# Partial Fraction Decomposition (Linear Denominators)

## Key Questions

Transforming a rational polynomial into a sum of simpler rational polynomials due to the factorization of the denominator.

#### Explanation:

Let me try and explain this to you. :)

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What is a partial-fraction decomposition?

We are all familiar with the process of adding and subtracting fractions, e.g.

$\frac{1}{x + 1} + \frac{2}{x + 3} = \frac{x + 3}{\left(x + 1\right) \left(x + 3\right)} + \frac{2 \left(x + 1\right)}{\left(x + 1\right) \left(x + 3\right)} = \frac{3 x + 5}{{x}^{2} + 4 x + 3}$

Now, partial-fraction decomposition does exactly the reverse thing.

It takes a rational polynomial, so a fraction like

$\frac{3 x + 5}{{x}^{2} + 4 x + 3}$

from my example, and tries to decompose it into a sum of "simpler" fractions (to be more precise: into a sum of fractions which denominators are factors of the original fractions's denominator.)

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How to compute a partial-fraction decomposition?

1) Linear and unique factors

Let's stick with my example:

$\frac{3 x + 5}{{x}^{2} + 4 x + 3}$

The first thing to do is a always to find a complete factorization of the denominator:

${x}^{2} + 4 x + 3 = \left(x + 3\right) \left(x + 1\right)$

Here, all the factors are linear and unique, this is the simple case.

The goal is to to find $A$, $B$ so that the following equation holds:

$\frac{3 x + 5}{\left(x + 3\right) \left(x + 1\right)} = \frac{A}{x + 3} + \frac{B}{x + 1}$

To do so, first we should multiply both sides with the denominator:

$\iff 3 x + 5 = A \left(x + 1\right) + B \left(x + 3\right)$

Now, in order to solve this equation, we need to "group" the $\textcolor{red}{x}$ terms and the terms $\textcolor{b l u e}{\text{without } x}$ (more precisely: ${x}^{0}$ terms):

$\textcolor{red}{3 x} + \textcolor{b l u e}{5} = \textcolor{red}{A \cdot x} + \textcolor{b l u e}{A} + \textcolor{red}{B \cdot x} + \textcolor{b l u e}{3 B}$

Thus, we can form an equation based on the "red" terms and an equation based on the "blue" terms:

$\left\{\begin{matrix}3 = A + B \\ 5 = A + 3 B\end{matrix}\right.$

The solution of this linear equation system is inded $A = 2$ and $B = 1$ which leads us to the following partial-fraction decomposition:

$\frac{3 x + 5}{\left(x + 3\right) \left(x + 1\right)} = \frac{A}{x + 3} + \frac{B}{x + 1} = \frac{2}{x + 3} + \frac{1}{x + 1}$

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2) Non-unique factors

The computation is more complicated in case that the complete factorization of the original denominator doesn't consist just of linear and unique factors.

Just a short example:

Let's say we could factorize our rational polynomial like follows:

x/(x^2(x+1)^2(x-1)

As the factors $x$ and $x + 1$ occur twice in the denominator, we need to account for that by building fractions with increasing powers of those factors.
In this case, we would need to find $A$, $B$, $C$, $D$ and $E$ so that

$\frac{x}{{x}^{2} {\left(x + 1\right)}^{2} \left(x - 1\right)} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x + 1} + \frac{D}{x + 1} ^ 2 + \frac{E}{x - 1}$

holds.

++++++++++++++++++++++

3) Non-linear factors

Last but not least, there might be non-linear factors in your factorized denominator.

Example:

$\frac{1}{{x}^{3} + x} = \frac{1}{x \left({x}^{2} + 1\right)}$

Unfortunately, you can't factorize ${x}^{2} + 1$ further (at least, for real numbers).

In this case, your partial-fraction decomposition needs to look like follows:
Find $A$, $B$, $C$ so that the following equation holds:

$\frac{1}{x \left({x}^{2} + 1\right)} = \frac{A}{x} + \frac{B x + C}{{x}^{2} + 1}$

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When can a partial-fraction decomposition be useful?

Finally, I'd like to note that a very common application of this method is integration of rational polynomials:

If you need to compute

$\int \frac{3 x + 5}{{x}^{2} + 4 x + 3} \text{d} x$

and know that $\frac{1}{x + 1} + \frac{2}{x + 3} = \frac{3 x + 5}{{x}^{2} + 4 x + 3}$ holds,

it's easy to integrate by "splitting" the original fraction into a sum of simpler fractions and integrating each one of them:

$\int \frac{3 x + 5}{{x}^{2} + 4 x + 3} \text{d"x = int 1/(x+1) "d"x + int 2/(x+3) "d} x$