# Ask Uncle Colin: My partial fractions decompose funny

Dear Uncle Colin,

I recently had to decompose $\frac{3+4p}{9p^2 - 16}$ into partial fractions, and ended up with $\frac{\frac{25}{8}}{p-\frac{4}{3}} + \frac{\frac{7}{8}}{p-\frac{4}{3}}$. Apparently, that’s wrong, but I don’t see why!

-- Drat! Everything Came Out Messy. Perhaps Other Solution Essential.

Hi, there, DECOMPOSE, and thanks for your message - and your mess!

Your problem here comes with the denominator - you’ve incorrectly factorised $9p^2 - 16$ as $\left(p-\frac 43\right)\left(p+\frac 43\right)$, leaving out a factor of 9.

It’s a great deal simpler to use the more natural factorisation of $(3p-4)(3p+4)$ - this has the same zeroes as your factorisation, but has the advantage of multiplying out to what you started with.

Now you can split everything up as:

$\frac{3 + 4p}{9p^2 - 16} \equiv \frac{A}{3p-4} + \frac{B}{3p+4}$.

The right-hand side is also equivalent to $\frac{A(3p+4) + B(3p-4)}{9p^2 - 16}$. Strictly speaking, it’s not defined for $p=\pm \frac43$, but it happens to work if you do use those values ((you’re really evaluating a limit rather than an equality, but it all comes out in the wash.))

Rather than incur the wrath of @realityminus3, let’s match coefficients instead: looking at the units, $3 = 4A - 4B$; looking at the $p$ terms, $4 = 3A + 3B$.

Solving simultaneously: $12A = 9 + 12B = 16 - 12B$, so $24B = 25$ and $B = \frac{7}{24}$, while $A = \frac{25}{24}$. Your answers are out by the factor of 9 you lost in the factorisation.

Hope that helps!

-- Uncle Colin

* Edited 2017-05-17 to correct a fraction - Thanks to @BuryMathsTutor for spotting it.