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In a bid to check my understanding I have this case: Decompose: $ \dfrac{x^4-8}{x^2+2x}$ Here I see it is an improper factions- degree of $x$ is higher on the numerator than on denominator. Using long division to divide, I get $$x^2-2x+4+ \dfrac{(-8x-8)}{x^2+2x}$$

this is also same as $$x^2-2x+4+ \dfrac{(-8x-8)}{x(x+2)}$$ How do I proceed to derived the decomposition template fraction?

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    $\begingroup$ Let the last fraction term be $\dfrac{A}{x}+\dfrac{B}{x+2}$ $\endgroup$ – peterwhy Oct 3 '13 at 7:45
  • $\begingroup$ @petewhy Where does that leave $x^2-2x+4$ ? $\endgroup$ – Sylvester Oct 3 '13 at 7:54
  • $\begingroup$ Leave these 3 terms outside fraction - like how you leave integer parts out of proper fractions. $\endgroup$ – peterwhy Oct 3 '13 at 7:58
  • $\begingroup$ Partial fractions only deals with the "proper" part of a mixed fraction. $\endgroup$ – steven gregory Jul 6 '18 at 15:12
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Let the last fraction term be $\dfrac{A}{x}+\dfrac{B}{x+2}$, then

$$\begin{align} \frac{x^4-8}{x^2+2x} =& x^2 - 2x + 4 + \frac{-8x-8}{x^2+2x}\\ =& x^2 - 2x + 4 + \frac{A}{x}+\frac{B}{x+2}\\ =& x^2 - 2x + 4 + \frac{A(x+2)+Bx}{x(x+2)}\\ \end{align}$$

By comparing coefficients on the numerator of the fraction term, $$A=-4,\ B=-4$$

So $$\frac{x^4-8}{x^2+2x} = x^2 - 2x + 4 - \frac{4}{x} - \frac{4}{x+2}$$

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  • $\begingroup$ I follow you everywhere except when you declared $A=-4, B=-4$. I also don't understand which numerator coefficients you are comparing. $\endgroup$ – Sylvester Oct 3 '13 at 8:08
  • $\begingroup$ Compare the numerators on the first and third lines: $\dfrac{-8x-8}{x^2+2x} = \dfrac{A(x+2)+Bx}{x^2+2x}$. $\endgroup$ – peterwhy Oct 3 '13 at 8:11
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    $\begingroup$ Oh Ok. so, it is because we have common denominator.....$$\therefore -8x-8=(A+B)x+2A $$ so $A+B=-8$ and $2A=-8$ $\endgroup$ – Sylvester Oct 3 '13 at 8:28
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for numerical working there is a useful rule which applies when the denominator polynomial is a product of distinct linear factors. suppose $f(x)=\frac{P(x)}{Q(x)}$ where $deg(Q)=n \gt deg(P)$ and $$Q(x) = \prod_{k=1}^n (x-\alpha_k) $$ where the $\alpha_k$ are all different. then define $Q_k(x) = \frac{Q(x)}{(x-\alpha_k)}$

in these happy circumstances we may write: $$ \frac{P(x)}{Q(x)} = \sum_{k=1}^n \frac{P(\alpha_k)}{Q_k(\alpha_k)} (x-\alpha_k)^{-1} $$

in the example given we have $\alpha_1=0$ and $\alpha_2=-2$ giving $Q_1(x)=x+2$, $Q_1(\alpha_1) = 2$ and $Q_2(x)=x$, $Q_2(\alpha_2) = -2$ with $P(\alpha_1)=P(0)=-8$ and $P(\alpha_2)=P(-2)=8$. you will see that this gives the answer already obtained.

don't be put off by the explication - with a little practice this is a very straightforward procedure where it is applicable. for example this can be written straight down:

$$ \frac{x^2+x+1}{(x-1)(x-2)(x-3)} = \frac{\frac32}{(x-1)} - \frac7{(x-2)} + \frac{\frac{13}2}{(x-3)} $$

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