# Partial fraction decomposition(small question)

1. Aug 7, 2014

### kevinshen18

I always see people when doing partial fraction decomposition just plug in arbitrary values of x to cancel out some constant terms like A,B,C in order to solve it. I just want to know how this works. I've heard that since A,B, and C are constants it has to hold true for any values of x(like a function). How does this work? I get confused when I see people changing the value of x. I though x was established and you can't change it. Is this true?

2. Aug 7, 2014

### gopher_p

Just to make it easier to talk about, let's focus on the specific case of finding the PFD of $\frac{1}{x(x-1)}$. The first step of the problem would dictate assuming the existence of constants $A$ and $B$ such that $$\frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$$ for all $x$ in the domain of the original expression (i.e. $x\neq 0,1$). As a typical student of calculus, you're kinda meant to take that part as a fact whose explanation is beyond the scope of the course. The only good explanation that I'm aware of is based on concepts from a branch of math called Abstract Algebra, a topic studied by a very small percentage of the students who learn PFD. The rest of the explanation, however, is very accessible to a calculus student learning PFD as a strategy for finding antiderivatives of rational functions.

For $x\neq 0,1$, the previous equation is equivalent to $$1=A(x-1)+Bx$$ Since this equation is true for all $x\neq 0,1$, $$1=\lim_{x\rightarrow 0}\big(A(x-1)+Bx\big)$$ and $$1=\lim_{x\rightarrow 1}\big(A(x-1)+Bx\big)$$

I'll leave it to you to fill in the rest and convince yourself why the "root-substitution method" for finding the coefficients in a PFD problem is valid in the general case.

3. Aug 7, 2014

### kevinshen18

Thanks for the response. But does plugging in x values change the equation in any ways. People do that because they want certain constants like A,B or C to disappear to make solving easier.

PS: I'm not actually studying calculus. This is part of Algebra 2. I was browsing through some videos on khan academy when this occurred to me).

4. Aug 7, 2014

### gopher_p

You're welcome.

I'm not sure what you mean. An equation is a kind of mathematical statement. Many times in math, we're concerned with whether or not (or when) a statement is true. In this case, we're saying that no matter what $x$ is the statement $1=A(x-1)+Bx$ is a true statement. In this case, plugging in a value for $x$ doesn't change the truth of the equation. So in a sense, the equation doesn't change. In more obvious/elementary ways - e.g. there are different symbols with different meanings - the equation does change when you plug in specific values for $x$.

Yes. It's the easiest way, in practice, to find the coefficients in many (though not all) cases. There are other ways to get the job done though.

Ah, ok. Sorry for any confusion with the $\lim$ stuff. The hand-wavy version of that part of the explanation is that since the equation is true everywhere except possibly $0$ and $1$ and everything in the equation is "nice" near $0$ and $1$, the equation is still true at $0$ and $1$. You'll learn more about what "nice" means in the first week or so of calculus when you get there.

From my experience, Khan academy is good for learning problem-solving algorithms to get you through a typical homework set or exam, but not so great on the explanations or "theory" behind those algorithms. In my opinion it shouldn't be used as a primary source for learning. It's mostly OK as a supplement, though.