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Can someone help explain the steps to solving this problem? I can't seem to figure out how to find the answer when the denominator is unfactorable or has a cubed exponent for the nominator.

1.) $$\frac{x^2+ x}{x^2 + x + 1}$$

2.) $$\frac{2x^3 -x^2 + x +5}{x^2 + 3x + 2}$$

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    $\begingroup$ Both of those are improper fractions; the degree of the numerator is not strictly smaller than that of the denominator. We must use long division to write them as a polynomial plus a proper fraction. $\endgroup$ – Alamos Apr 22 '15 at 1:07
  • $\begingroup$ For instance, in the first question following your instructions, should I get 1+ -2x-1 divided by (x^2 +x +1)? Now with that, how would I write the next equation? Or am I just completely wrong? $\endgroup$ – FoolishNumber Apr 22 '15 at 1:10
  • $\begingroup$ For the first one the quotient is $1$ and the remainder is $-1$. We get that it is equal to $1+\frac{-1}{x^2+x+1}$ and this is its partial fraction decomposition. Make sure you know how the algorithm of long division goes. $\endgroup$ – Alamos Apr 22 '15 at 1:14
  • $\begingroup$ Ah. I see. Thanks! $\endgroup$ – FoolishNumber Apr 22 '15 at 1:20
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We have $$\dfrac{x^2+x}{x^2+x+1} = 1 - \dfrac1{x^2+x+1}$$ and \begin{align} \dfrac{2x^3-x^2+x+5}{x^2+3x+2} & = \dfrac{2x^3 + 6x^2 + 4x - 7x^2 - 3x + 5}{x^2+3x+2} = 2x - \dfrac{7x^2+3x-5}{x^2+3x+2}\\ & = 2x - \dfrac{7x^2+21x+14 - 18x - 19}{x^2+3x+2} = 2x-7 + \dfrac{18x+19}{x^2+3x+2}\\ & = 2x-7 + \dfrac{18x+18}{x^2+3x+2} + \dfrac1{x^2+3x+2} = 2x-7 + \dfrac{18}{x+2} + \dfrac1{(x+1)(x+2)}\\ & = 2x-7 + \dfrac{18}{x+2} + \dfrac1{x+1} - \dfrac1{x+2} = 2x-7 + \dfrac{17}{x+2} + \dfrac1{x+1} \end{align}

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