Post #4008964

2026-07-21 08:57 UTC

@SmartmanApps@dotnet.social it makes sense that the textbook should have very dry material and should rely on a teacher to elucidate its contents to their students. Though, many teachers have a pretty uninvolved style of teaching, where they hand out the material without much comment. As you suggested, I had a particular problem in mind when I wrote my post. We are told that when splitting a complex polynomial fraction, we are to split the denominator into its factors, and that for each of these factors we can create a new fraction with a placeholder variable as its numerator and the given factor as its denominator, and we are given to believe that the sum of these new fractions can be equated to the original fraction, so long as we find the right values for the numerators. This is all fairly intuitive, but I have some trouble when considering the scenario where one of the resulting denominators is an irreducible quadratic. We are told that in such a case, the numerator should be a prototype of linear form, with two place holder variables in the form Ax + B. I've managed to get my hands on something of an explanation for why this is necessary, which is that we want to express as many polynomials as we can in the numerator, but any polynomial of order two or higher will make the fraction representable as the sum of a polynomial and a fraction with an order one or fewer numerator. Supposedly this means we don't need to bother with more than order one. What I don't understand is why we can't just use a single placeholder variable? After all, a variable could well come to be defined as being a polynomial in some other variable, so why doesn't this work?

Replies (1)

@FishFace@ioc.exchange 2026-07-22 09:45

@bemmesr@mathstodon.xyz you don't want to pay attention to @SmartmanApps@dotnet.social on this. Check out any of his "#MathsMonday" pinned posts that have replies from others and you'll see what I mean. In this case, you are right that you *could* just use a single variable that is equal to a polynomial expression, but this is missing the point of the technique: when we create the placeholder variables A, B, etc, there is in fact an assumption that these are not dependent on x (or whatever the unknown is), and this assumption makes the technique work, together with the assumption that the equation you set up holds for all values of x. Because they don't vary with x, if you know that, for example, Ax + B = 3x + 5, it must be that A = 3, B = 5. For particular values of x, you could pick other values of A,B, but these are the only ones which work for all values of x. Unless, that is, A and B may themselves depend on x! For example, A=5/x, B=3x would work if they could. In general it is better to make the dependencies of variables on other variables explicit, by using functional notation, so that "A" always stands for an independent variable, with "A(x)" meaning some quantity which depends on x. But this is not how mathematics is taught in practice, so you end up with these situations where you may not realise what all the assumptions are. Sorry you had to read the charlatan's answer before getting a proper explanation.

Open ##4008962