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?


  • 3
    $\begingroup$ 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

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

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


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.


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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.