Partial fraction expansion is a nice trick to simplify certain kinds of problems.
The two main applications are integral problems and inverse Laplace transforms.
Consider adding these numerical fractions:
$$\frac{1}{2} + \frac{2}{3} = \frac{1\cdot 3 + 2 \cdot 2}{2 \cdot 3} = \frac{7}{6}$$
From this you can see that a fraction can be written as a sum of smaller fractions.
This is also true of polynomial fractions, and partial fraction expansion is the method used to find these smaller fractions.
Looking at our numerical example, the partial fraction expansion is similar to finding that 1/2 and 1/3 add together to make 7/6.
For just numbers it isn’t anything special but with polynomials it can greatly simplify a problem.
Contents
Simple Case
Let’s try one with a quadratic denominator. First, the denominator must be factored.
$$\frac{2x + 1}{x^2 + 5x + 6} = \frac{2x + 1}{(x + 2)(x + 3)}$$
We can now write it as a sum of terms, one per factor, with an unknown coefficient.
$$\frac{2x + 1}{(x + 2)(x + 3)} = \frac{A}{x + 2} + \frac{B}{x + 3}$$
The next trick is to find the coefficients and there are two popular methods.
Matching Coefficients
Multiply out all of the terms to eliminate the fractions, then group terms into a polynomial form.
$$2x + 1 = A(x + 3) + B(x + 2)$$
$$2x + 1 = Ax + 3A + Bx + 2B$$
$$2x + 1 = (A + B)x + (3A + 2B)$$
By comparing the left and right sides you can match the coefficients, giving these two equations:
$$2 = A + B$$
$$1 = 3A + 2B$$
Subtracting twice the first equation from the second gives:
$$1 – 2(2) = 3A – 2A + 2B – 2B$$
$$A = -3$$
And plugging it back in gives:
$$B = 5$$
Therefore our partial fraction expansion is:
$$\frac{2x + 1}{x^2 + 5x + 6} = \frac{-3}{x + 2} + \frac{5}{x + 3}$$
Plug in Roots
An alternate method is to plug in the roots of the denominator.
To recap, the roots are the solution to the equation:
$$x^2 + 5x + 6 = 0$$
$$(x + 2)(x + 3) = 0$$
$$x + 2 = 0 \rightarrow x = -2$$
$$x + 3 = 0 \rightarrow x = -3$$
Now we plug those one at a time into this equation from the previous section:
$$2x + 1 = A(x + 3) + B(x + 2)$$
Start with -2:
$$2(-2) + 1 = A(-2 + 3) + B(-2 + 2)$$
$$-3 = A(1) + B(0)$$
$$A = -3$$
Then use -3:
$$2(-3) + 1 = A(-3 + 3) + B(-3 + 2)$$
$$-5 = A(0) + B(-1)$$
$$B = 5$$
This is the same answer as before.
Repeated Roots
What made the earlier case simple was that the denominator had 2 distinct, real roots.
Consider a different equation which has a real root that repeats:
$$\frac{x + 5}{x^2 – 2x + 1} = \frac{x + 5}{(x – 1)(x – 1)}$$
The trick here is that you just need to add a power to the repeated term, as shown below:
$$\frac{x + 5}{(x – 1)(x – 1)} = \frac{A}{(x – 1)^2} + \frac{B}{x – 1}$$
(If it repeated three times you would have a cube, square and linear term).
Now continue as before by multiplying out the fractions and match the coefficients.
$$x + 5 = A + B(x – 1)$$
$$x + 5 = Bx + (A – 1)$$
$$A = 6, B = 1$$
Which gives the expansion:
$$\frac{x + 5}{(x – 1)(x – 1)} = \frac{6}{(x – 1)^2} + \frac{1}{x – 1}$$
Complex Roots
When the denominator has complex or imaginary roots it cannot be factored.
Then, the next best thing is to leave it as a more complicated fraction as in the following example:
$$\frac{3x + 7}{x^3 + 3x^2 + 5x} = \frac{3x + 7}{x(x^2 + 3x + 5)}$$
In this case, the numerator of the quadratic root needs to be a linear term.
$$\frac{3x + 7}{x(x^2 + 3x + 5)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 3x + 5}$$
(In general the numerator is always 1 order lower than the denominator when solving for coefficients, so a cubic denominator needs a quadratic numerator)
Proceeding as normal by multiplying out the fractions:
$$3x + 7 = A(x^2 + 3x + 5) + (Bx + C)(x)$$
$$3x + 7 = Ax^2 + 3Ax + 5A + Bx^2 + Cx$$
$$3x + 7 = (A + B)x^2 + (3A + C)x + 5A$$
Giving the equations:
$$0 = A + B$$
$$3 = 3A + C$$
$$7 = 5A$$
Solving the last one for A and then plugging back in:
$$A = \frac{7}{5} \rightarrow B = -\frac{7}{5} \rightarrow C = -\frac{6}{5}$$
Therefore the expansion is:
$$\frac{3x + 7}{x^3 + 3x^2 + 5x} = \frac{\frac{7}{5}}{x} + \frac{-\frac{7}{5}x – \frac{6}{5}}{x^2 + 3x + 5}$$
Caveats
Partial fraction expansion does not work in all cases.
The following rules apply:
- The numerator and the denominator must be polynomials.
- The denominator must be at least one order higher than the numerator. That is, x over x^2 is ok, but x^2 over x will not work.