# Ask Uncle Colin: About Partial Fractions

Dear Uncle Colin,

When I express $\frac{x^2}{(x-1)(x-4)^2}$ in partial fractions, why do I need to use three separate fractions? I accept that that’s how it works, I just want to know why!

Pupils Absolutely Raging: “Two Is Adequate, Like!”

Hi, PARTIAL, and thanks for your message!

I’ve got a handwavy explanation and one that goes into a bit more depth.

### Wave your hands in the air like you just don’t care

As a general principle, you need the top of a partial fraction to be one degree lower than the bottom. The $(x-4)^2$ part is quadratic, so it needs to have a linear expression – something like $ax+b$ – on the top.

However, you can rewrite this as $a(x-4) + (b+4a)$ or $a(x-4) + c$, which gives you the three fractions you were expecting.

### Take your hands out of the air and seek a small black square

The top of any (proper) fraction with $(x-1)(x-4)^2$ as a denominator could be *any* quadratic expression.

An arbitrary quadratic expression has three coefficients, so we will need to solve for three constants.

More specifically, we’re trying to *change the basis* in our quadratic. At the moment, it’s written as $x^2$, but we’d ideally like to write it in a way that’s more convenient for cancelling down fractions.

The space of quadratic expressions is three-dimensional, so we need three *linearly independent* expressions, picked as conveniently as possible – by which I mean, so that they will cancel with things on the bottom to leave relatively nice fractions.

Two obvious candidates are $(x-1)$ and $(x-4)^2$, which leave $(x-4)^2$ and $(x-1)$ respectively – but we need a third to span the space.

I stress that we can pick whatever we like here, so long as it’s not a linear combination of the other two, but the most convenient choice is $(x-1)(x-4)$, which just leaves an $x-4$ on the bottom.

So, we can rewrite *any* quadratic expression as $a(x-1) + b(x-4)^2 + c(x-4)(x-1)$, and anything in that form, when divided through by the original bottom, is relatively simple to integrate, differentiate or apply the binomial expansion to.

Hope that helps!

- Uncle Colin