How do you divide $\frac{- 7 x^{3} + 5 x^{2} + 2 x + 5}{x - 4}$ $(-7x^3+5x^2+2x+5)/(x-4)$ ?

Question

How do you divide $\frac{- 7 x^{3} + 5 x^{2} + 2 x + 5}{x - 4}$ $(-7x^3+5x^2+2x+5)/(x-4)$ ?

1 Answer

Impact of this question

1398 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-19T22:00:13

Make the denominator appear in the numerator.

Explanation:

$- 7 x^{3} + 5 x^{2} - 2 x + 5 = (x - 4) (- 7 x^{2} - 33 x - 130) - 520 + 5$ $-7x^3 + 5 x^2 - 2x + 5 = (x-4)(-7x^2 - 33x - 130) - 520 + 5$
So the division by $x - 4$ $x-4$ yields
$- 7 x^{2} - 33 x - 130 - \frac{515}{x - 4}$ $-7x^2 - 33x - 130 - 515/(x-4)$

Science

Math

Humanities

... and beyond

How do you divide $\frac{- 7 x^{3} + 5 x^{2} + 2 x + 5}{x - 4}$ $(-7x^3+5x^2+2x+5)/(x-4)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you divide (-7x^3+5x^2+2x+5)/(x-4) −7x3+5x2+2x+5x−4(-7x^3+5x^2+2x+5)/(x-4) ?

1 Answer

Explanation:

Related questions

Impact of this question

How do you divide $\frac{- 7 x^{3} + 5 x^{2} + 2 x + 5}{x - 4}$ $(-7x^3+5x^2+2x+5)/(x-4)$ ?