How do you divide $\frac{2 x^{4} + 2 x^{3} - 10 x - 9}{x^{3} + x^{2} - 5}$ $(2x^4+2x^3-10x-9)/(x^3+x^2-5 )$ ?

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How do you divide $\frac{2 x^{4} + 2 x^{3} - 10 x - 9}{x^{3} + x^{2} - 5}$ $(2x^4+2x^3-10x-9)/(x^3+x^2-5 )$ ?

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Answer 1 · 2017-05-17T09:19:47

A = $\frac{- 6 x - 9}{5}$ $(-6x -9)/5$

Explanation:

All you really need to do is simplify:

= $\frac{2 x^{4} + 2 x^{3} - 10 x - 9}{x^{3} + x^{2} - 5}$ $(2x^4 + 2x^3 - 10x - 9)/(x^3 + x^2 - 5)$
= $\frac{2 x + 2 x - 10 x - 9}{- 5}$ $(2x + 2x - 10x - 9)/-5$
= $\frac{4 x - 10 x - 9}{5}$ $(4x - 10x -9)/5$
= $\frac{- 6 x - 9}{5}$ $(-6x - 9)/5$

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How do you divide $\frac{2 x^{4} + 2 x^{3} - 10 x - 9}{x^{3} + x^{2} - 5}$ $(2x^4+2x^3-10x-9)/(x^3+x^2-5 )$ ?

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How do you divide 2x4+2x3−10x−9x3+x2−5(2x^4+2x^3-10x-9)/(x^3+x^2-5 )?

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How do you divide $\frac{2 x^{4} + 2 x^{3} - 10 x - 9}{x^{3} + x^{2} - 5}$ $(2x^4+2x^3-10x-9)/(x^3+x^2-5 )$ ?