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

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

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Answer 1 · 2017-05-20T15:24:06

The quotient is $= (x + 2)$ $=(x+2)$ and the remainder is $= (- 3 x - 6)$ $=(-3x-6)$

Explanation:

Let's perform the long division

$a a a a$ $color(white)(aaaa)$ $x^{2} + 1$ $x^2+1$ $a a a a$ $color(white)(aaaa)$ $∣$ $|$ $x^{3} + 2 x^{2} - 2 x - 4$ $x^3+2x^2-2x-4$ $a a a a$ $color(white)(aaaa)$ $∣$ $|$ $x + 2$ $x+2$

$a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaa)$ $x^{3} + 00 + x$ $x^3+00+x$

$a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaa)$ $0 + 2 x^{2} - 3 x - 4$ $0+2x^2-3x-4$

$a a a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaaaa)$ $+ 2 x^{2} - 00 + 2$ $+2x^2-00+2$

$a a a a a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaaaaaa)$ $+ 0 - 3 x - 6$ $+0-3x-6$

Therefore,

$\frac{x^{3} + 2 x^{2} - 2 x - 4}{x^{2} + 1} = x + 2 + \frac{- 3 x - 6}{x^{2} + 1}$ $(x^3+2x^2-2x-4)/(x^2+1)=x+2+(-3x-6)/(x^2+1)$

The quotient is $= (x + 2)$ $=(x+2)$ and the remainder is $= (- 3 x - 6)$ $=(-3x-6)$

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

1 Answer

Explanation:

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