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I can do the process fine by memorizing the steps.

But one step I do not understand is why the fraction of the quadratic factor in the form $x^2+a$ has a numerator $bx +c$ where $a$, $b$, and $c$ are some constants. Why would the numerator not be just a constant?

Thanks

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    $\begingroup$ How would you express $\frac x{x^2+1}$ in the form you'd prefer? It's important that the numerators contain all possible remainders upon division by the denominator. $\endgroup$ – lulu May 31 '19 at 18:19
  • $\begingroup$ Essentially the same as this question. $\endgroup$ – Bill Dubuque May 31 '19 at 18:41
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Take $$\frac{1}{(x+e)(x^2+a)}= \frac{E}{x+e} + \frac{bx+c}{x^2+a}.$$

Now, by giving these a common denominator

$$\frac{E}{x+e} + \frac{bx+c}{x^2+a}=\frac{Ex^2 +Ea + bx^2 +(e+c)x+ec}{(x+e)(x^2+a)}$$

$$=\frac{(E+b)x^2+(e+c)x+ec+Ea}{(x+e)(x^2+a)}.$$

So, what you'll notice here is that if you didn't have the $bx+c$, that is, if it were only a constant, then you wouldn't be able to cancel off the $Ex^2$ term introduced by giving these a common denominator.

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When you "pull" a certain denominator out of a fraction,

$$\frac ND=\frac{N}{M\,(x^2+px+q)}=\frac QM+\frac R{x^2+px+q}$$

you actually perform a long division, solving

$$N=Q\,(x^2+px+q)+R.$$

The remainder $R$ is a polynomial of degree less than the divisor, i.e. in general of degree $1$.

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