# quadratic factors in partial fractions

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

• 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. – lulu May 31 '19 at 18:19
• Essentially the same as this question. – 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

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

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

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.

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 $$1$$.