# Confusion About Partial Fractions In Pre-Calculus

I'm confused about partial fractions. Let's take the partial fraction $$\frac{8x+4}{x^2-4}.$$ We can break it up into $$\frac{8x+4}{x^2-4} = \frac{a}{x+2} + \frac{b}{x-2}.$$ Once we multiply both sides by $(x+2)$ and $(x-2)$, we get $8x+4=a(x-2)+b(x+2)$.

My confusion comes on the two steps after all this. We first assume that $x=2$, making $a(x-2)=0$. Then, we solve for $b$. After that, we assume $x=-2$, making $b(x-2)=0$ and solve for $a$. My question is, how can we just assume that $x=2$, or -2, or whatever that makes the roots =0? Why would it be logical to make such an assumption here?

EDIT: I think my confusion comes from 2 parts in particular. First, how do we know that $8x+4=a(x-2)+b(x+2)$ applies for all real values of x? What about the way this equation is written ensures that to be true? Second, how do we know that the values of a and b won't fluctuate, as we sub in different x-values? How do we know that a and b will be constant no matter what x-value we sub in? If we can know both, I think I can be convinced that the step above is legitimate.

First off, a bit of technical pedantry: the expression $$\frac{8x+4}{x^2-4}$$ is not a partial fraction. This expression is a rational expression, which we would like to decompose into partial fractions. In this problem, the partial fractions will be the rational expressions which compose the decomposition.

Getting to the problem itself, I think that you might be thinking about it backwards. Here is how I like to think about partial fraction decomposition: we have this nasty rational expression, i.e. $$\frac{8x+4}{x^2-4},$$ and we ask

Is it possible that this rational expression can be decomposed into simpler rational expressions with more tractable denominators?

In the case of the problem given, we note that $$\frac{8x+4}{x^2-4} = \frac{8x+4}{(x-2)(x+2)}.$$ The denominator on the right "looks like" the common denominator of two fractions, one of which has denominator $x-2$, and the other of which has denominator $x+2$. Because it looks like this fraction might be the result of adding two fractions together, we make a leap of faith and assume that there are two fractions $\frac{a(x)}{x-2}$ and $\frac{b(x)}{x+2}$ such that $$\frac{8x+4}{x^2-4} = \frac{a(x)}{x-2} + \frac{b(x)}{x+2}, \tag{1}$$ where $a(x)$ and $b(x)$ are functions which depend on $x$. A priori, this assumption is not entirely justified (nevermind that it could be justified with some more theory; we don't need to justify it right now). However, if we really can do this, then we should be able to work out what $a$ and $b$ have to be.

If we assume that (1) holds, then the next step might be to clear the denominators (i.e. multiply both sides of the equation by $x^2-4$ in order to simplify things a bit). If we do this, we obtain $$(1) \implies 8x + 4 = a(x)(x+2) + b(x)(x-2) = (a(x)+b(x))x + 2(a(x)-b(x)).$$ It could be that $a$ and $b$ are very complicated functions of $x$, but let's make another assumption: let's assume that $a$ and $b$ are fairly simple functions. In some sense, the simplest assumption that we can make is that both $a$ and $b$ are polynomials. Then, since the left-hand side is a polynomial of degree 1, it follows that $a(x) + b(x)$ can be at most degree 0 (because, if not, the first term on the right would have degree greater than 1, which is problematic). This further implies that both $a$ and $b$ are constant functions which do not depend on $x$. Simplifying things a bit, we can write $$8x+4 = (a+b)x + 2(a-b), \tag{2}$$ where both $a$ and $b$ are constants (pardon the abuse of notation—they were functions earlier).

The question now becomes

What are the possible values of $a$ and $b$ that will satisfy this equation?

One possible approach is to notice that two polynomials are equal to each other if and only if their coefficients are equal. In other words, we need $$8 = a+b \qquad\text{and}\qquad 4 = 2(a-b).$$ This is a system that we can solve in a reasonably simple manner: \begin{align*} 8 = a+b &\implies a = 8-b \\ &\implies 4 = 2(a-b) = 2(8-b-b) = 16 - 4b \\ &\implies 4b = 12 \\ &\implies b = 3 \\ &\implies a = 8-b = 8-3 = 5. \end{align*}

Alternatively, if we go back to (1) and leave it in the form $$\underbrace{8x + 4}_{=f(x)} = \underbrace{a(x+2) + b(x-2)}_{=g(x)},$$ we can think of the left- and right-hand sides as functions which must be equal for all values of $x$. But when $x=2$, the term involving $b$ dies, and when $x=-2$, the term involving $a$ dies. That is $$g(2) = a(2+2) + b(2-2) = 4a,$$ which must be equal to $f(2) = 20$. Since $$4a = g(2) = f(2) = 20,$$ we have $a = 5$. Similarly, $$-4b = g(-2) = f(-2) = -12 \implies b = 3.$$ Note that we didn't have to evaluate these two functions at $x=\pm 2$, it just turns out that this choice makes life easier. We could have evaluated at any two values of $x$ and obtained a system of equations which we could have worked with.

In either case we conclude that $a=5$ and $b=3$. Then, carefully examining the chain of implications, we can conclude that if the original expression allows a decomposition of the form given in (1), then we must have $a(x) = 5$ and $b(x) = 3$, i.e. it must be the case that $$\frac{8x+4}{x^2-4} = \frac{5}{x-2} + \frac{3}{x+2}.$$ We can confirm that this works by multiplying out the right-hand side.

Notice that I have told a story here about first making an assumption, following that assumption to its logical conclusion, then checking that the logical conclusion actually works. This is the way to develop the mathematical idea. In practice, it is possible to prove that these kind of decompositions always work out (with appropriate assumptions imposed on the numerators). So, while I made a big deal out of not having a justification for the first assumption we made, it turns out that things will always work out nicely for us in this setting.

Reason why you can assume such it is because we know

$$8x + 4 = a(x-2) + b(x+2)$$

is true for every $x$ that is a real number. In particular, it must be true for $x=2$ or $x=-2$, just as it is true that

$$8 \pi + 4 = a( \pi -2 ) + b (\pi + 2)$$

• But how do we know that the values of a and b won't fluctuate, as we sub in different x-values? How do we know that a and b will be constant no matter what x-value we sub in? Jun 27, 2018 at 4:36

An alternative way to solve this problem is to expand out the equation, compare coefficients, and get a system of equations to solve. However, this is somewhat tedious. The reason the method you described works is that $a$ and $b$ are constants, so their value is independent of the value if $x$. Thus, as that equation holds for all $x$, you may simply choose the most opportune value.

• The thing is though, how do we know that they are constants such that they are independent of the value x? That's the crux of what I'm asking here. Jun 28, 2018 at 9:09