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:


Can anyone please help me out?

  • 14
    $\begingroup$ This is not an equation. $\endgroup$ – Dietrich Burde May 25 at 16:02
  • 1
    $\begingroup$ An equation "equates" a (left hand side) LHS to a RHS with an equal sign. $\endgroup$ – Jean Marie May 25 at 16:20
  • $\begingroup$ I think neuron wants to find the partial fractions as he has stated in the body of the question, probably misunderstood about equation $\endgroup$ – 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, $$\frac{9x^2+48x+18}{(2x+1)(x^2+8x+3)} = \frac{5}{2x+1} + \frac{2x+3}{x^2+8x+3}$$

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

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