x+4)/(3-x) = color(magenta)(3x^2+11x+20-56/(3-x))

To match Ching's answer consider -56/(3-x)

Not that 3-x is the same as -(+x-3)

So we have " "-(56/(-(x-3))) ->+56/(x-3)

Giving: " "3x^2+11x+20+56/(x-3)