How do you divide $( -3x^3+ x^2-x-72 )/(x - 2 )$ ?

Question

How do you divide $( -3x^3+ x^2-x-72 )/(x - 2 )$ ?

1 Answer

Impact of this question

1513 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-27T13:08:05

$(-3x^3+x^2-x-72)/(x-2)$
$=(-3x^2(x-2)-6x^2+x^2-x-72)/(x-2)$
$=(-3x^2(x-2)-5x(x-2)-10x-x-72)/(x-2)$
$=(-3x^2(x-2)-5x(x-2)-11(x-2)-22-72)/(x-2)$
$=-3x^2-5x-11-94/(x-2)$

Science

Math

Humanities

... and beyond

How do you divide $( -3x^3+ x^2-x-72 )/(x - 2 )$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you divide ( -3x^3+ x^2-x-72 )/(x - 2 )( -3x^3+ x^2-x-72 )/(x - 2 )?

1 Answer

Related questions

Impact of this question

How do you divide $( -3x^3+ x^2-x-72 )/(x - 2 )$ ?