Partial Fractions

Adding rational expressions and simplifying is relatively easy. For example:

\begin{eqnarray} \frac{2}{x-2}-\frac{1}{x+1}&=&\frac{2(x+1)}{(x-2)(x+1)}-\frac{x-2}{(x-2)(x+1)}\\ &=&\frac{2x+2-x+2}{(x-2)(x+1)}\\ &=&\frac{x+4}{(x-2)(x+1)}. \end{eqnarray}

However, going the other way round--expressing a single rational function as the sum of two or more simpler ones--is much more difficult. This reverse process is known as resolution into partial fractions.

The process is simplest if the denominator consists entirely of distinct linear factors. There are two possible complications: the denominator might contain a repeated (that is, squared, cubed or whatever) linear factor, or it might contain a quadratic factor that can't be factorised.

Example 1: denominator consists of distinct linear factors

Consider the rational expression

$$\frac{4x^2+3x-16}{(x-2)(x+1)(2x-3)}.$$

It can be shown that this can be expressed in the form

$$\frac{A}{x-2}+\frac{B}{x+1}+\frac{C}{2x-3}$$ (we can treat this fact as known). Our problem becomes finding the values of the unknown constants A, B and C.

The calculation goes in stages.

1. Multiply up
   
   We know that
   $$\frac{4x^2+3x-16}{(x-2)(x+1)(2x-3)}=\frac{A}{x-2}+\frac{B}{x+1}+\frac{C}{2x-3};$$ we start by multiplying both sides by (x-2)(x+1)(2x-3). This gives $$4x^2+3x-16=A(x+1)(2x-3)+B(x-2)(2x-3)+C(x-2)(x+1).$$
2. Choose values
   
   This is true for all values of x, so it's true for any values of x we care to choose. The strategy is to choose values that make most of the terms vanish. In this case, those values are x=2, x=-1 and x=3/2.
   
   Starting with x=2, we obtain
   $$4\times2^2+3\times2-16=A\times3\times1+B\times0+C\times0$$ and therefore
   $$6=3A$$ giving A=2.
   
   Setting x=-1, we get
   $$4\times(-1)^2+3\times(-1)-16=A\times0+B\times(-3)\times(-5)+C\times0$$ and therefore
   $$-15=15B$$ giving B=-1.
   
   Finally, setting x=3/2 gives
   $$4\times\frac{3^2}{2^2}+3\times{3}{2}-16= A\times0+B\times0+C\times\left(-\frac{1}{2}\right)\times\left(\frac{5}{2}\right)$$ and thus
   $$-\frac{5}{2}=-\frac{5}{4}C$$ giving C=2
   
   
3. Substitute in
   
   We now know the values of A, B and C, and have that
   
   $$\frac{4x^2+3x-16}{(x-2)(x+1)(2x-3)}=\frac{2}{x-2}-\frac{1}{x+1}+\frac{2}{2x-3}.$$

Example 2: what happens if a factor is repeated?

Problem: resolve into partial fractions

$$\frac{3x^2+6x+5}{(x+2)^2(x-3)}.$$

Two things change in these circumstances: first, the form of the partial fractions is altered, and secondly, our "choose values'' technique from above will no longer take us all the way on its own.

The partial fractions form of this expression is actually

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

note the way the repeated factor appears, once on its own and once squared.

Once again, we use a staged process: one that begins in much the same way as in the last example, but then becomes somewhat different.

1. Multiply up
   
   We know that $$\frac{3x^2+6x+5}{(x+2)^2(x-3)}=\frac{A}{x+2}+\frac{B}{(x+2)^2}+\frac{C}{x-3};$$ we start by multiplying both sides by (x+2)^2(x-3). This gives
   $$3x^2+6x+5=A(x+2)(x-3)+B(x-3)+C(x+2)^2.$$
   
   
2. Choose values
   
   Again, we try choose values that make most of the terms vanish. In this case, those values are x=-2 and x=3.
   
   Starting with x=-2, we obtain
   $$3\times(-2)^2+6\times(-2)+5=A\times0+B\times(-5)+C\times0^2$$ and therefore $$5=-5B$$ giving B=-1.
   Setting x=3, we get $$3\times3^2+6\times3+5=A\times0+B\times0\times(-5)+C\times5^2$$ and therefore $$50=25C$$ giving C=2.
   
   Unfortunately, this is where ``choosing values'' runs out of steam, and we don't yet know the value of A. We turn to...
   
   
3. Compare like terms
   
   The quickest way to calculate the value of A is to compare the x^2 terms in $$3x^2+6x+5$$ and$$A(x+2)(x-3)+B(x-3)+C(x+2)^2.$$ This gives us 3=A+C, from which we get that A=3-C=3-2=1.
   
   
4. Substitute in
   
   We now know the values of A, B and C, and have that $$\frac{3x^2+6x+5}{(x+2)^2(x-3)}=\frac{1}{x+2}-\frac{1}{(x+2)^2}+\frac{2}{x-3}.$$

Example 3: what happens if there is a quadratic factor that can't be factorised?

Problem: resolve into partial fractions $$\frac{2x^2+7}{(x+2)(x^2+1)}.$$

Here, too, the form is different and ``choosing values'' only takes us part of the way. Once again, we need to compare like terms; this time, we need to start doing that slightly earlier. The partial fractions form of this expression is

$$\frac{A}{x+2}+\frac{Bx+C}{x^2+1};$$ note that the second fraction's numerator is not constant but linear.

1. Multiply up
   
   We know that $$\frac{2x^2+7}{(x+2)(x^2+1)}=\frac{A}{x+2}+{Bx+C}{x^2+1};$$ we start by multiplying both sides by (x+2)(x^2+1). This gives $$2x^2+7=A(x^2+1)+(Bx+C)(x+2).$$
2. Choose values
   
   Again, we try choose values that make most of the terms vanish. In this case, the only candidate is x=-2.
   Setting x=-2, we obtain $$2\times(-2)^2+7=A\times((-2)^2+1)+(Bx+C)\times0$$ and therefore $$15=5A$$ giving A=3.
   
   We can take ``choosing values'' no further.
   
   
3. Compare like terms
   
   We start by comparing the x^2 terms in $$2x^2+7$$ and $$A(x^2+1)+(Bx+C)(x+2).$$ This gives us 2=A+B, from which we get that B=2-A=2-3=-1.
   
   We can then compare constant terms, obtaining 7=A+2C, meaning 2C=7-A=7-3=4 and thus C=2.
   
   
4. Substitute in
   
   We now know the values of A, B and C, and have that $$\frac{2x^2+7}{(x+2)(x^2+1)}=\frac{3}{x+2}+\frac{2-x}{x^2+1}.$$
   
   Note that it's entirely possible for both a repeated linear factor and an unfactorisable quadratic factor to appear in the numerator, as in $$\frac{5x^2+8x+9}{(x-2)(x+1)^2(x^2+1)},$$ in which case these techniques are combined. The partial fractions form of this example, for instance, is $$\frac{A}{x-2}+\frac{B}{x+1}+\frac{C}{(x+1)^2}+\frac{Dx+E}{x^2+1}.$$ This fairly complicated example is left as an exercise to the keen reader (answer: A=1, B=-1, C=-1, D=0, E=-2).
   
   Note also that in all these cases, the degree of the numerator is less than that of the denominator. If the degrees are equal, the partial fractions form of the expression has a constant term, and if that of the numerator is one more than that of the denominator, there's also an x-term; etc. For example, the partial fractions form of $$\frac{x^3+2x^2+1}{(x+2)(x+1)}$$ is $$Ax+B+\frac{C}{x+2}+\frac{D}{x+1}.$$ This is also left as an exercise (answer: A=1, B=-1, C=-1, D=2).