# How to split using partial fractions: $\frac{9x^2+48x+18}{(2x+1)(x^2+8x+3)}$?

I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics", but am stuck on a problem from the "Partial Fractions" chapter.

I've been running around in circles trying to solve it, and just can't seem to get it right (although I understand the general mechanics around the solution process).

The expression that I need to split into partial fractions is:

$$\frac{9x^2+48x+18}{(2x+1)(x^2+8x+3)}$$

• This is not an equation. – Dietrich Burde May 25 at 16:02
• An equation "equates" a (left hand side) LHS to a RHS with an equal sign. – Jean Marie May 25 at 16:20
• I think neuron wants to find the partial fractions as he has stated in the body of the question, probably misunderstood about equation – Ak19 May 25 at 16:23

(Assuming you want to split it into partial fractions)

Let $$\frac{9x^2+48x+18}{(2x+1)(x^2+8x+3)} = \frac{a}{2x+1} + \frac{bx+c}{x^2+8x+3}$$

By inspection,

At $$x=-\frac{1}{2}$$,

$$\frac{9/4-48/2+18}{1/4-8/2+3} = a\implies a = 5$$

At $$x=0$$

$$\frac{0+0+18}{(0+1)(0+0+3)} = \frac{5}{0+1}+\frac{c}{0+0+3}\implies c =3$$

At $$x=1$$

$$\frac{9+48+18}{3(1+8+3)} = \frac{5}{3}+\frac{b+3}{1+8+3}\implies b=2$$

So, $$\frac{9x^2+48x+18}{(2x+1)(x^2+8x+3)} = \frac{5}{2x+1} + \frac{2x+3}{x^2+8x+3}$$

• You should explain how you know the thing has a partial fractions expansion of that form! – David C. Ullrich May 25 at 18:00
• For example, $\frac{x}{(x-1)(x^2-1)}=\frac a{x-1}+\frac{bx+c}{x^2-1}$ will not work... – David C. Ullrich May 25 at 18:03