How do you divide $\frac{x^{4} + 2 x^{3} + 3 x - 1}{x^{2} + 2}$ $(x^4 + 2x^3 +3x -1) /(x^2+2)$ ?

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How do you divide $\frac{x^{4} + 2 x^{3} + 3 x - 1}{x^{2} + 2}$ $(x^4 + 2x^3 +3x -1) /(x^2+2)$ ?

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Answer 1 · 2018-04-29T04:07:32

Using the distribution method, you can expand the expression.

Explanation:

$\frac{x^{4} + 2 x^{3} + 3 x - 1}{x^{2} + 2}$ $(x^4+2x^3+3x-1)/(x^2+2)$

distribute the $(x^{4} + 2 x^{3} + 3 x - 1)$ $(x^4+2x^3+3x-1)$ to $x^{2}$ $x^2$ and $2$ $2$ and now we get:

$\frac{x^{4} + 2 x^{3} + 3 x - 1}{x^{2}} + \frac{x^{4} + 2 x^{3} + 3 x - 1}{2}$ $(x^4+2x^3+3x-1)/x^2 + (x^4+2x^3+3x-1)/2$

we can further distribute $x^{2}$ $x^2$ and $2$ $2$ to $(x^{4} + 2 x^{3} + 3 x - 1)$ $(x^4+2x^3+3x-1)$ :

$(\frac{x^{4}}{x^{2}}$ $(x^4/x^2$ $+$ $+$ $\frac{2 x^{3}}{x^{2}}$ $(2x^3)/x^2$ $+$ $+$ $\frac{3 x}{x^{2}}$ $(3x)/x^2$ $-$ $-$ $\frac{1}{x^{2}})$ $1/x^2)$ $+$ $+$ $(\frac{x^{4}}{2}$ $((x^4)/2$ $+$ $+$ $\frac{2 x^{3}}{2}$ $(2x^3)/2$ $+$ $+$ $\frac{3 x}{2}$ $(3x)/2$ $-$ $-$ $\frac{1}{2})$ $1/2)$

This is the simplified version:
$x^{2} + 2 x + \frac{3}{x} - \frac{1}{x^{2}} + \frac{x^{4}}{2} + x^{3} + 1.5 x - \frac{1}{2}$ $x^2+2x+3/x-1/x^2+x^4/2+x^3+1.5x-1/2$

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How do you divide $\frac{x^{4} + 2 x^{3} + 3 x - 1}{x^{2} + 2}$ $(x^4 + 2x^3 +3x -1) /(x^2+2)$ ?

1 Answer

Explanation:

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How do you divide x4+2x3+3x−1x2+2(x^4 + 2x^3 +3x -1) /(x^2+2)?

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How do you divide $\frac{x^{4} + 2 x^{3} + 3 x - 1}{x^{2} + 2}$ $(x^4 + 2x^3 +3x -1) /(x^2+2)$ ?