# Is there a simple explanation for calculus classes of why partial fractions work?

I'd be happy even with an explanation in the simplest case: an explanation of why expressions of the form $\frac{ax + b}{(x - c)(x - d)}$ with $c \neq d$ can always be rewritten in the form $\frac{A}{x - c} + \frac{B}{x - d}$. I'm following Stewart's book, and his explanation, in its entirety, is "A theorem in algebra guarantees that it is always possible to do this".

In some sense, that is enough, because the thing to do is to try to rewrite it as $\frac{A}{x - c} + \frac{B}{x - d}$, solve for $A$ and $B$, and the "theorem in algebra" says that you will always be able to. But can one be more convincing?

My students will not have seen linear algebra, so that explicit discussion of linear independence will be unhelpful, even though this is the proper language in which to formulate a proof. They will also not have seen number theory, so that comparisons with modular arithmetic and the Chinese Remainder Theorem are also likely to be unhelpful.

• I think I understood that better myself after my first complex analysis class, but that won't help for a calculus class... – Benoît Kloeckner Sep 5 '14 at 19:20
• A more serious observation: it is a more universally useful ability to know how to deal with $\frac{ax+b}{cx+d}$, which is close to many other expression simplification. It can make for a soft introduction to the case of primary interest. – Benoît Kloeckner Sep 5 '14 at 19:29
• Just looking at this, I see two slightly different ideas. One is that the combined fraction can be written in the separated form. That's not really a `Theorem of Algebra' - the proof would be to solve it in the general case for $A$ and $B$ in terms of $a$, $b$, $c$, $d$. The second idea is that the method typically taught for finding $A$ and $B$ (equating coefficients) will generate the correct answer. That is what uses linear independence. Maybe for that aspect it would satisfy students to use the 'plug in values of $x$' technique. At least doing so would teach them something useful. – Jessica B Oct 24 '15 at 8:49
• Related: McDowell, E.L. (2002). A Numerical Introduction to Partial Fractions. College Mathematics Journal, 400-403. – Benjamin Dickman Nov 7 '15 at 20:46

Here is what I might do, if I were teaching a class which taught some abstract proof skill (but not a true proof class) or if I was happy reaching only the top half of a standard Calculus class here at Michigan.

Suppose we want to expand $$\frac{f(x)}{(x-r_1)(x-r_2)(x-r_3)(x-r_4)}$$ where all the $r_i$ are distinct, and $\deg f \leq 3$. I claim that $$\begin{array}{rcl} \frac{f(x)}{(x-r_1)(x-r_2) \cdots (x-r_4)} &=&\phantom{+} \frac{f(r_1)}{(r_1-r_2)(r_1-r_3)(r_1-r_4)} \frac{1}{x-r_1}\\ & &+ \frac{f(r_2)}{(r_2-r_1)(r_2-r_3)(r_2-r_4)} \frac{1}{x-r_2}\\ & &+ \frac{f(r_3)}{(r_3-r_1)(r_3-r_2)(r_3-r_4)} \frac{1}{x-r_3}\\ & &+ \frac{f(r_4)}{(r_4-r_1)(r_4-r_2)(r_4-r_3)} \frac{1}{x-r_4}\\ \end{array}.$$

Of course, you might wonder how on earth I found that formula, but let's first take it as handed down by magic and show that it works.

Clearing denominators, the claim is that $$\begin{array}{rcl} f(x) &=& \phantom{+}\frac{f(r_1)}{(r_1-r_2)(r_1-r_3)(r_1-r_4)} (x-r_2)(x-r_3)(x-r_4)\\ & & +\frac{f(r_2)}{(r_2-r_1)(r_2-r_3)(r_2-r_4)} (x-r_1)(x-r_3)(x-r_4)\\ & & +\frac{f(r_3)}{(r_3-r_1)(r_3-r_2)(r_3-r_4)} (x-r_1)(x-r_2)(x-r_4)\\ & & +\frac{f(r_4)}{(r_4-r_1)(r_4-r_2)(r_4-r_3)}(x-r_1)(x-r_2)(x-r_3)\\ \end{array}.$$

Now remember: $f$ and the $r$'s are all constants that we know. In the last problem we did, they were $8x-13$, $1$ and $2$ (or whatever). The supposed claim is that this is a true equation for all $x$.

Notice that both sides of this equation are polynomials in $x$. What is the degree of that polynomial? Right, $3$, or possibly less if the coefficients of $x^3$ cancel on the right hand side. A degree $3$ polynomial is determined by $4$ values, just like a linear polynomial is determined by $2$ values. Which $4$ values might we plug in for $x$? Let's try the $r_i$.

Pause for at least $10$ seconds. Do the computation. Pause for at least $10$ seconds again.

Okay, how did I find this formula? This approach is much more worth memorizing than the formula is. Suppose that there were a formula like $$\frac{f(x)}{(x-r_1)(x-r_2)(x-r_3)(x-r_4)} = \frac{c_1}{x-r_1} + \frac{c_2}{x-r_2} + \frac{c_3}{x-r_3} + \frac{c_4}{x-r_4}.$$ So the $c_i$ are the constants we want to find: in the last problem, they turned out to be $3$ and $5$ (or whatever). Here is the trick:

Multiply both sides by $x-r_i$ and take the limit as $x \to r_i$.

Pause for at least $10$ seconds. Do the limit. Pause for at least $10$ seconds again.

As a general note, I've found that sum and product notation, and using variables for discrete quantities like the degree of a polynomial, are huge stumbling blocks. I pretty much always use specific values of $n$, like $4$, whenever working with non-math majors.

• I was taught this in my high school calculus class. We called it the "Heaviside method". Also, if you have a term like $\frac{c_{1,1}}{(x-r_1)^2}$ you can clear denominators, differentiate once, and take a limit. – Steven Gubkin Sep 5 '14 at 15:56
• Yup! And I think there would be no trouble teaching the rule and having everyone understand it, the question is how many people would understand the proof. In particular, the distinction between "if there is a formula, it is this one" and "this formula works" is subtle. – DES-SupportsMonicaAndTransfolk Sep 5 '14 at 16:26
• It seems the $\cdots$ early on just stand for $(x-r_3)$? – Benjamin Dickman Nov 7 '15 at 20:48

It is a matter of taste, but I find the following approach easier to understand for beginners (and thus more convincing) than trying to rewrite the fraction in a good form and solving for coefficients.

It might be good to start with stating that if $ax+b$ can be written in the form $A(x-d)+B(x-c)$, then $\frac{ax + b}{(x - c)(x - d)}$ can be written in the form $\frac{A}{x - c} + \frac{B}{x - d}$; don't expect that all students will think of this unless told. It is quite simple to explain that this is possible if for all $a,b$ there are $A,B$ so that $A+B=a$ and $Ad+Bc=-b$. This last statement can be tricky to prove (if one does not have more advanced tools and there is not much time to spend on it), but little experimentation will certainly make it plausible. Proving the last claim by hand can be given as an extra exercise for those that are interested in order not to disturb the flow of the course.

I tried to organize the above argument so that the problem is reduced to a more tractable one in a couple of steps and the first step is (hopefully) the simplest one to grasp. I assume your goal is not to give a full proof, but to give a justification for why it works so that all students can understand at least a little. If the problem is linked to some other problems (formally equivalent does not mean equivalent in all students' minds), it suffices that the students believe that at least one of them can be solved; students will probably disagree with each other about the simplest formulation.

Trying to give a fuller proof before appropriate machinery is available might — depending on the exact situation — be a waste of time.

• I think this is a good answer to the question "How can I show students how to perform partial fraction decomposition," but I don't think this at all answers the question posed. I guess at the end you seem to answer the question with a "no." – Chris Cunningham Sep 4 '14 at 21:03
• @ChrisCunningham, my calculations indeed show how to perform partial fraction decomposition. But I believe they can be presented in such a way that the emphasis is on why it works. I understood that what is needed is a justification, not necessarily a proof. For an early calculus class, I think it is enough that the partial fraction decomposition turns out to work. If a proof is deemed necessary, I'd save it for later in the OP's scenario. – Joonas Ilmavirta Sep 4 '14 at 21:13
• I believe I have now offered both proof of the method as well as proof of your assertion it is a waste of time to show students. Proof of the later is implicit within the length of my answer and the nature of the algebra required ;) – James S. Cook Sep 5 '14 at 2:21
• @JamesS.Cook, thanks, your answer is a convincing proof of both assertions. I think it turns my claim (that it is a waste of time) into a theorem, no? :-) – Joonas Ilmavirta Sep 5 '14 at 6:03
• Some things which might make the computation more followable: $\frac{ax+b}{(x-r)(x-s)} = \frac{u}{x-r} + \frac{v}{x-s}$. The $7$ variables here are of three different sorts: $a$, $b$, $r$ and $s$ are known constants "in our last example, they were $8$, $-13$, $1$ and $2$"; $u$ and $v$ are what we are solving for "in our last example, they were $3$ and $5$"; and we want the equation to hold for all $x$. So use variables from different parts of the alphabet, and talk explicitly about the different roles. – DES-SupportsMonicaAndTransfolk Sep 5 '14 at 14:47

I have always found it instructive and fun to address this by thinking graphically. Students in Calculus II will know the graphs of the rational functions $f(x)=\frac{ax+b}{(x-c)(x-d)}$ and $g(x)=\frac{A}{x-c}+\frac{B}{x-d}$ have the same vertical asymptotes and same horizontal asymptote. Varying $a$ and $b$ moves the $x$ and $y$ intercepts of $y=f(x)$. It is also clear that varying $A$ and $B$ moves the $x$ and $y$ intercepts of $y=g(x)$.

Fixing $c$ and $d$, we have a two-parameter family of curves $y=f(x)$ and similarly for $y=g(x)$. The key point here is that two families are identical--we can find $A$ and $B$ so that the $g$-curve matches any given $f$-curve. It is straightforward to illustrate this in class with animations generated by software--Geometers Sketchpad or Maple for example. As you vary the two parameters of one family, the curves fill out the curves from the other family.

This is not a proof, but rather a conceptual understanding that calculus students can appreciate. The proof that Stewart alludes to really is straightforward and mundane, and is covered in a precalculus course--it reduces to the existence of a solution to two linear equations in two unknowns.

• very pretty, if I was to teach calculus II sometime soon I think I might try this. – James S. Cook Sep 6 '14 at 13:17

As in @user52817's graphically-oriented answer, I think it is fair to secretly borrow some ideas nowadays labelled "complex analysis" to help explain why this should work: that is, while thinking in terms of the residue at a simple pole $x_1$, one can explain it (perhaps quasi-geometrically...) as giving the "asymptotics" of the rational function as $x\to x_1$. That is, for low-degree numerator, $f(x)/(x-x_1)\ldots(x-x_n)\sim {1\over x-x_1}\cdot {f(x_1)\over (x_1-x_2)\ldots(x_1-x_n)}$ near $x_1$, and so on. Subtracting all these "polar parts" (or whatever one chooses to call them for calculus students) leaves a rational function that has no poles, and does goes to $0$ for large $x$, so has to be $0$ (a tiny prototype for Liouville's theorem, perhaps).

• "a rational function that has no poles, and does goes to $0$ for large $x$". Like $(1+x^2)^{-1}$? It seems that one has to enter into the details of the definition of a pole to be convincing here, which is a serious issue with non-math majors. – Benoît Kloeckner Sep 16 '14 at 20:29
• @BenoîtKloeckner, well, perhaps you're right, but, also, I find this as much an argument in favor of introducing complex numbers much earlier, since otherwise the behavior of so many things is obscure. E.g., why the power series of exactly $1/(1+x^2)$ has radius of convergence just $1$, while it behaves so well on the real line. I really do confess such when I teach calculus, although I don't "test on it". Anyway, one can restrict the discussion to real-rooted things for the plausibility argument, also. – paul garrett Sep 16 '14 at 21:06

Given they have no algebra background, the simplest thing is just to do the algebra. $$\frac{ax + b}{(x - c)(x - d)} = \frac{A}{x-c}+\frac{B}{x-d}$$ Multiplying by $(x-c)(x-d)$ yields (for $x \neq c,d$) $$ax+b= A(x-d)+B(x-c) = (A+B)x-(Ad+Bc)$$ From which we read $a=A+B$ and $b = -Ad-Bc$. Multiplying by $d$ we obtain $ad=Ad+Bd$ and adding equations yields $b+ad=Bd-Bc=B(d-c)$ and as $d \neq c$ we find $B = \frac{b+ad}{d-c}$. It follows that $A=a-B = a-\frac{b+ad}{d-c} = \frac{a(d-c)-b-ad}{d-c} = \frac{-ac-b}{d-c} = -\frac{ac+b}{d-c}$. Now, I will reverse engineer the algebra: \begin{align} \frac{-\frac{ac+b}{d-c}}{x-c}+\frac{\frac{b+ad}{d-c}}{x-d} &= \frac{1}{d-c} \left[\frac{-(ac+b)}{x-c}\cdot\frac{x-d}{x-d}+\frac{b+ad}{x-d}\cdot\frac{x-c}{x-c} \right] \\ &= \frac{1}{d-c} \left[\frac{-(ac+b)(x-d)+(b+ad)(x-c)}{(x-c)(x-d)}\right] \\ &= \frac{1}{d-c} \left[\frac{(d-c)ax-bx+bx-acd+acd+b(d-c)}{(x-c)(x-d)}\right] \\ &= \frac{ax + b}{(x - c)(x - d)}. \end{align} Ok, so, you can start with the last expression, multiply by $1$ and add $0$ until you obtain the partial fractions decomposition $\displaystyle \frac{-\frac{ac+b}{d-c}}{x-c}+\frac{\frac{b+ad}{d-c}}{x-d}$. If the student thirsts for formulas, here they are: $$A = -\frac{ac+b}{d-c} \qquad \& \qquad B=\frac{b+ad}{d-c}.$$ Of course, for specific examples, it is far easier to simply work out the algebra. That said, I think what I have written above can be formalized into a proof. You just have to add zero and multiply by $1$ creatively.

• It might be enough to say that the algebra can be worked out and show the final formulas. But it should not be forgotten that this is only a very special case of partial fractions (albeit a common one), and this method of proof will lead to a very messy proof for more complicated cases (and the general case). Anyway, I think something like your answer is indeed the best approach to the special case. – Joonas Ilmavirta Sep 5 '14 at 6:04
• @JoonasIlmavirta Indeed, moreover, the process of integration is educated guessing, so, while the proof of the existence of a partial fractal decomposition is mathematically interesting, the level of justification seems unwarranted. The real proof for any antiderivative ultimately falls on the derivative of the proposed answer. Either it works or it doesn't. It is doubtful most teachers have even worked out $\int x dx$ directly from a Riemann sum with variable bound. Yet, the totally reasonable thing is to guess $x^2/2$... – James S. Cook Sep 5 '14 at 6:52
• @JamesS.Cook in this case, we can use the geometry of triangles. – Steven Gubkin Sep 5 '14 at 14:13
• @Steven Gubkin Fair enough, but, I intend to imply direct calculation of the Riemann sum. – James S. Cook Sep 5 '14 at 15:05
• @BenjaminDickman I do think symmetry arguments of this sort can be very powerful, however, for me, I usually find them in the retrospective light of some more brutish solution. Neat idea. – James S. Cook Nov 7 '15 at 13:44

If you see this in calculus, there is a rather natural approach: to use the methods of calculus to study it!

Significant features of $f(x) = \frac{ax + b}{(x - c)(x - d)}$ happen around $x=c$ and $x=d$. Suppose we study the $x \approx c$ case.

One basic thing is to observe that $x-c$ is the only part of the expression that behaves interestingly; consequently, we should expect

$$x \approx c \implies f(x) \approx \frac{ac+b}{(x-c)(c-d)}$$

Following the pattern of subtracting off features, you write

$$f(x) = \frac{ac+b}{(x-c)(c-d)} + g(x)$$

and have some expectation that $g(x)$ doesn't have a 'feature' at $c$, or at least, has a simpler feature. Then, either you do a few examples to see that this really does eliminate the pole at $c$, or maybe you do something with Taylor/Laurent series, or you estimate it with asymptotic analysis.

A more generic lemma is probably simpler to use, though:

$$\frac{r(x)}{x-c} = \frac{r(c)}{x-c} + \frac{r(x)-r(c)}{x-c}$$

and we recognize the second term as a difference quotient, which should remain well-behaved at $x=c$, as it converges to $r'(c)$.

The remaining clever idea to convey (and it is a clever idea until you're used to such things) is that once you know the end form of the manipulation, you can skip straight there and solve for the unknowns by hook or by crook. (or even by the formula indicated by the lemma above)

• The difference quotient appearance here is neat; I'd not seen it before. – Chris Cunningham Mar 12 '18 at 19:24

As long as your goal is not proof, but just helping people get a general feel for it, there are several ways you can go.

The first is to say that, even if there were no proof, it is a worthwhile thing to try. That is, we can imagine not having the proof that it will work, but still attempting to do it anyway, just like we do with a lot of other methods. We can see that adding some numerator over $(x - c)$ and some numerator over $(x - d)$ will give us a sum that has a denominator of $(x - c)(x - d)$. We also know that it would be a heck of a lot easier to do the integral if we could find some isolated number to put in the numerator. So, why not see if there is one there.

The procedure of partial fractions is fairly straightforward if we just see if something will turn up. If we proceed just on the hunch that maybe we'll find something appropriate, then we can move forward. If we succeed we have found something we can integrate, and if we fail then, just like any other attempt, we'll have to find some other way.

In fact (I don't know if this is the case here), sometimes in math this is the way it works. You have a method that you know can work sometimes, and then, you just kind of notice that it works always. You don't know if it is "always always" or just "always for my problems", so then you search to see if you can prove it one way or the other. In any case, seeing that it could work is the first step to seeing how it works.

Now, given that it could work, it isn't too hard to see why it should always work. To see this, imagine what would happen if $(x - c)$ or $(x - d)$ had an $x$ in the numerator. When cross-multiplying, that would bring the degree up to $x^2$! Since we don't have any $x^2$ to work with, we can say for certain that either neither of them or both of them have $x$ (they could both have $x$ if one were a negative of the other to cancel them out). However, if they both had an $x$ in the numerator, then they could both be simplified through polynomial long division, which would get rid of the $x$ and leave a $1$ and a $-1$ by themselves, and a fraction without the $x$ in the numerator. When the $1$ and the $-1$ cancel out, that leaves just the fractions without the $x$ in the numerator!

So, we identified two possible cases - either both have an $x$ or neither do, and realized that the only way that both of them had an $x$ is if it were in a way that they canceled out and reduced to the case where neither one had an $x$. Thus, we can always assume that neither one has an $x$.

It's kind of convoluted, and not entirely a proof, but I think it goes a long way to showing why this method can work, and why they should trust that a proof exists.

1. I suspect the issue is more one for the teacher than the students. Having a proof (or at a lighter level, a derivation) is nice motivation, but the key thing to getting confidence is just drilling the method itself. I suspect teacher is implicitly unhappy with teaching non-proved topics and that the student questions come from some implicit presentation (alluding to the proof gap by teacher).

2. If I wanted to show a little derivation for motivation, I would do something like this (showing the backwards derivation):

a. Show them a specific quadratic reciprocal integral (with numbers, not general). Make a joke about how scary and evil it looks--we don't know what the heck to do with it.

b. hey, y'all...remember in grade school, when you combined fractions? Well let's do one for practice.

[work through $\frac23 + \frac45 = \frac{22}{15}$]

c. OK, then. Well now we aren't in grade school any more, we need to do stuff with variables and all! Remember in Algebra 2, when we learned about partial fractions. "Oh....don't you groan now...because it is finally going to have a use!"

[Work through combination of $\frac{A}{x-c} + \frac{B}{x-d}$. In other words, GO BACKWARDS.]

d. Do the algebraic derivation forwards. (from the quadratic on the bottom to the two fractions.)

e. Solve the specific "scary looking" integral using partial fractions. Make a joke about how we kicked the butt of that scary looking integral using grade school fractions and high school algebra. Not so scary any more!

f. Have the kids do one in class (simple, simple version).

g. "Ok, guys, this is just a bunch of algebra. It's not as hard as that darned integration by parts or those evil related rate word problems. But you gotta practice it. If you don't, you will mess up the grade school fractions and high school algebra. So, I want each of you to do 10 of these problems tonight. And we will have a one question quiz tomorrow. And next week on the test, I guarantee you at least one problem will be a partial fraction integral. And remember, you need to kick the integral's butt. Not let it kick yours!"

• -1, This answer is aimed at students far below the level the question is looking to help. – Chris Cunningham Mar 12 '18 at 6:51
• Sounds like they are in second semester freshman calc (or AP BC) since they are learning partial fractions integration and have not had LA yet. Do you know something else about the level of the trainees (e.g. how competitive a school they are at)? – guest Mar 12 '18 at 12:51
• I think that being down to earth is more practical to teach people (including really smart people). – guest Mar 12 '18 at 12:52
• Plus the author said " I'd be happy EVEN WITH AN EXPLANATION IN THE SIMPLEST CASE..." – guest Mar 12 '18 at 12:55
• A little research found this: "I am due to teach Calculus II in the fall at an American state university. Our calculus sequence is somewhat slow, due to the fact that many of our students come with limited backgrounds." – guest Mar 12 '18 at 12:58