# Introduction to Partial Fractions Decomposition

Partial fractions decomposition is an algebraic technique that can be used to decompose (break down) a product of rational expressions into a sum of simpler rational expressions. A rational expression is one in which both the numerator and denominator are polynomials. A proper rational expression is one in which the degree of the numerator is strictly less than the degree of the denominator.

One important use of partial fractions decomposition is convert a difficult integration problem into two or more easier integration problems. However, in this Insights tutorial, we won’t look at the calculus applications, but will focus only on the algebraic aspects of this technique.

Before going any further, a fully worked example might be helpful.

Example: Decompose ##\frac{x}{(x – 1)(x + 1)}## into the sum of two rational expressions.

Solution and discussion: Decomposing this rational expression means finding constants A and B so that ##\frac{x}{(x – 1)(x + 1)} = \frac{A}{x – 1} + \frac{B}{x + 1}.##

Note: The equation here is really an identity — an equation that is true for all values of x for which all terms are defined. That is, other than x = 1 or x = -1.

##\frac{x}{(x – 1)(x + 1)} = \frac{A}{x – 1} + \frac{B}{x + 1}##

Multiplying both sides of the equation by ##(x – 1)(x + 1)## yields this equation:

##x = A(x + 1) + B(x – 1)##

Expanding the right side, we get:

##x = Ax + Bx + A – B = (A + B)x + (A – B)##

To make things completely obvious, we can write the equation above as

##0 + 1x = (A – B) + (A + B)x##

If this equation is to hold for all values of x, the coefficients on both sides have to be identically equal. As a result we must have 0 = A – B and 1 = (A + B). From these equations and a bit of algebra, we see that ##A = \frac 1 2## and ##B = \frac 1 2.##

So ##\frac{x}{(x – 1)(x + 1)} = \frac{\frac 1 2}{x – 1} + \frac{\frac 1 2}{x + 1}.##

For the remainder of this Insights article, I’ll show the form of the decomposition, but won’t show the work in solving for the constants.

Decomposition problems fall into one of the following categories, based on the factors in the denominator:

• Distinct linear factors
• Repeated linear factors
• Mixed factors

Each of these types is discussed with an example in the following sections.

## Distinct Linear Factors

In this case there are no repeated factors (factors raised to power 2 or higher) in the denominator.

Rational expression: ##\frac x {(x – 1)(x + 1)(x – 2)}##

Decomposition: ##\frac x {(x – 1)(x + 1)(x – 2)} = \frac{A}{x – 1} + \frac{B}{x + 1} + \frac{C}{x – 2}##

Multiply both sides of the equation above by ##(x – 1)(x + 1)(x – 2)##, and then solve for the constants A, B, and C. Although the equation above is undefined if x = 1, x = -1, or x = 2, the new equation (with denominators cleared) has no restrictions. You can set x to, respectively, 1, -1, and 2 to get three equations that can be easily solved for A, B, and C.

## Repeated Linear Factors

For this case, at least one of the factors in the denominator occurs to a power of two or higher.

Rational expression: ##\frac 5 {(x – 1)^2(x + 1)}##

Decomposition: ##\frac 5 {(x – 1)^2(x + 1)} = \frac{A}{x – 1} + \frac{B}{(x – 1)^2} + \frac{C}{x + 1}##

The middle term on the right side above, is counterintutive, but this is the decomposition that works. Inexperienced students often are tempted to try the following decomposition, which does not work.

Incorrect! ##\frac 5 {(x – 1)^2(x + 1)} = \frac{A}{x – 1} + \frac{B}{x – 1} + \frac{C}{x + 1}##

By irreducible, I mean that the factors have coefficients that are not real. For example, ##x^2 + 4 = (x – 2i)(x + 2i)##, so is irreducible, but ##x^2 – 4## can be factored into (x – 2)(x + 2).

Rational expression: ##\frac {x^2 – 3x} {(x^2 + 1)(x^2 + 9)}##

Decomposition: ##\frac {x^2 – 3x} {(x^2 + 1)(x^2 + 9)} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{x^2 + 9}##

Rational expression: ##\frac {x^2 – 3x} {(x^2 + 1)^2(x^2 + 9)}##

Decomposition: ##\frac {x^2 – 3x} {(x^2 + 1)^2(x^2 + 9)} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2} + \frac{Ex + D}{x^2 + 9}##

Note that ##x^2 + 1## occurs to the second power in the denominator, making it a repeated quadratic factor.

## Mixed factors

For example, ##\frac{3x + 7}{(x^2 + 4x + 4)(x^2 + 1)}## would be decomposed as ##\frac{A}{x + 2} + \frac{B}{(x + 2)^2} + \frac{Cx + D}{x^2 + 1} ##.

## One for you to try:

Decompose ##\frac{2}{x^3 – x^2 + x – 1}## into two simpler rational expressions.

Hint: Start by factoring the denominator.

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3 replies
1. symbolipoint says:

[QUOTE=”SammyS, post: 5335194, member: 295898″]Nice and complete and to the point ![/QUOTE]
It LOOKED good, but I only LOOKED at it; did not read it. I assume it said what is was supposed to. Some good college algebra and precalculus books have similar sections or two on this, just as well.