Precalculus by Richard Wright

But I tell you, love your enemies and pray for those who persecute you. Matthew 5:44 NIV

Summary: In this section, you will:

- Decompose rational expressions into partial fractions with:
- Distinct linear factors
- Repeated linear factors
- Distinct quadratic factors
- Repeated quadratic factors

SDA NAD Content Standards (2018): PC.6.1, PC.6.5

Partial fractions takes a fraction and splits into two or more fractions that add together to make the original fraction. This is sort of like \(\frac{5}{12} = \frac{1}{6} + \frac{1}{4}\). This is very useful for analyzing rational functions when using calculus.

Finding partial fractions can use systems of linear equations. Hence partial fractions are in the chapter on systems of equations.

- Factor the denominator.
- Use each factor as the denominator of separate fractions. If a factor is repeated, use it the same number of times as it is repeated, but square the second time, cube the third, etc.
- For each linear factor of the denominator use the form \(\frac{A}{px + q} + \frac{B}{\left(px + q\right)^2} + \cdots\)
- For each quadratic factor of the denominator use the form \(\frac{Ax + B}{ax^2 + bx +c} + \frac{Cx + D}{\left(ax^2 + bx + c\right)^2} + \cdots\)
- Solve for
*A*,*B*,*C*, etc. - Multiply by the LCD
- Choose convenient values of
*x*to find*A*,*B*,*C*, etc. - Or create a system of linear equations based on the coefficients of
*x*.

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

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*.

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

Find the partial fractions for \(\frac{x - 16}{x^2 - 2x - 8}\)

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

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

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.

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

Find the partial fractions for \(\frac{3x + 13}{x^2 + 8x + 16}\)

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

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

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 2*A*.

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

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

Eliminate the −2*B* 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.

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

Find the partial fractions for \(\frac{3x^2 - 3x - 7}{x^3 - 2x^2 - 5x + 10}\)

\(\frac{1}{x-2} + \frac{2x+1}{(x^2-5)}\)

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

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.

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

Find the partial fractions for \(\frac{x^3 - 2x + 1}{(x^2 - 3)^2}\)

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

- Factor the denominator.
- Use each factor as the denominator of separate fractions. If a factor is repeated, use it the same number of times as it is repeated, but square the second time, cube the third, etc.
- For each linear factor of the denominator use the form \(\frac{A}{px + q} + \frac{B}{\left(px + q\right)^2} + \cdots\)
- For each quadratic factor of the denominator use the form \(\frac{Ax + B}{ax^2 + bx +c} + \frac{Cx + D}{\left(ax^2 + bx + c\right)^2} + \cdots\)
- Solve for
*A*,*B*,*C*, etc. - Multiply by the LCD
- Choose convenient values of
*x*to find*A*,*B*,*C*, etc. - Or create a system of linear equations based on the coefficients of
*x*.

Helpful videos about this lesson.

- What are partial fractions?
- \(\frac{x}{x^2 + 3x + 2}\)
- \(\frac{4}{x^2 - 6x + 9}\)
- \(\frac{x^2 + 4}{x^3 + x^2 + 5x + 5}\)
- \(\frac{x + 13}{x^2 - 4x - 21}\)
- \(\frac{8x + 20}{x^2 + 6x + 5}\)
- \(\frac{9x + 1}{2x^2 + x}\)
- \(\frac{2x + 5}{x^2 + 2x + 1}\)
- \(\frac{-7x + 33}{x^2 - 10x + 25}\)
- \(\frac{5x^2 - 21x + 32}{x^3 - 8x^2 + 16x}\)
- \(\frac{4x^2 + 3x + 4}{x^3 + x}\)
- \(\frac{3x^2 + x - 6}{x^3 + x^2 - 3x - 3}\)
- \(\frac{3x^2 - x + 1}{(x - 2)(x^2 + 2x + 3)}\)
- \(\frac{4x^3 - 20x + 1}{(x^2 - 5)^2}\)
- \(\frac{2x^4 + 3x^3 + 17x^2 + 11x + 32}{x(x^2 + 4)^2}\)
- (8-03) Solve by Gaussian Elimination \(\left\{\begin{align} x + y + z &= 5 \\ 2y - z &= -2 \\ -3y + 2z &= 5 \end{align}\right.\)
- (8-03) Is this in Row Echelon Form? If not, why? \(\left\{\begin{align} x + 3y - z &= 7 \\ 2y + 7z &= 3 \\ 0 &= 0 \end{align}\right.\)
- (8-02) Is (−1, 2) a solution to \(\left\{\begin{align} 2x + 3y &= 4 \\ -x + 2y &= 5 \end{align}\right.\)?
- (8-01) Solve by graphing \(\left\{\begin{align} y &= 2x + 1 \\ y &= -3x + 6 \end{align}\right.\)
- (7-09) Classify the conic \(r = \frac{2}{2 + 3 \sin \theta}\)

Write the partial fractions, but do not solve for *A*, *B*, *C*, etc.

Find the partial fractions.

Mixed Review

- Splitting a rational expression into a sum of smaller rational expressions each with a single factor in the denominator.
- \(\frac{A}{x + 2} + \frac{B}{x + 1}\)
- \(\frac{A}{x - 3} + \frac{B}{(x - 3)^2}\)
- \(\frac{A}{x + 1} + \frac{Bx + C}{x^2 + 5}\)
- \(\frac{2}{x - 7} + \frac{-1}{x + 3}\)
- \(\frac{3}{x + 1} + \frac{5}{x + 5}\)
- \(\frac{7}{2x + 1} + \frac{1}{x}\)
- \(\frac{2}{x + 1} + \frac{3}{(x + 1)^2}\)
- \(\frac{-7}{x - 5} + \frac{-2}{(x - 5)^2}\)
- \(\frac{3}{x - 4} + \frac{7}{(x - 4)^2} + \frac{2}{x}\)
- \(\frac{4}{x} + \frac{3}{x^2 + 1}\)
- \(\frac{2}{x + 1} + \frac{x}{x^2 - 3}\)
- \(\frac{1}{x - 2} + \frac{2x + 1}{x^2 + 2x + 3}\)
- \(\frac{4x}{x^2 - 5} + \frac{1}{(x^2 - 5)^2}\)
- \(\frac{2}{x} + \frac{3}{x^2 + 4} + \frac{x - 1}{(x^2 + 4)^2}\)
- (0, 1, 4)
- No, the leading coefficient of the 2nd equation is not a 1.
- Yes
- (1, 3)
- Hyperbola