Partial Fractions

Section 4.3 Partial Fractions

After long division and factoring, this section introduces a new algebraic technique for understanding rational functions. This technique is the basis for integrating rational function, but by itself is just an algebraic technique. It is essentially doing common denominator in reverse. Therefore, let me start by reviewing common denominator with polynomials.

Subsection 4.3.1 Common Denominators

When adding or subtracting polynomials, it is necessary to take the polynomials to a common denominator. In general, this can by done by simply multiplying the denominators together to form the common denominator, then multiplying the numerators by the appropriate factors to preserve the fractions. (A fraction is preserved when the numerator and denominator are multiplied by the same term.) Here is a simple example that just multiplies the two denominators together.

\begin{equation*} \frac{1}{x-3} + \frac{1}{x+4} = \frac{(x+4) + (x-3)}{(x-3)(x+4)} = \frac{2x+1}{x^2+x-12} \end{equation*}

However, like with fractions, its often wise to make use of common factors already existing in the denominators. To add the fractions \(\frac{7}{16}\) and \(\frac{5}{24}\text{,}\) it makes more sense to use the common denominator \(48\text{,}\) since \(3 \times 16 = 48\) and \(2 \times 24 = 48\text{.}\) (For those of you who remember your algebra, this \(48\) is the least common multiple or LCM of \(16\) and \(24\text{.}\)) Certainly, it is nice to use \(48\) rather than the product of the two terms \(384\text{.}\) The same thing works for rational functions. Consider this example, where the two polynomials share a factor. I only need to multiply the first polynomial by \((x-1)\) and the second polynomial by \(x^2\) to form the common denominator of \(x^2(x-1)(x-3)\text{.}\)

\begin{align*} \amp \frac{x^2-2}{x^2(x-3)} + \frac{1}{(x-1)(x-3)}\\ \amp = \frac{(x-1)(x^2-2)}{x^2(x-1)(x-3)} + \frac{x^2}{x^2(x-1)(x-3)} = \frac{(x-1)(x^2-2) + x^2}{x^2(x-1)(x-3)}\\ \amp = \frac{x^3-2x+2}{x^2(x-1)(x-3)} \end{align*}

Subsection 4.3.2 Undoing Common Denominator

Common denominator takes the sum or difference of two or more rational functions and puts them together into one fraction. What is I want to do the opposite? Given a rational function, can I pull it apart into its various pieces? Can I do common denominator in reverse?

I can, and the technique is known as the partial fractions decomposition or, more commonly, just partial fractions. It applies to proper fractions with factored denominators, explaining the need to review these techniques in Section 4.2. Assume that \(\frac{p(x)}{q(x)}\) is a proper rational function and \(q(x) = q_1(x) q_2(x) \ldots q_r(x)\) is a factorization into linear and irreducible quadratic factors. Then I want to find new numerators \(p_i\) to undo the common denominator, as in the following expression.

\begin{equation*} \frac{p(x)}{q(x)} = \frac{p_1(x)}{q_1(x)} + \frac{p_2(x)}{q_2(x)} + \frac{p_3(x)}{q_3(x)} + \ldots + \frac{p_r(x)}{q_r(x)} \end{equation*}

Partial fractions tries to calculate these numerators. It is often a long and complicated process. Instead of trying to explain it abstractly, it’s much more reasonable to explain it through examples.

Example 4.3.1.

\begin{equation*} \frac{x+2}{(x+1)(x-6)} \end{equation*}

The partial fraction decomposition should write this as two fractions, each of which have only one of the two factors of the denominator.

\begin{equation*} \frac{x+2}{(x+1)(x-6)} = \frac{a}{x-6} + \frac{b}{x+1} \end{equation*}

The problem of partial fractions is to calculate the numerators \(a\) and \(b\text{.}\) This is the general technique: I write the desired form with unknown numerators. All linear factor (of multiplicity one) will have constant numerators (since all the pieces have to proper as well, and a linear denominator for a proper fraction means a constant numerator). After writing the desired form with unknown numerators, I’ll then simply construct the common denominator on the right side.

\begin{equation*} \frac{x+2}{(x+1)(x-6)} = \frac{a}{x-6} + \frac{b}{x+1} = \frac{a(x+1) + b(x-6)}{(x-6)(x+1)} = \frac{(a+b)x + (a-6b)}{(x+1)(x-6)} \end{equation*}

Taking the right side to common denominator makes the denominators the same. Since this is an equality, this means the numerators must also be the same. I write that fact as an equation.

\begin{equation*} x+2 = (a+b)x + (a-6b) \end{equation*}

Now this is the next general step: I compare coefficients in the equation of numerators. This will give a system of equations in the unknowns. In this case, the coefficient of \(x\) and the constant coefficient must be the same. I can write this as two equations.

\begin{align*} 1 \amp = a+b\\ 2 \amp = a-6b \end{align*}

I have to solve these two equations for \(a\) and \(b\text{.}\) I do this by isolating and replacing. (For students with linear algebra, know that the system in partial fractions will always be a linear system. Feel free to use the matrix techniques of linear algebra to solve these systems if you wish.) The first equation gives \(a = 1-b\text{,}\) which I substitute in the second equation to get \(2 = (1-b) -6b\) which is \(1 = -7b\) or \(b = \frac{-1}{7}\text{.}\) The I substitute back to get \(a = 1-b = 1 - \frac{-1}{7} = \frac{8}{7}\text{.}\) Now I’ve sovled the systems and I know the unknowns from the numerators. I can now write the original partial fraction decomposition with these values in the numerators.

\begin{equation*} \frac{x+2}{(x+1)(x-6)} = \frac{\frac{8}{7}}{x-6} + \frac{\frac{-1}{7}}{x+1} \end{equation*}

That’s the end of the process. The right side is the partial fraction composition: it is common denominator in reverse. The rational function is separated in a sum of the simplest possible fractions.

After one example, I can summarize the steps of the algorithm. I assume that I am starting with a proper rational function with a factored denominator.

Write the partial fraction decomposition with unknowns in the numerators.
Take the partial fraction decomposition with the unknowns to common denominator.
Compare this to the original fraction, geting an equation of numerators.
Compare the coefficients of this equation of numerators. The comparison of each coefficient becomes a new individual equation. The produces a system of equations.
Solve the linear system of equations to calculate the unknown from the numerators.
Use the solution of the system to write the partial fraction decomposition with known numerators.

Example 4.3.2.

\begin{equation*} \frac{x^2+1}{(x+1)(x-2)(x-3)} \end{equation*}

There are three linear factors in the denominator. I expect the partial fractions decomposition to split up into three pieces, one for each of these linear factors. The numerators will be constant. This is the form I expect:

\begin{equation*} \frac{x^2+1}{(x+1)(x-2)(x-3)} = \frac{a}{x+1} + \frac{b}{x-2} + \frac{c}{x-3} \end{equation*}

Now I need to determine the unknowns \(a\text{,}\) \(b\) and \(c\text{.}\) I do this by taking the partial fractions decomposition with the unknowns and creating a common denominator. For each term with one linear factor in the denominator, I have to multiply by the other two linear factors to produce the common denominator. Then I simplify and group the resulting numerator.

\begin{align*} \frac{x^2+1}{(x+1)(x-2)(x-3)} \amp = \frac{a}{x+1} + \frac{b}{x-2} + \frac{c}{x-3}\\ \amp = \frac{a(x-2)(x-3) + b(x+1)(x-3) + c(x+1)(x-2)} {(x+1)(x-2)(x-3)}\\ \amp = \frac{ax^2 -5ax + 6a + bx^2 - 2bx - 3b + cx^2 - cx - 2c} {(x+1)(x-2)(x-3)}\\ \frac{x^2+1}{(x+1)(x-2)(x-3)} \amp = \frac{(a+b+c)x^2 + (-5a-2b-c)x + (6a-3b-2c)} {(x+1)(x-2)(x-3)} \end{align*}

In the above equation, the denominators are the same, so the numerators must be as well. I write the equation of the numerators.

\begin{equation*} x^2 + 1 = x^2 + 0x + 1= (a+b+c)x^2 + (-5a-2b-c)x + (6a-3b-2c) \end{equation*}

Two polynomials are only equal if all their coefficients are equal, so I get three equations, one for each coefficient. I write the system of equations.

\begin{align*} a+b+c \amp = 1\\ -5a-2b-c \amp = 0\\ 6a-3b-2c \amp = 1 \end{align*}

I solve the system by several steps of isolating and replacing variables.

\begin{align*} a \amp = 1-b-c \end{align*}

I substitute for \(a\) into the second equation.

\begin{align*} -5(1-b-c) - 2b - c \amp = 0\\ -5+5b+5c - 2b -c \amp = 0\\ 3b + 4c \amp = 5\\ b \amp = \frac{5-4c}{3} \end{align*}

I substitute for \(a\) and \(b\) into third equation.

\begin{align*} 6a - 3b -2c \amp = 1\\ 6(1-b-c) - 3b - 2c \amp = 1\\ 6-6b-6c-3b-2c \amp = 1\\ -9b - 8c \amp = -5\\ 9 \frac{5-4c}{3} + 8 c \amp = 5\\ 45 - 36c + 24 c \amp = 15\\ -12 c \amp = -30 \implies c = \frac{-30}{-12} = \frac{5}{2} \end{align*}

Having solved for \(c\text{,}\) I substitute back in to the previous equations to determine \(a\) and \(b\text{.}\)

\begin{align*} b \amp = \frac{5 - 4 \frac{5}{2}}{3} = \frac{5-10}{3} = \frac{-5}{3}\\ a = 1 - b - c \amp = 1 + \frac{5}{3} - \frac{5}{2} = \frac{6 + 10 - 15}{6} = \frac{1}{6} \end{align*}

The values \(a = \frac{1}{6}\text{,}\) \(b = \frac{-5}{3}\) and \(c = \frac{5}{2}\) are the appropriate numerators. This lets me finish the partial fraction decomposition.

\begin{equation*} \frac{x^2+1}{(x+1)(x-2)(x-3)} = \frac{\frac{1}{6}}{x+1} - \frac{\frac{5}{3}}{x-2} + \frac{\frac{5}{2}}{x-3} \end{equation*}

Example 4.3.3.

Here is an example with an irreducible factor.

\begin{equation*} \frac{x^2-x-1}{(x-1)(x^2-2x+3)} \end{equation*}

The discriminant of the quadratic \(x^2 - 2x + 3\) is \(-8\text{,}\) so it is irreducible and cannot be factored any further. Therefore, I am looking for a partial fraction decomposition of the following form.

\begin{equation*} \frac{x^2- x - 1}{(x-1)(x^2-2x+3)} = \frac{a}{x-1} + \frac{bx+c}{x^2-2x+3} \end{equation*}

I take the right side to common denominator.

\begin{align*} \frac{x^2- x - 1}{(x-1)(x^2-2x+3)} \amp = \frac{a}{x-1} + \frac{bx+c}{x^2-2x+3}\\ \amp = \frac{a(x^2 - 2x + 3) + (bx+c)(x-1)}{(x-1)(x^2-2x+3)}\\ \amp = \frac{ax^2 - 2ax + 3a + bx^2 - bx + cx -c}{(x-1)(x^2-2x+3)}\\ \amp = \frac{(a+b)x^2 + (-2a-b+c)x + (3a-c)}{(x-1)(x^2-2x+3)} \end{align*}

By comparing the coefficients the polynomials in the numerators, I get a system with \(a+b=1\text{,}\) \(-2-b+c=-1\) and \(3a-c=-1\text{.}\) I can use the first and third equations to write everything in terms of \(a\text{.}\) That gives \(b = 1-a\) and \(c = 3a+1\text{.}\) Then I replace both \(b\) and \(c\) in the second equation.

\begin{align*} -2a + (1-a) + (3a + 1) \amp = -1\\ -2a -1 + a + 3a + 1 \amp = -1\\ 2a \amp = -1 \implies a = \frac{-1}{2}\\ b \amp = 1 - \frac{-1}{2} = \frac{3}{2}\\ c \amp = 3 \frac{-1}{2} + 1 = \frac{-1}{2} \end{align*}

This finishes the partial fraction decomposition.

\begin{equation*} \frac{x^2-x-1}{(x-1)(x^2-2x+3)} = \frac{\frac{-1}{2}}{x-1} + \frac{\frac{3}{2} x - \frac{1}{2}}{x^2 - 2x +3} \end{equation*}

Subsection 4.3.3 Multiple Factors

So far, I’ve done examples with linear factors an irreducible quadratic factors. The only thing I haven’t done is examples that have factors with multiplicity. For the needs of this course, I’m only going to worry about linear factor with multiplicity. (There is an equivalent process for irreducible quadratic factors with multiplicity, but seeing integrals with multiple quadratic factors in the denominator is rare enougth that I feel it’s reasonable to skip this piece.)

For, let me assume I have a denominator factor \((x-\alpha)^n\) for some multiplicity \(n \geq 2\text{.}\) To turn this into partial fractions, I need a decomposition that includes a sum of terms with various powers of this linear denominator.

\begin{equation*} \frac{a_1}{(x-\alpha)} + \frac{a_2}{(x-\alpha)^2} + \frac{a_3}{(x-\alpha)^3} + \ldots + \frac{a_n}{(x-\alpha)^n} \end{equation*}

This gives enough piece to put together the original rational function. If I don’t have all these terms, I won’t have enough information to solve the system of equations. Let me illustrate this method with an example.

Example 4.3.4.

\begin{equation*} \frac{2x-3}{(x-1)(x+2)^2} \end{equation*}

To account for the repeated root, I look for a partial fraction decomposition of the following form.

\begin{align*} \frac{2x-3}{(x-1)(x+2)^2} \amp = \frac{a}{x-1} + \frac{b}{x+2} + \frac{c}{(x+2)^2} \end{align*}

As always, I take the right side to common denominator, so that I can compare numerators and make a system of equations.

\begin{align*} \frac{2x-3}{(x-1)(x-2)^2}\amp = \frac{a(x+2)^2 + b(x+2)(x-1) + c(x-1)} {(x-1)(x-2)^2}\\ \amp = \frac{ax^2 + 4ax + 4a + bx^2 + bx - 2b + cx - c} {(x-1)(x-2)^2}\\ \amp = \frac{(a+b)x^2 + (4a+b+c)x + (4a-2b-c)} {(x-1)(x-2)^2} \end{align*}

I compare the coefficients of the numerators to get the system of equations.

\begin{align*} a+b \amp = 0\\ 4a + b + c \amp = 2\\ 4a-2b-c \amp = -3 \end{align*}

I solve the system. Start with the first equation is nice, because I can replace all the \(b\) coefficients quickly.

\begin{align*} b \amp = -a \end{align*}

Then I substitute this into the second equation.

\begin{align*} 4a + (-a) + c \amp = 2\\ 3a + c \amp = 2\\ a \amp = \frac{2-c}{3} \end{align*}

Then I substitute this into the third equation. After solving for \(c\text{,}\) I can replace in previous equations to find \(b\) and finally \(a\text{.}\)

\begin{align*} 4a - 2b - c \amp = -3\\ 4 \frac{2-c}{3} - 2 \frac{c-2}{3} - c \amp = -3\\ 8 - 4x - 2c + 4 - 3c \amp = -9\\ -9c \amp = -21\\ c \amp = \frac{7}{3}\\ a \amp = \frac{2 - \frac{7}{3}}{3} = \frac{-1}{9}\\ b \amp = -a = \frac{1}{9} \end{align*}

The system is solved, so I can write the partial fraction decomposition.

\begin{equation*} \frac{2x-3}{(x-1)(x+2)^2} = \frac{\frac{-1}{9}}{x-1} + \frac{\frac{1}{9}}{x+2} + \frac{\frac{7}{3}}{(x+2)^2} \end{equation*}

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