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 values1

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.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. you're really evaluating a limit rather than an equality, but it all comes out in the wash. []

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5 comments on “Ask Uncle Colin: My partial fractions decompose funny

  • Mark Ritchings

    Hi Colin, I’m sure you noticed the 7,24,25 in the answer.

    In fact

    $ \frac{a+bx}{a^2x^2-b^2} \equiv \frac{b^2 – a^2}{2 a b (a x + b)} + \frac{a^2 + b^2}{2 a b (a x – b)} $

    and of course

    $ (b^2-a^2)^2 + (2ab)^2 \equiv (a^2+b^2)^2 $

    so you always get a Pythagorean triple.

  • Mark Ritchings

    By the way, your \pm 3/4 should be \pm 4/3.

    • Colin

      Thanks, Mark, corrected now!

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