How do you divide ( -x^3 - 6x^2+3x+4 )/(2x^2 - x )?

1 Answer
marfre
Jul 7, 2018

-1/2x - 13/4 + (-x/4 + 4)/(2x^2 - x)

Explanation:

Given: (-x^3 - 6x^2 + 3x += 4)/(2x^2 - x)

Using long division:

2x^2 - x|bar(-x^3 -6x^2+3x+4)

What times 2x^2 = -x^3? " "-1/2x

" "-1/2x
2x^2 - x|bar(-x^3 -6x^2+3x+4)
" "ul (-x^3 + 1/2x^2)
" "-13/2 x^2 + 3x

What times 2x^2 = -13/2x^2? " "-13/4

" "-1/2x -13/4
2x^2 - x|bar(-x^3 -6x^2+3x+4" ")
" "ul (-x^3 + 1/2x^2)
" "-13/2 x^2 + 3x
" "ul(-13/2x^2 + 13/4x
" "-1/4x + 4 This is the remainder

(-x^3 - 6x^2 + 3x += 4)/(2x^2 - x) = -1/2x - 13/4 + (-x/4 + 4)/(2x^2 - x)

The remainder can be simplified if desired.

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