#readwise
# Partial Fractions - Linear Factors
## Metadata
- Author: [[brilliant.org]]
- Full Title: Partial Fractions - Linear Factors
- URL: https://brilliant.org/wiki/partial-fractions-linear-factors/
## Summary
The text explains how to find the partial fraction decomposition of rational expressions. It shows how to factor the denominator and set up equations to solve for constants. Examples are provided to illustrate the process. The method can be generalized for various rational expressions.
## Highlights
[[Readwise/Articles/Partial Fractions|Partial Fractions]] is a technique used to write a [rational function](https://brilliant.org/wiki/rational-functions/) as the sum of simpler rational expressions. In certain cases, a rational function can be expressed as the sum of fractions whose denominators are [linear binomials](https://brilliant.org/wiki/polynomials/) For example,
$
\frac{2}{x^2 - 1} \implies \frac{1}{x - 1} - \frac{1}{x + 1}.
$
([View Highlight](https://read.readwise.io/read/01jftd318haeqsabx2s15pynh8))
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Finding the partial fraction decomposition form is only part of the goal of partial fraction decomposition. The ultimate goal is to calculate the values of the numerators so that the partial fraction decomposition is equivalent to the original expression.
Returning to the example introduced in the previous section:
Find the partial fraction decomposition of the rational expression
$
\frac{2}{x^2 + 5x + 6}.
$
Recall from the previous section that the partial fraction decomposition form is
$
\frac{2}{x^2 + 5x + 6} = \frac{A}{x + 2} + \frac{B}{x + 3}.
$
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The goal now is to find the values of $A$ and $B$ so that these expressions are equivalent. Begin by combining the fractions on the right-hand side of the equation:
$
\frac{A}{x + 2} + \frac{B}{x + 3} = \frac{A(x + 3) + B(x + 2)}{(x + 2)(x + 3)} = \frac{(A + B)x + 3A + 2B}{x^2 + 5x + 6}.
$
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Note that the denominator of this expression is the same as the denominator of the original expression. This means that the numerators must be equal:
$
2 = (A + B)x + 3A + 2B.
$
There is no $x$ term in the original numerator. Therefore,
$
A + B = 0.
$
The other part of the numerator must be equal to 2:
$
3A + 2B = 2.
$
Solving this system of equations gives the values of $A$ and $B$ that will cause the partial fraction decomposition to be equivalent to the original expression: $A = 2$ and $B = -2$. Then the partial fraction decomposition is
$
\frac{2}{x^2 + 5x + 6} = \frac{2}{x + 2} - \frac{2}{x + 3}.
$
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([View Highlight](https://read.readwise.io/read/01jftd7syyw9xbdrnwgvjrw2gn))
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