#readwise
# Partial Fractions - Irreducible Quadratics
## Metadata
- Author: [[brilliant.org]]
- Full Title: Partial Fractions - Irreducible Quadratics
- URL: https://brilliant.org/wiki/partial-fractions-irreducible-quadratics/
## Summary
Partial fraction decomposition simplifies a rational function into simpler parts. When the denominator has irreducible quadratic factors, the method differs from linear factors. Each irreducible quadratic factor has a linear binomial as its numerator. To find the coefficients, set up an equation based on equal numerators after combining fractions.
## Highlights
A partial fraction has irreducible quadratic factors when one of the denominator factors is a quadratic with [irrational](https://brilliant.org/wiki/irrational-numbers/) or [complex](https://brilliant.org/wiki/complex-numbers/) roots:
$
\frac{1}{x^3 + x} \implies \frac{1}{x(x^2 + 1)} \implies \frac{1}{x} - \frac{x}{x^2 + 1}.
$
The process for irreducible quadratic factors is slightly different than the process for [[Partial Fractions - Linear Factors|linear factors]]and [[Partial Fractions - Repeated Factors|repeated factors]]. ([View Highlight](https://read.readwise.io/read/01jfwah40695804t1wh4ewtx3g))
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For each irreducible quadratic factor, write a rational expression with that factor as the denominator and a linear binomial as the numerator. ([View Highlight](https://read.readwise.io/read/01jfwaj6zn5t2q26h1nacmypqe)) ^y2w85n
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Example:
Find the partial fraction decomposition of the following rational expression:
$
\frac{x^2 + 8}{x^3 + 8}.
$
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The denominator can be factored as a sum of cubes:
$
x^3 + 8 = (x + 2)(x^2 - 2x + 4).
$
The quadratic factor has complex roots, so it cannot be factored any further. In the partial fraction decomposition, the $(x+2)$ denominator will have a constant numerator, and the $(x^2 - 2x + 4)$ denominator will have a linear binomial numerator:
$
\frac{x^2 + 8}{x^3 + 8} = \frac{A}{x+2} + \frac{Bx + C}{x^2 - 2x + 4}.
$
^d0oajk
To solve for the coefficients, combine the fractions on the right hand side of the equation:
$
\frac{x^2 + 8}{x^3 + 8} = \frac{A(x^2 - 2x + 4) + (Bx + C)(x + 2)}{(x + 2)(x^2 - 2x + 4)}.
$
Since the denominators are equal, the numerators must also be equal:
$
x^2 + 8 = A(x^2 - 2x + 4) + (Bx + C)(x + 2).
$
Setting $x = -2$ in this equation gives $A = 1$. Group the remaining terms by degree:
$
\begin{aligned}
x^2 + 8 &= 1(x^2 - 2x + 4) + (Bx + C)(x + 2) \\
&= x^2 - 2x + 4 + Bx^2 + 2Bx + Cx + 2C \\
&= (B + 1)x^2 + (2B + C - 2)x + (2C + 4).
\end{aligned}
$
This gives the system of equations
$
\begin{aligned}
1 &= B + 1 \\
0 &= 2B + C - 2 \\
8 &= 2C + 4.
\end{aligned}
$
Solving this system gives $B = 0$ and $C = 2$. Then the partial fraction decomposition is
$
\frac{x^2 + 8}{x^3 + 8} = \frac{1}{x+2} + \frac{2}{x^2 - 2x + 4}.
$
([View Highlight](https://read.readwise.io/read/01jfwakh5945g3h32fyrz0qaj9))
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