# Ask Uncle Colin... About Partial Fractions

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

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