8-04 Partial Fractions

Example 1: Partial Fractions with Linear Factors

Find the partial fractions for \(\frac{x + 5}{x^2 + 4x + 3}\)

Solution

Factor the denominator.

$$ \frac{x + 5}{(x + 1)(x + 3)} $$

Write the partial fractions.

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

Multiply by the Lowest Common Denominator (LCD) which will always be the denominator of the original fraction.

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

Simplify.

$$ x + 5 = A(x + 1) + B(x + 3) $$

Because we are looking for A and B, we can choose any values of x. It would be convenient if we could choose a value of x that would make A or B disappear. For example, if x = −3, then B will be multiplied by 0 and disappear.

Let \(x = -3\)
$$ \color{blue}{-3} + 5 = A(\color{blue}{-3} + 1) + B(\color{blue}{-3} + 3) $$

$$ 2 = A(-2) $$

$$ -1 = A $$

Let \(x = -1\) to eliminate A
$$ \color{blue}{-1} + 5 = A(\color{blue}{-1} + 1) + B(\color{blue}{-1} + 3) $$

$$ 4 = B(2) $$

$$ 2 = B $$

Write the answer by plugging in A and B.

$$ \frac{x + 5}{(x + 1)(x + 3)} = -\frac{1}{x + 1} + \frac{2}{x + 3} $$

Example 2: Partial Fraction with Repeated Linear Factor

Find the partial fractions for \(\frac{-x^2 + 3x + 4}{x^3 - 4x^2 + 4x}\)

Solution

Factor the denominator.

$$ \frac{-x^2 + 3x + 4}{x(x - 2)(x - 2)} $$

Write as partial fractions. The second time (x − 2) is used, square it.

$$ \frac{-x^2 + 3x + 4}{x(x - 2)(x - 2)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2} $$

Multiply by the LCD (which will be the denominator of the original fraction) and then simplify.

$$ \frac{(-x^2 + 3x + 4)x(x - 2)(x - 2)}{x(x - 2)(x - 2)} = \frac{Ax(x - 2)(x - 2)}{x} + \frac{Bx(x - 2)(x - 2)}{x - 2} + \frac{Cx(x - 2)(x - 2)}{(x - 2)^2} $$

$$ -x^2 + 3x + 4 = A(x - 2)^2 + Bx(x - 2) + Cx $$

Convenient values would be 0 and 2; however, two values will not give three answers. So, another way to solve for A, B, and C is to multiply everything out and create a system of linear equations.

$$ -x^2 + 3x + 4 = A(x^2 - 4x + 4) + B(x^2 - 2x) + Cx $$

$$ -x^2 + 3x + 4 = Ax^2 - 4Ax + 4A + Bx^2 - 2Bx + Cx $$

Collect the right side into groups by powers of x.

$$ -x^2 + 3x + 4 = (Ax^2 + Bx^2) + (-4Ax - 2Bx + Cx) + (4A) $$

Factor out the x values from the groups.

$$ -x^2 + 3x + 4 = (A + B)x^2 + (-4A - 2B + C)x + (4A) $$

Create the system of equations by setting corresponding coefficients equal to each other.

$$ \color{blue}{-1}x^2 + \color{red}{3}x + \color{purple}{4} = \color{blue}{(A + B)}x^2 + \color{red}{(-4A - 2B + C)}x + \color{purple}{(4A)} $$

$$ \left\{\begin{align} \color{blue}{-1} &= \color{blue}{A + B} \\ \color{red}{3} &= \color{red}{-4A - 2B + C} \\ \color{purple}{4} &= \color{purple}{4A} \end{align}\right. $$

Now solve the system. Interchanging the first two equations will set the system up to be similar to row echelon form.

$$ \left\{\begin{align} -4A - 2B + C &= 3 \\ A + B \qquad &= -1 \\ 4A \qquad \qquad &= 4 \end{align}\right. $$

Solve the 3rd equation for A

$$ 4A = 4 $$

$$ A = 1 $$

Back substitute this into the 2nd equation to find B.

$$ A + B = -1 $$

$$ 1 + B = -1 $$

$$ B = -2 $$

Back substitute both A and B into the 1st equation to find C.

$$ -4A - 2B + C = 3 $$

$$ -4(1) - 2(-2) + C = 3 $$

$$ C = 3 $$

Now write the answer by filling in the partial fractions.

$$ \frac{-x^2 + 3x + 4}{x(x - 2)(x - 2)} = \frac{1}{x} + \frac{-2}{x - 2} + \frac{3}{(x - 2)^2} $$

Example 3: Partial Fraction with Quadratic Factor

Find the partial fractions for \(\frac{5x^2 + 3x + 4}{x^3 + x^2 + 2x + 2}\).

Solution

Start by factoring the denominator. Since there are 4 terms try factoring by grouping.

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

Write the partial fractions. Remember linear factors just have A and quadratic factors have \(Ax + B\). Choose different letters for each fraction.

$$ \frac{5x^2 + 3x + 4}{(x + 1)(x^2 + 2)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 2} $$

Multiply by the LCD (which is the denominator of the original fraction) and simplify.

$$ \frac{(5x^2 + 3x + 4)(x + 1)(x^2 + 2)}{(x + 1)(x^2 + 2)} = \frac{A(x + 1)(x^2 + 2)}{x + 1} + \frac{(Bx + C)(x + 1)(x^2 + 2)}{x^2 + 2} $$

$$ 5x^2 + 3x + 4 = A(x^2 + 2) + (Bx + C)(x + 1) $$

Special values of x will not help us here, so multiply everything out and create a system of equations.

$$ 5x^2 + 3x + 4 = Ax^2 + 2A + Bx^2 + Bx + Cx + C $$

Collect like term with powers of x.

$$ 5x^2 + 3x + 4 = (A + B)x^2 + (B + C)x + (2A + C) $$

Create equations with the corresponding coefficients of x.

$$ \color{blue}{5}x^2 + \color{red}{3}x + \color{green}{4} = \color{blue}{(A + B)}x^2 + \color{red}{(B + C)}x + \color{green}{(2A + C)} $$

$$ \left\{\begin{align} \color{blue}{5} &= \color{blue}{A + B} \\ \color{red}{3} &= \quad \color{red}{B + C} \\ \color{green}{4} &= \color{green}{2A \quad + C} \end{align}\right. $$

Solve the system of equations using Gaussian Elimination. Add −2 times the 1st equation to the 3rd equation and replace the 3rd equation to eliminate the 2A.

$$ \left\{\begin{align} \color{blue}{-2}(5) &= \color{blue}{-2}(A + B) \\ 3 &= \quad B + C \\ 4 &= 2A \quad + C \end{align}\right. $$

$$ \left\{\begin{align} 5 &= A + B \\ 3 &= \quad B + C \\ \color{blue}{-6} &= \color{blue}{-2B + C} \end{align}\right. $$

Eliminate the −2B by adding 2 time the 2nd equation to the 3rd equation and replacing the 3rd equation.

$$ \left\{\begin{align} 5 &= A + B \\ \color{blue}{2}(3) &= \color{blue}{2}(B + C) \\ -6 &= -2B + C \end{align}\right. $$

$$ \left\{\begin{align} 5 &= A + B \\ 3 &= \quad B + C \\ \color{blue}{0} &= \color{blue}{\qquad 3C} \end{align}\right. $$

$$ \left\{\begin{align} 5 &= A + B \\ 3 &= \quad B + C \\ \color{blue}{0} &= \color{blue}{\qquad C} \end{align}\right. $$

The 3rd equation says that \(C = 0\). Back substitute to find B.

$$ 3 = B + C $$

$$ 3 = B + 0 $$

$$ 3 = B $$

Back substitute to find A.

$$ 5 = A + B $$

$$ 5 = A + 3 $$

$$ 2 = A $$

Write the answer by filling the partial fractions.

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

Example 4: Partial Fraction with Repeated Quadratic Factor

Find the partial fractions for \(\frac{2x^3 + 8x}{\left(x^2 + 2\right)^2}\).

Solution

Start by factoring the denominator. Notice, however, that it is already factored.

Next, write the partial fractions. Remember linear factors just have A and quadratic factors have \(Ax + B\). Choose different letters for each fraction.

$$ \frac{2x^3 + 8x}{\left(x^2 + 2\right)^2} = \frac{Ax + B}{x^2 + 2} + \frac{Cx + D}{\left(x^2 + 2\right)^2} $$

Multiply by the LCD (which is the denominator of the original fraction) and simplify.

$$ \frac{(2x^3 + 8x)(x^2 + 2)^2}{\left(x^2 + 2\right)^2} = \frac{(Ax + B)(x^2 + 2)^2}{x^2 + 2} + \frac{(Cx + D)(x^2 + 2)^2}{\left(x^2 + 2\right)^2} $$

$$ 2x^3 + 8x = (Ax + B)(x^2 + 2) + (Cx + D) $$

Special values of x will not help us here, so multiply everything out and create a system of equations.

$$ 2x^3 + 8x = Ax^3 + 2Ax + Bx^2 + 2B + Cx + D $$

Collect like term with powers of x.

$$ 2x^3 + 8x = Ax^3 + Bx^2 + (2A + C)x + (2B + D) $$

Create equations with the corresponding coefficients of x.

$$ \color{blue}{2}x^3 + \color{purple}{0}x^2 + \color{red}{8}x + \color{green}{0} = \color{blue}{A}x^3 + \color{purple}{B}x^2 + \color{red}{(2A + C)}x + \color{green}{(2B + D)} $$

$$ \left\{\begin{align} \color{blue}{2} &= \color{blue}{A} \\ \color{purple}{0} &= \color{purple}{\quad B} \\ \color{red}{8} &= \color{red}{2A \quad + C} \\ \color{green}{0} &= \color{green}{\quad 2B \quad + D} \end{align}\right. $$

Usually, this would be solved using Gaussian Elimination, but this can be solved by simple substitution. Start by substituting \(A = 2\) into the 3rd equation to find C.

$$ 8 = 2A + C $$

$$ 8 = 2(2) + C $$

$$ 4 = C $$

Now substitute \(B = 0\) into the 4th equation to find D.

$$ 0 = 2B + D $$

$$ 0 = 2(0) + D $$

$$ 0 = D $$

Write the answer by filling the partial fractions.

$$ \frac{2x^3 + 8x}{\left(x^2 + 2\right)^2} = \frac{2x}{x^2 + 2} + \frac{4x}{\left(x^2 + 2\right)^2} $$