# Thread: Is this correct? (Partial fractions).

1. ## Is this correct? (Partial fractions).

I was asked to resolve $\frac{2x+1}{(2x+6)(x^2+4)}$ into partial fractions. I got: $\frac{2x+1}{(2x+6)(x^2+4)} \equiv \frac{-11}{(13)(2x+6)}+\frac{11x+19}{(26)(x^2+4)}$. Is it right?

2. Actually it's not correct. If you fill in some number for x, you can verify that those equations are not the same.

3. No.
You can check by assigning a value to x; use x = 1 (easier)

4. A correct approach would be:
$\frac{2x+1}{(2x+6)(x^{2}+4)} = \frac{2x+1}{(2x+6)} \cdot \frac{1}{x^{2}+4}$
$\frac{2x+1}{(2x+6)} \cdot \frac{1}{x^{2}+4} = \frac{1}{2} \cdot \frac{2x+1}{x+3} \cdot \frac{1}{x^{2}+4}$

here is a start, I haven't tried to solve it further.

5. Ah, little silly mistake.

6. ## Thought

I believe you're missing a term on the far R.H.S.

7. Originally Posted by wonderboy1953
I believe you're missing a term on the far R.H.S.
I wrote the wrong thing down. It was $\frac{4x+1}{(2x+6)(x^2+4)}$ and I got $\frac{4x+1}{(2x+6)(x^2+4)} \equiv \frac{-11}{13(2x+6)}+\frac{33x+57}{78(x^2+4)}$