Write the partial-fraction form
Pieces that hug the curve
The chalk curve is the fraction \(\dfrac{3x - 1}{(x + 1)(x - 3)}\). The dashed vertical lines are at \(x = -1\) and \(x = 3\), where the denominator is 0.
The two faint curves are the simpler fractions \(\dfrac{1}{x + 1}\) and \(\dfrac{2}{x - 3}\).
Drag the orange handle on the x-axis. At each x, the two orange bars are the heights of the faint curves. Together they reach the chalk curve, so the two simpler fractions add up to the given one.
Press the buttons to turn one faint curve orange. Near \(x = -1\), the chalk curve follows \(\dfrac{1}{x + 1}\), and near \(x = 3\) it follows \(\dfrac{2}{x - 3}\), so each dashed line belongs to one simpler fraction.
One piece for each factor
Take a proper fraction, one whose numerator has smaller degree than its denominator. Its denominator is a product of distinct linear factors: each factor has degree 1, such as x − a, and no two factors are the same.
The fraction is a sum of simple pieces, one for each factor, and each piece has a constant on top and one factor below.
Move a slider to change the numerator. The pieces change size and can change sign, but the dashed lines do not move, because they come from the denominator.
Each piece rises or falls without bound at its own dashed line and at no other, so each factor needs its own piece.
Notation
Start with a proper fraction. Its denominator is a product of distinct linear factors, such as \((x - a)(x - b)\).
\[\frac{P(x)}{(x - a)(x - b)}\]Write one term for each factor. The numerator of each term is an unknown constant.
\[\begin{aligned} &\frac{P(x)}{(x - a)(x - b)} \\ &\quad = \frac{\cy{A}}{x - a} + \frac{\cy{B}}{x - b} \end{aligned}\]Each term has a constant on top because each term must itself be proper. A denominator of degree 1 needs a numerator of degree 0.
Three factors give three terms.
\[\begin{aligned} &\frac{P(x)}{(x - a)(x - b)(x - c)} \\ &\quad = \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c} \end{aligned}\]Copy each factor, as it is, into the denominator of its term. The factor \(2x + 1\) gives the term \(\dfrac{A}{2x + 1}\).
This page writes the form only. The next page finds the constants.
When it fails
The form needs distinct factors. Make the two factors equal, so the denominator is \((x - 1)^2\). One term \(\dfrac{A}{x - 1}\) would need \(A(x - 1) = 3x - 1\) for every \(x\). At \(x = 1\) the left side is 0 and the right side is 2, so no constant \(A\) works.
The red curve is \(\dfrac{A}{x - 1}\) with \(A = 3\). It stays below the chalk curve, because \(\dfrac{3x - 1}{(x - 1)^2} = \dfrac{3}{x - 1} + \dfrac{2}{(x - 1)^2}\) and the extra term is positive. A repeated factor needs that second term, and AP questions use distinct factors.
Example 1: Two distinct factors
Write the form of \(\dfrac{6x + 1}{(x - 2)(x + 5)}\).
The numerator has degree 1 and the denominator has degree 2, so the fraction is proper.
The denominator is already factored. The factors \(x - 2\) and \(x + 5\) are distinct, and the dashed lines are at \(x = 2\) and \(x = -5\).
Write one term for each factor, with a constant on top.
\[\begin{aligned} &\frac{6x + 1}{(x - 2)(x + 5)} \\ &\quad = \frac{A}{x - 2} + \frac{B}{x + 5} \end{aligned}\]
Example 2: Read the factor, not the number in it
Write the form of \(\dfrac{4x - 3}{x(x + 3)}\).
The fraction is proper. The factors are \(x\) and \(x + 3\), and they are distinct.
A tempting wrong move: The factor \(x + 3\) contains the number 3, so write \(x - 3\) in the denominator.
\[\frac{4x - 3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x - 3}\]The piece \(\dfrac{B}{x - 3}\) has its dashed line at \(x = 3\), shown in red, where the chalk curve has none. At \(x = -3\), where the chalk curve has its dashed line, the red curve stays finite, so it cannot follow the chalk curve.
Write the factor itself. The factor \(x + 3\) is zero at \(x = -3\), which is where the chalk curve has its dashed line.
\[\frac{4x - 3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3}\]
Practice
Write the form of \(\dfrac{6x - 5}{(3x - 1)(x + 4)}\).
Drag the point to each place where the curve rises or falls without bound. Each place comes from one factor of the denominator, and each factor has one term.
Solution
The fraction is proper. The factors \(3x - 1\) and \(x + 4\) are distinct, and they are zero at \(x = \tfrac13\) and \(x = -4\).
Write one term for each factor. Keep \(3x - 1\) as it is.
\[\begin{aligned} &\frac{6x - 5}{(3x - 1)(x + 4)} \\ &\quad = \frac{A}{3x - 1} + \frac{B}{x + 4} \end{aligned}\]
Questions
The figure shows the curve of a proper fraction with two dashed vertical lines. The faint curves are its two pieces.
Which is the partial-fraction form of this fraction?
- \(\dfrac{A}{x + 2} + \dfrac{B}{x - 4}\)
- \(\dfrac{A}{x - 2} + \dfrac{B}{x + 4}\)
- \(\dfrac{A}{(x + 2)(x - 4)}\)
Answer
\(\dfrac{A}{x + 2} + \dfrac{B}{x - 4}\)
Just right of the dashed line at \(x = 4\), the chalk curve falls steeply. Which piece does it follow?
- The piece with the denominator \(x - 4\)
- The piece with the denominator \(x + 2\)
- Both pieces equally
Answer
The piece with the denominator \(x - 4\)
What is the partial-fraction form of \(\dfrac{x^2 - 3}{x(x + 1)(x - 4)}\)?
- \(\dfrac{A}{x} + \dfrac{B}{x + 1} + \dfrac{C}{x - 4}\)
- \(\dfrac{A}{x} + \dfrac{B}{x - 1} + \dfrac{C}{x + 4}\)
- \(\dfrac{A}{x} + \dfrac{B}{x + 1}\)
- \(\dfrac{A}{x(x + 1)(x - 4)}\)
Answer
\(\dfrac{A}{x} + \dfrac{B}{x + 1} + \dfrac{C}{x - 4}\)
Summary
You write one term for each distinct linear factor of the denominator, with a constant on top.
This works because each term is a proper fraction on its own, and each piece rises or falls without bound at its own dashed line.
Use it after you divide and factor, when the fraction is proper.
This is one of four skills in topic 6.12 of Which & Why: AP Calculus.