9. Partial Fractions

b1. Finding Coefficients - Linear

On the previous page, we saw examples of finding partial fraction decompositions and finding the unknown coefficients. We here summarize the techniques which are useful for finding the coefficients. There are two types of terms and corresponding techniques for each type:

1. Non-Repeated and Repeated Linear Denominators (discussed on this page)
2. Non-Repeated and Repeated Quadratic Denominators (discussed on the next page)

Non-Repeated and Repeated Linear Denominators

As we discuss the techniques, we will refer to the example: \[ \dfrac{6x-10}{(x-3)(x-1)^3} =\dfrac{A}{x-3}+\dfrac{B}{x-1}+\dfrac{C}{(x-1)^2}+\dfrac{D}{(x-1)^3} \] This has one non-repeated and one repeated linear factor in the denominator.

The first step is always:

This means to multiply both sides by the denominator of the left side. On the left this leaves the numerator. For each term on the right, we cancel the denominator of that term. Here is what this does to our example: \[\begin{aligned} 6x-10=&A(x-1)^3+B(x-3)(x-1)^2 \\ &+C(x-3)(x-1)+D(x-3) \qquad (*) \end{aligned}\]

The second step is to find the coefficients. There are several methods which may be mixed and matched:

This always works and gives all the coefficients, but it is not necessarily the most efficient method. Let's look at our example. We expand each term and collect terms with the same power of \(x\): \[\begin{aligned} 6x-10 =&A(x^3-3x^2+3x-1)+B(x^3-5x^2+7x-3) \\ &+C(x^2-4x+3)+D(x-3) \\ =&(A+B)x^3+(-3A-5B+C)x^2 \\ &+(3A+7B-4C+D)x+(-A-3B+3C-3D) \end{aligned}\] We then equate the coefficients of each power of \(x\). This gives \(4\) equations in \(4\) unknowns: \[\begin{aligned} A+B&=0 \qquad (**) \\ -3A-5B+C&=0 \\ 3A+7B-4C+D&=6 \\ -A-3B+3C-3D&=-10 \end{aligned}\] Solve the first equation for \(B\) in terms of \(A\). Then, the second equation for \(C\) in terms of \(A\), and then, the third equation for \(D\) in terms of \(A\). Next, plug all this into the fourth equation to find \(A\). Finally, substitute back to get the solution: Here are the details. \[ A=1 \qquad B=-1 \qquad C=-2 \qquad D=2 \]

So the partial fraction expansion is \[ \dfrac{6x-10}{(x-3)(x-1)^3} =\dfrac{1}{x-3}-\dfrac{1}{x-1}-\dfrac{2}{(x-1)^2}+\dfrac{2}{(x-1)^3} \]

Although the above method always works, it can be unwieldly if there are lots of coefficients. (A course on linear algebra teaches how to solve a system of linear equations with a large number of variables.) So it is useful to have a way to find some or all coefficients easily. Luckily, there is a straightforward method to find the numerators of terms whose denominators are linear or linear to a power.

Since the equations \((*)\) are true for all values of \(x\), we can plug in any number of values for \(x\) and get equations relating the coefficeints. The obvious values to plug in are the roots of the linear denominators. Here is what happens for our example. We plug the numbers \(x=3\) and \(x=1\) into \[\begin{aligned} 6x-10=&A(x-1)^3+B(x-3)(x-1)^2 \\ &+C(x-3)(x-1)+D(x-3) \qquad (*) \end{aligned}\]

• \(x=3\): (root of the non-repeated linear denominator) \[ 8=A(2)^3 \qquad \Longrightarrow \qquad A=1 \] This determines the numerator of the non-repeated linear denominator.
• \(x=1\): (root of the repeated linear denominator) \[ -4=D(-2) \qquad \Longrightarrow \qquad D=2 \] This determines the numerator of the highest power of the repeated linear denominator.

Using the values of \(A\) and \(D\) greatly simplifies the equations \((**)\). However, solving equations \((**)\) can still be tedious. So you can get additional coefficients more quickly using:

Once we have plugged in the obvious values of \(x\), we can plug in any other convenient values. For our example, we plug in the numbers \(x=0\) and \(x=2\) into\((*)\) and use the values of \(A\) and \(D\):

• \(x=0\): (always an easy number to plug in) \[\begin{aligned} -10&=A(-1)^3+B(-3)(-1)^2+C(-3)(-1)+D(-3) \\ &=-1-3B+3C-6 \\ &\Longrightarrow \qquad B-C=1 \end{aligned}\]
• \(x=2\): \[\begin{aligned} 2&=A(1)^3+B(-1)(1)^2+C(-1)(1)+D(-1) \\ &=1-B-C-2 \\ &\Longrightarrow \qquad B+C=-3 \end{aligned}\]

Together these equations say \(B=-1\) and \(C=-2\). (Why? Just add and subtract the two equations.)

Although it worked this time, there is no guarantee that plugging in randomly chosen values of \(x\) will give the necessary equations. So there is an alternate second step which is guaranteed to give the remaining numerators for denominators for repeated linear factors.

Since the left and right sides of \((*)\) are equal for all values of \(x\), so are their derivatives. After differentiating, we can again plug in values for \(x\) and get equations for the coefficients. This can be repeated until all coefficients are found. Recall equation \((*)\): \[\begin{aligned} 6x-10=&A(x-1)^3+B(x-3)(x-1)^2 \\ &+C(x-3)(x-1)+D(x-3) \qquad (*) \end{aligned}\] Its derivative is \[\begin{aligned} 6=A3(x-1)^2&+B\left[(x-1)^2+(x-3)2(x-1)\right] \\ &+C[(x-1)+(x-3)]+D \end{aligned}\]

Notice that when we did the derivative, we used the product rule instead of multiplying out and we don't simplify! This makes it easier to plug in the obvious root:

• \(x=1\): \[ 6=C(-2)+D=-2C+2 \qquad \Longrightarrow \qquad C=-2 \]

The second derivative of \((*)\) is \[ 0=A6(x-1)+B[2(x-1)+2(x-1)+(x-3)2]+C(2) \] We plug in the root.

• \(x=1\): \[ 0=B(-2)2+C(2)=-4B+2C=-4B-4 \qquad \Longrightarrow \qquad B=-1 \]

Each derivative gave one more coefficient for the repeated linear denominator.

Although the derivative method is guaranteed to give all the coefficients, it is harder to differentiate than plug in numbers. So we frequently just plug in and see what happens.

That completes the solution process for linear factors in the denominator. On the next page we will discuss the techniques for quadratic factors.

You can also practice finding the Coefficients in a Partial Fraction Expansion by using the following Maplet (requires Maple on the computer where this is executed):

Partial Fractions: Finding Coefficients   Rate It

At this point you can only do fractions whose denominators have only linear factors.