#readwise
# Partial Fractions - Repeated Factors
## Metadata
- Author: [[brilliant.org]]
- Full Title: Partial Fractions - Repeated Factors
- URL: https://brilliant.org/wiki/partial-fractions-repeated-factors/
## Summary
Partial fractions can have repeated factors when a denominator factor appears more than once. For repeated linear factors, each rational expression will have a constant numerator. The partial fraction decomposition involves expressing a rational function as a sum of simpler fractions. To find the coefficients, you combine and equate the numerators of the fractions.
## Highlights
A partial fraction has repeated factors when one of the denominator factors has [multiplicity](https://brilliant.org/wiki/fundamental-theorem-of-algebra/#factoring) greater than 1:
$
\frac{1}{x^3 - x^2} \implies \frac{1}{x^2(x - 1)} \implies \frac{1}{x - 1} - \frac{1}{x} - \frac{1}{x^2}.
$
The process for repeated factors is slightly different than the process for [[Partial Fractions - Linear Factors|linear, non-repeated factors]]. ([View Highlight](https://read.readwise.io/read/01jfw05h5cysmqqdtzqnsw766f))
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If $k$ is the multiplicity of the repeated factor, write $k$ rational expressions, each of which has that factor raised to a different power in the denominator. ([View Highlight](https://read.readwise.io/read/01jfw06scd9kq8pk7bg02cvw45)) ^0nqmdq
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Once the form is down, you can follow the same process as with linear factors to [[Partial Fractions - Linear Factors|solve for the coefficients]]. ([View Highlight](https://read.readwise.io/read/01jfw076bdg5e6gn5dssws71d0))
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Example:
Find the partial fraction decomposition of the following rational expression:
$
\frac{x^2 + x + 1}{x^3 + 3x^2 + 3x + 1}.
$
^wvdtf7
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The denominator can be factored as a perfect cube:
$
x^3 + 3x^2 + 3x + 1 = (x + 1)^3.
$
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The multiplicity of this factor is 3. Therefore, 3 rational expressions are needed in the partial fraction decomposition, each of which has $(x + 1)$ raised to a different positive integer power up to 3. Since $(x + 1)$ is a linear factor, each of these rational expressions will have a constant numerator coefficient:
$
\frac{x^2 + x + 1}{x^3 + 3x^2 + 3x + 1} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{(x + 1)^3}.
$
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To solve for the coefficients, combine the fractions on the right hand side of the equation:
$
\frac{x^2 + x + 1}{x^3 + 3x^2 + 3x + 1} = \frac{A(x + 1)^2 + B(x + 1) + C}{(x + 1)^3}.
$
Since the denominators are equal, the numerators must also be equal:
$
x^2 + x + 1 = A(x + 1)^2 + B(x + 1) + C.
$
Setting $x = -1$ in this equation gives $C = 1$.
Group the remaining terms by degree:
$
x^2 + x + 1 = A(x + 1)^2 + B(x + 1) + 1 = A x^2 + 2A x + A + B x + B + 1 = A x^2 + (2A + B) x + (A + B + 1).
$
This gives the following system of equations:
$
\begin{aligned}
A &= 1 \\
2A + B &= 1.
\end{aligned}
$
Solving this system of equations gives $A = 1$ and $B = -1$. Then the partial fraction decomposition is
$
\frac{x^2 + x + 1}{x^3 + 3x^2 + 3x + 1} = \frac{1}{x + 1} - \frac{1}{(x + 1)^2} + \frac{1}{(x + 1)^3}.
$
([View Highlight](https://read.readwise.io/read/01jfw08dv289vpwkk09cd1s2xm))
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