How to know if partial fractions have been done incorrectly?

Question

Say you start with a set of fractions already broken up:

$$ 2 + \frac{3}{x-1} + \frac{1}{x-3} $$

These can be combined into a single fraction by cross multiplying them:

$$ \frac{2(x-1)(x-3) + 3(x-3) + (x-1)}{(x-1)(x-3)} = \frac{2x^2 - 2x - 6x + 6 + 3x - 9 + x - 1}{(x-1)(x-3)} = \frac{2x^2 - 4x - 4}{(x-1)(x-3)} $$

Now I will incorrectly apply the method of partial fraction expansion to split this back into its partial fractions:

$$ \frac{2x^2 - 4x - 4}{(x-1)(x-3)} = \frac{A}{x-1} + \frac{B}{x-3} $$ $$ 2x^2 - 4x - 4 = A(x-3) + B(x-1) $$ $$ \text{let } x = 1 \qquad 2(1)^2 - 4(1) - 4 = A(1-3) + B(1-1) \qquad -6 = -2A \qquad A = 3 $$ $$ \text{let } x = 3 \qquad 2(3)^2 - 4(3) - 4 = A(3-3) + B(3-1) \qquad 2 = 2B \qquad B = 1 $$ $$ \frac{2x^2 - 4x - 4}{(x-1)(x-3)} = \frac{3}{x-1} + \frac{1}{x-3} $$

And of course when you compare this to the initial set of fractions you get garbage:

$$ \frac{2x^2 - 4x - 4}{(x-1)(x-3)} = 2 + \frac{3}{x-1} + \frac{1}{x-3} = \frac{3}{x-1} + \frac{1}{x-3} \quad \therefore \quad 2 = 0$$

I understand that in order to correctly apply the method of partial fractions I would need to have another constant on its own, since without it when multiplying out the denominators of the fractions there would be no way to obtain an $x^2$ term on the RHS, as is shown below with the constant term being required for getting the $x^2$ term:

$$ \frac{2x^2 - 4x - 4}{(x-1)(x-3)} = \frac{A}{x-1} + \frac{B}{x-3} + C $$ $$ 2x^2 - 4x - 4 = A(x-3) + B(x-1) + C(x-3)(x-1) $$

But my question is why does the initial method not work, all the steps look reasonable to me and I can't seem to see why that method fails. I have a feeling that I am inadvertently dividing by $0$ somewhere since that is normally how "proofs" like this arise, but I cannot seem to find where this could be.

Answer 1

The method you used correctly shows that if $$ \frac{2x^2-4x-4}{(x-1)(x-3)} = \frac{A}{x-1} + \frac{B}{x-3} $$ then it must be the case that $A = 3$ and $B = 1$. So this shows that those values of $A$ and $B$ are the only ones that have any chance of working; it doesn't show that those values actually work.

Answer 2

It is clear that the equality $2x^2-4x-4=A(x-3)+B(x-1)$ isn't an identity as the coefficients corresponding to the term of the degree $2$ aren't the same (you have 2 on the left-hand side and $0$ on the right hand side).

When you set $x=1$ and $x=3$ and you solve in $A$ and $B$, you find the values of $A$ and $B$ for which the equation $2x^2-4x-4=A(x-3)+B(x-1)$ has $1$ and $3$ as solutions. Geometrically, you find the only line of the family $y=A(x-3)+B(x-1)$ with intersects the parable $y=2x^2-4x-4$ at the points $(1,-6)$ and $(3,2)$. But obviously, the fact that the line and the parable have two points in common is not enough to say that they $\textit{coincide}$.

When you introduce the third constant $C$, you get an equation of the form $2x^2-4x-4=A(x-3)+B(x-1)+C(x-3)(x-1)$. Then you take 3 values of $x$ and you find $A,B,C$ such that the relations hold. In that case, you are forcing two parables to have 3 common points and it is a general fact that two $n$-th degree polynomials with $n+1$ points in common are the same, so your initial expression becomes an identity for the values of $A,B,C$ that you found.