- Break down fractions with polynomials when the denominator can be factored into different linear terms
- Break down fractions with polynomials when the denominator includes quadratic terms that can’t be factored
Linear Factors
Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression.
- Factor the numerator and denominator.
- Find the LCD of the expressions.
- Multiply the expressions by a form of 1 that changes the denominators to the LCD.
- Add or subtract the numerators over the common denominator.
- Simplify.
In this section, we will look at partial fraction decomposition, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified rational expression to the original expressions, called the partial fractions. Some types of rational expressions require solving a system of equations in order to decompose them.
[latex]\dfrac{2}{x - 3}+\dfrac{-1}{x+2}[/latex]
We would first need to find a common denominator,
[latex]\left(x+2\right)\left(x - 3\right)[/latex].
Next, we would write each expression with this common denominator and find the sum of the terms.
[latex]\begin{align}&\frac{2}{x - 3}\left(\frac{x+2}{x+2}\right)+\frac{-1}{x+2}\left(\frac{x - 3}{x - 3}\right) \\[2mm] &=\frac{2x+4-x+3}{\left(x+2\right)\left(x - 3\right)} \\[2mm] &=\frac{x+7}{{x}^{2}-x - 6} \end{align}[/latex]
Partial fraction decomposition is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.
[latex]\begin{align} \dfrac{x+7}{{x}^{2}-x - 6}&=\dfrac{2}{x - 3}+\dfrac{-1}{x+2} \\[2mm]\text{Simplified sum}&\hspace{6mm}\text{Partial fraction decomposition} \end{align}[/latex]
When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of [latex]{x}^{2}-x - 6[/latex] are [latex]\left(x - 3\right)\left(x+2\right)[/latex], the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition.
partial fraction decomposition: nonrepeated linear factors
The partial fraction decomposition of [latex]\dfrac{P\left(x\right)}{Q\left(x\right)}[/latex] when [latex]Q\left(x\right)[/latex] has nonrepeated linear factors and the degree of [latex]P\left(x\right)[/latex] is less than the degree of [latex]Q\left(x\right)[/latex] is
[latex]\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\dfrac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\dfrac{{A}_{3}}{\left({a}_{3}x+{b}_{3}\right)}+\cdot \cdot \cdot +\dfrac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}[/latex].
- Use a variable for the original numerators, usually [latex]A,B,[/latex] or [latex]C[/latex], depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use [latex]{A}_{n}[/latex] for each numerator
[latex]\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\dfrac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\cdots \text{+}\dfrac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}[/latex]
- Multiply both sides of the equation by the common denominator to eliminate fractions.
- Expand the right side of the equation and collect like terms.
- Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
[latex]\dfrac{3x}{\left(x+2\right)\left(x - 1\right)}[/latex]