# When you are decomposing fractions into constants EX: 1/(x-1)(x-2)^2 = A/(x-1) + B/(x-2) +...

## Question:

When you are decomposing fractions into constants EX: {eq}\frac 1{(x-1)(x-2)^2} =\frac A{x-1} +\frac B{x-2} + \frac C{(x-2)^2} {/eq} Why do you have to repeat (x-2), instead of just putting {eq}\frac B{(x-2)^2} {/eq}?

## Partial Fraction Decomposition

When given a complex rational expression, partial fraction decomposition breaks down the expression into smaller and simpler rational expressions that add or subtract to get to the original, complex expression. When the denominator, specifically a binomial in the denominator, is raised to a power such as {eq}\left(x+4\right)^2 {/eq}, the equivalent term in partial fraction decomposition is {eq}\frac{A}{\left(x+4\right)} + \frac{B}{\left(x+4\right)^2} {/eq}, where {eq}A {/eq} and {eq}B {/eq} are constants based on the degree of the numerator of the original expression.

## Answer and Explanation:

Given: {eq}\frac{1}{(x-1)(x-2)^2} =\frac A{x-1} +\frac B{x-2} + \frac C{(x-2)^2} {/eq}

The reason {eq}(x-2) {/eq} is repeated as a factor in two separate expressions for the partial fraction decomposition is because {eq}(x-2) {/eq} as a factor in the denominator is a repeated factor (or have a multiplicity greater than 1). When this occurs, the general for partial decomposition is that the number of the multiplicity of the repeated factor is the same number of rational expressions that are needed to be expressed in the partial decomposition involved this repeated factor, with each expression having a factor raised to a different power in the denominator.

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from High School Algebra I: Help and Review

Chapter 3 / Lesson 26