Decomposing P(x) / Q(x), Where Q(x) Has a Nonrepeated Irreducible Quadratic Factor
So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators [latex]A,B[/latex], or [latex]C[/latex] representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as [latex]Ax+B,Bx+C[/latex], etc.
A General Note: Decomposition of [latex]\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)[/latex] Has a Nonrepeated Irreducible Quadratic Factor
The partial fraction decomposition of [latex]\frac{P\left(x\right)}{Q\left(x\right)}[/latex] such that [latex]Q\left(x\right)[/latex] has a nonrepeated irreducible quadratic factor and the degree of [latex]P\left(x\right)[/latex] is less than the degree of [latex]Q\left(x\right)[/latex] is written as[latex]\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}x+{B}_{1}}{\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\right)}+\frac{{A}_{2}x+{B}_{2}}{\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\right)}+\cdot \cdot \cdot +\frac{{A}_{n}x+{B}_{n}}{\left({a}_{n}{x}^{2}+{b}_{n}x+{c}_{n}\right)}[/latex]
The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator: [latex]A,B,C[/latex], and so on.
How To: Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it.
- Use variables such as [latex]A,B[/latex], or [latex]C[/latex] for the constant numerators over linear factors, and linear expressions such as [latex]{A}_{1}x+{B}_{1},{A}_{2}x+{B}_{2}[/latex], etc., for the numerators of each quadratic factor in the denominator.
[latex]\frac{P\left(x\right)}{Q\left(x\right)}=\frac{A}{ax+b}+\frac{{A}_{1}x+{B}_{1}}{\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\right)}+\frac{{A}_{2}x+{B}_{2}}{\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\right)}+\cdot \cdot \cdot +\frac{{A}_{n}x+{B}_{n}}{\left({a}_{n}{x}^{2}+{b}_{n}x+{c}_{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.
Example 3: Decomposing [latex]\frac{P\left(x\right)}{Q\left(x\right)}[/latex] When Q(x) Contains a Nonrepeated Irreducible Quadratic Factor
Find a partial fraction decomposition of the given expression.[latex]\frac{8{x}^{2}+12x - 20}{\left(x+3\right)\left({x}^{2}+x+2\right)}[/latex]
Solution
We have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus,[latex]\frac{8{x}^{2}+12x - 20}{\left(x+3\right)\left({x}^{2}+x+2\right)}=\frac{A}{\left(x+3\right)}+\frac{Bx+C}{\left({x}^{2}+x+2\right)}[/latex]
We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator.
[latex]\begin{array}{l}\left(x+3\right)\left({x}^{2}+x+2\right)\left[\frac{8{x}^{2}+12x - 20}{\left(x+3\right)\left({x}^{2}+x+2\right)}\right]=\left[\frac{A}{\left(x+3\right)}+\frac{Bx+C}{\left({x}^{2}+x+2\right)}\right]\left(x+3\right)\left({x}^{2}+x+2\right)\hfill \\ \text{ }8{x}^{2}+12x - 20=A\left({x}^{2}+x+2\right)+\left(Bx+C\right)\left(x+3\right)\hfill \end{array}[/latex]
Notice we could easily solve for [latex]A[/latex] by choosing a value for [latex]x[/latex] that will make the [latex]Bx+C[/latex] term equal 0. Let [latex]x=-3[/latex] and substitute it into the equation.
[latex]\begin{array}{l}\text{ }8{x}^{2}+12x - 20=A\left({x}^{2}+x+2\right)+\left(Bx+C\right)\left(x+3\right)\hfill \\ \text{ }8{\left(-3\right)}^{2}+12\left(-3\right)-20=A\left({\left(-3\right)}^{2}+\left(-3\right)+2\right)+\left(B\left(-3\right)+C\right)\left(\left(-3\right)+3\right)\hfill \\ \text{ }16=8A\hfill \\ \text{ }A=2\hfill \end{array}[/latex]
Now that we know the value of [latex]A[/latex], substitute it back into the equation. Then expand the right side and collect like terms.
[latex]\begin{array}{l}\hfill \\ 8{x}^{2}+12x - 20=2\left({x}^{2}+x+2\right)+\left(Bx+C\right)\left(x+3\right)\hfill \\ 8{x}^{2}+12x - 20=2{x}^{2}+2x+4+B{x}^{2}+3B+Cx+3C\hfill \\ 8{x}^{2}+12x - 20=\left(2+B\right){x}^{2}+\left(2+3B+C\right)x+\left(4+3C\right)\hfill \end{array}[/latex]
Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations.
[latex]\begin{array}{ll}\text{ }2+B=8\hfill & \text{(1)}\hfill \\ 2+3B+C=12\hfill & \text{(2)}\hfill \\ \text{ }4+3C=-20\hfill & \text{(3)}\hfill \end{array}[/latex]
Solve for [latex]B[/latex] using equation (1) and solve for [latex]C[/latex] using equation (3).
[latex]\begin{array}{ll}\text{ }2+B=8\hfill & \text{(1)}\hfill \\ \text{ }B=6\hfill & \hfill \\ \hfill & \hfill \\ 4+3C=-20\hfill & \text{(3)}\hfill \\ \text{ }3C=-24\hfill & \hfill \\ \text{ }C=-8\hfill & \hfill \end{array}[/latex]
Thus, the partial fraction decomposition of the expression is
[latex]\frac{8{x}^{2}+12x - 20}{\left(x+3\right)\left({x}^{2}+x+2\right)}=\frac{2}{\left(x+3\right)}+\frac{6x - 8}{\left({x}^{2}+x+2\right)}[/latex]
Q & A
Could we have just set up a system of equations to solve Example 3?
Yes, we could have solved it by setting up a system of equations without solving for [latex]A[/latex] first. The expansion on the right would be:[latex]\begin{array}{l}\begin{array}{l}\\ 8{x}^{2}+12x - 20=A{x}^{2}+Ax+2A+B{x}^{2}+3B+Cx+3C\end{array}\hfill \\ 8{x}^{2}+12x - 20=\left(A+B\right){x}^{2}+\left(A+3B+C\right)x+\left(2A+3C\right)\hfill \end{array}[/latex]
So the system of equations would be:
[latex]\begin{array}{l}\text{ }A+B=8\hfill \\ A+3B+C=12\hfill \\ \text{ }2A+3C=-20\hfill \end{array}[/latex]
Try It 3
Find the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic factor.[latex]\frac{5{x}^{2}-6x+7}{\left(x - 1\right)\left({x}^{2}+1\right)}[/latex]
Solution
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