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?

**Include more fractions in Step 3....**

__More factors?____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)

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

__Repeated binomial factors?____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)

**If the denominator is quadratic, make the numerator of the fraction a binomial...**

__A quadratic factor?____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.

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