PARTIAL FRACTIONS (Revisited)

WHY DO WE STUDY PARTIAL FRACTIONS IN ALGEBRA AND PRECALCULUS?

This one is for Emma.  A "strange" question appeared on her Precalc review worksheet.  It didn't appear to be related to the transformation of graphs concept that was the focus of the homework.  The problem looked something like this:
         Solve for A and B:                    x             =       A    +    B   
                                       x^2 + 3x + 2      (x + 1)  (x + 2)       I wondered how many of the students didn't even know that this is a partial fractions question.  This is an honors class, but I don't recall the topic being covered in previous courses.  For the benefit of everyone in the class, here's the lesson we shared in our tutorial session.  (It's a repeat of the blog entry from June, 2013)

In the early days of Algebra study, we covered “collect like terms” so that the process would be a simple step once we got to FOIL (also called “double distributive” in some schools).  The same thinking encourages us to learn other basic algebraic steps before using them in more complicated mathematics.  In Calculus, we may need to decompose rational expressions before applying the rules of integration.  By doing the initial algebraic work of decomposition now, higher level concepts will be easier to learn and understand.

There are several situations which may be presented in Partial Fraction exercises.  We’ll start with the simple and work toward the more complex.

Simple rational expression:

               2x + 3       
          (x^2 - 7x +10)        Notice that the degree of the numerator is smaller than the degree of the denominator.  This is an important restriction that must be met.  (see below: “What if the Numerator is a higher degree than the Denominator?”)

    STEP 1 - factor the denominator

              2x + 3  
            (x-2)(x-5)

    STEP 2 - using the factors from the denominator, list separate fractions to form an equation. (We don’t know the numerators yet, so let’s just keep them blank until Step 3).

              2x + 3        ?          ?  
            (x-2)(x-5)= (x-2) + (x-5)

    STEP 3 - Substitute letters for the unknown numerators.

              2x + 3         A          B 
            (x-2)(x-5) = (x-2) + (x-5)

    STEP 4 - Expand the right side to have a common denominator.

              2x + 3        A(x-5) + B(x-2)
            (x-2)(x-5) =   (x-2)      (x-5)

    STEP 5 - Since denominators on both sides of the equation are the same, set the numerators equal and distribute A and B.
   
              2x + 3    =  Ax - A5 + Bx - 2B

    STEP 6 - Collect like variable terms.

              2x + 3   = (A + B)x - 5A - 2B

    STEP 7 - Create a system of equations by setting coefficients of the variables equal.

            2  = A + B
            3  = -5A -2B            The reason I prefer this ‘systems’ method to some others is that I can use the matrix function on my calculator to get answers with only a few key strokes and no manual arithmetic.
    STEP 8 - Solve the system.

            A = -7/3
            B = 13/3

    STEP 9 - Seems like a lot of steps so far, but this is the last one.  Substitute A and B values into the equation from Step 3.

              2x + 3         -7/3      13/3 
            (x-2)(x-5) =  (x-2) +   (x-5)


WHAT ABOUT MORE COMPLEX DENOMINATORS?             
   
    More factors?  Include more fractions in Step 3....
   
                 2x + 3              A         B         C  
            (x-2)(x-5)(x-1) = (x-2) + (x-5) + (x-1)

    .... and continue with Steps 4 - 9  (SOLUTION:  A = -7/3, B= 13/12, C= 5/4)

    Repeated binomial factors?  Consider the highest degree of the repeated binomial and EVERY SMALLER DEGREE to set up the equation in Step 3....

                2x + 3             A         B           C    
            (x-2)(x-5)^2   = (x-2) + (x-5) + (x-5)^2

    ....and continue with Steps 4 - 9.  (SOLUTION:  A= 7/9, B= -7/9, C=13/3)

A quadratic factor?  If the denominator is quadratic, make the numerator of the fraction a binomial...

          2x + 3             A        Bx + C   
    (x-2)(x^2 - 5)   = (x-2) + (x^2 - 5)

    ....and continue with Steps 4 - 9.  (SOLUTION: A=-7, B=7, C=16)

    What if the Numerator is a higher degree than the Denominator? 
AHA!!  Another example of how simple algebra concepts are eventually used again as steps within more complex algorithms!!  To satisfy the requirement that Partial Fractions can only be constructed if the Numerator is a lower degree than the Denominator, we need to break up the rational expression even more by LONG DIVISION.  The Partial Fraction is constructed from the REMAINDER, expressed as a fraction with the divisor as the Denominator.