Calc 2 Review: Partial Fractions

Introduction
To begin with, we'll return to the wonderful world of arithmetic. If we wanted to add to fractions, it is just a matter of finding a common denominator.

If we started with 31/35 and wanted to go back to two separate terms summed from 5th and 7th, how would we go about doing so?

To make our lives more simple, we will multiply by the least common denominator to clear the fractions and end up with 31 = 7A +5B. In this example, we get an underdetermined system of equations (two unknowns but only one equation). In fact there are an infinite number of A’s and B’s that would. We know that A=3 and B=2 works, but so does A = 8 and B= -5. Luckily, when dealing with polynomials, we get no such complications. If P(x) is a factorable polynomial, then the rational expression Q(x)/P(x) can be rewritten as the sum of polynomials with the denominators from the factors of P(x). That was a lot of math talk, let’s get into the examples.

Example One

We can factor the denominator and set up our equation like so

Multiplying both sides by the LCD we get:

1=A(x+1) +B(x-1) or 1= (A+B)x + (A-B)

Our goal is to solve for A and B. The first method is to match up the coefficients on the powers of x. On the right hand side we have 0 = A + B for the x term (or more specifically, “x to the one power” term) and 1 = A – B for the constant term (“x to the zero power”)

With two equations, we can solve the system using substitution, graphing, elimination, Gauss-Jordan elimination, Cramer’s Rule (and probably some other methods, too).

No matter how you do it, you end up with A = 1/2 B = -1/2.

So, 

If you want to double check your work using a technological aid, you can graph both equations on your calculator. As long as you are careful with your parenthesis, the graph of y1= 1/(x^2 – 1) and y2=(1/2)(1/(x-1)-1/(x+1) should come out the same.

Example Two

In this case, when we factor the denominator, we get a repeated factor. The rules for partial fractions tell us that we have to include each power of each factor. So we will start off with

Multiplying by the LCD gives us (x+3)=A(x+2)+B
We could solve for A and B by matching up the coefficients. Instead, we use will use a slightly different method. The equation has to be true for any value of x, so we will just pick strategic values of x. If x = -2, then the equation simplified down to 1 = 0 + B, so B = 1. Knowing this, we choose x = 0 and get the equation 3 = 2A + B. We know B = 1. Substituting and solving we get A = 1.

Example Three

When we factor the denominator, we get an irreducible/un-factorable quadratic. That’s ok. In this case, we set up the partial fraction like so:

Please note the numeration in the second term. Because we have an irreducible quadratic in the denominator, we have to have a linear term up top. 
To solve the problem, we follow the same process of multiplying by the least common denominator and solving for A, B, and C from there.

Rather than foil out the second term, I am going to pick values for x and build my equations that way. 
When x =-1, 9 = 2A
When x = 0, 8 = A + C
When x = 1, 9  = 2A + 2B + 2C
Back substituting, the numbers get a little bit ugly. We end up with 

Summary

Partial fractions only work with rational expressions. That is a polynomial divided by a polynomial. Also, they need to be proper fractions.If the degree of the numerator is the same or larger than the the degree of the denominator, first do polynomial long division and then use partial fractions on the remainder.
The most common mistake is initially setting up the fraction wrong. Remember the rules for linear factors (Example 1), repeated factors (Example 2) and quadratic factors (Example 3). The second most common mistake is to make an arithmetic error, so take your time and try to check your work.

That is the basics of partial fractions. They are not going away. Get good at partial fractions; they're coming back in Chapter 6 (and possibly before then, too).