Dear Uncle Colin,

I’m told that $f(x) = \frac{5x-7}{(x-1)(x-2)}, x\ne 1, x\ne 2$, and need to express it in partial fractions. My usual method would be to write it as $\frac{A}{x-1} + \frac{B}{x-2}$, multiply by $(x-1)(x-2)$ and substitute $x=1$ and $x=2$ to find $A$ and $B$ - but the definition says I can’t use $x=1$ or $x=2$! What do I do?

Pick A Random Trial Integer And Laugh?

Hi, PARTIAL, and thanks for your message. That’s a really good question!

My answer has two parts: firstly, a justification that you can still use 1 and 2; and some alternative methods in case you’re not convinced.

It still works and that’s ok

This is going to sound like sophistry. Roll with it.

We’re going to work… sort of backwards. I’m going to start from the contention that $5x-7$ is exactly the same thing as $3(x-1) + 2(x-2)$. You could put any number you like in there - rational, irrational, even complex or quaternion - and the two expressions would have exactly the same values.

A step backward: consider $5x-7 = A(x-1) + B(x-2)$. For any single value of $x$, there are infinitely many choices of $A$ and $B$ that would give the same value on both sides, but only one pair of choices ($A=3$ and $B=2$) that works for two or more. You could find $A$ and $B$ by substituting in any two values for $x$, and solving simultaneously.

These expressions don’t mind that you’re about to divide by stuff.

I’ll say that again: whatever pair of values you substitute for $x$, you get the same answers for $A$ and $B$. You would get answers of 3 and 2 whether you used $x=1$ and $x=2$, or $x=\pi$ and $x=3 + 4i$. (Some are easier to calculate than others).

The only reason the function excludes $x=1$ and $x=2$ is that the function is undefined there. The identity that determines the constants is perfectly well defined, and you can use it without worrying about it.

If it didn’t, you could…

You might not like that, and that’s fine. It’s going to cost you a bit more work, though.

There are two belt-and-braces methods you could use to find $A$ and $B$ without bothering 1 and 2.

  • You could pick two other numbers and solve simultaneously ((I often do this if the function is complicated and the denominators are non-trivial; I have lost enough minus signs doing it the ‘simple’ way that I prefer the other method.))
  • You could compare coefficients in the identity: if $5x - 7 \equiv A(x-1) + B(x-2)$, then $5 = A+B$ because of the $x$s, and $-7=-A-2B$ because of the numbers. You can solve this simultaneously, too.

I hope that helps! (I’m looking forward to Proper Mathematicians giving better explanations of this – I’ll add them as and when they arrive!

- Uncle Colin