# Introduction to Partial Fraction Expansion/Decomposition (PFE)

• PRACTICE (online exercises and printable worksheets)

You already know how to add fractions.
For example: $$\frac{1}{x-1} + \frac{3}{2x+1} \ =\ \frac{(2x+1) + 3(x-1)}{(x-1)(2x+1)} \ =\ \frac{5x-2}{2x^2-x-1}$$ The process of going backwards:

from   $\displaystyle \frac{5x-2}{2x^2-x-1}$         back to   $\displaystyle\frac 1{x-1} + \frac{3}{2x+1}$

is called either Partial Fraction Expansion (PFE) or Partial Fraction Decomposition.

Note that adding fractions takes you from two or more fractions to a single, more complicated fraction.
Partial Fraction Expansion, on the other hand, takes you from a single (complicated) fraction to two or more simpler pieces.

This section introduces partial fraction expansion, with review of needed concepts and a simple example.
There is more detail in subsequent sections.

## Which name: ‘Partial Fraction Expansion’ or ‘Partial Fraction Decomposition’?

Going from (say) $\,\color{green}{\displaystyle \frac{5x-2}{2x^2-x-1}}\,$ to the new name $\,\color{red}{\displaystyle\frac 1{x-1} + \frac{3}{2x+1}}\,$:

• ‘expands’ the starting fraction into simpler pieces (the ‘partial fractions’)
• ‘decomposes’ (breaks down) the starting fraction into simpler pieces (the ‘partial fractions’)
So, both names are appropriate.
Since ‘expansion’ seems a bit more optimistic than ‘decomposition’,
this author prefers the name Partial Fraction Expansion, and usually abbreviates it as PFE.

## What fractions can PFE be used for?

Recall that a rational function is a ratio of polynomials (with a nonzero denominator).
That is, a rational function is a fraction, with any polynomial in the numerator, and a nonzero polynomial in the denominator.

Theoretically, PFE can be used on any rational function.
(However, the second step below limits its usefulness, in practice.)

• The first step of PFE is to check that the degree of the numerator is strictly less than the degree of the denominator.

If not, then you'll do a long division first, to write the starting fraction $\,\displaystyle\frac{N(x)}{D(x)}\,$ as: $$\frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$$ By the Division Algorithm, either $\,R(x) = 0\,$ (in which case you won't need PFE),
or else the degree of $\,R(x)\,$ is strictly less than the degree of $\,D(x)\,$.

The bulk of the work of PFE is then done with the new fraction $\,\displaystyle\frac{R(x)}{D(x)}\,$.
• The second step of PFE is to completely factor the denominator into linear and irreducible quadratic factors.
So, partial fraction expansion is usually used only when the denominator has a low degree, so it can be easily factored.

## Which direction is easier: Adding Fractions or Partial Fraction Expansion?

Adding fractions is a lot easier than PFE!

Partial Fraction Expansion involves:

• factoring the denominator of the starting fraction
(which may be difficult once you move beyond quadratics)
• solving a system of equations (sometimes)
(the more factors the denominator has, the more complicated things can get)
The example below shows how to go from $\,\displaystyle \frac{5x-2}{2x^2-x-1}\,$ to $\,\displaystyle\frac 1{x-1} + \frac{3}{2x+1}\,$.

Even in this simple case, you'll see that it's a bit of work.

## Uses for Partial Fraction Expansion

PFE is valuable whenever you need to represent a complicated fraction of polynomials as a sum of simpler fractions.
Two areas where PFE is frequently used:

• calculus: integrating rational functions
• engineering: finding inverse Laplace transforms
(Don't worry if you have no idea what these applications are—yet!)

## Review of Concepts and Terminology Needed for PFE

Partial Fraction Expansion draws on lots of beautiful mathematical theory!

• QUADRATIC EXPRESSIONS:
Let $\,a\,$, $\,b\,$, and $\,c\,$ be real numbers with $\,a \ne 0\,$.
Then, $\,ax^2 + bx + c\,$ is a quadratic expression.
(Here, the variable is $\,x\,$; other variables can be used.)
• DISCRIMINANT:
The discriminant of the quadratic expression $\,ax^2 + bx + c\,$ is $\,b^2 - 4ac\,$.
• IRREDUCIBLE:
A quadratic expression is irreducible if and only if its discriminant is negative.
• Irreducible quadratics cannot be factored using real numbers.
• EXAMPLES of irreducible quadratics:
$\,x^2 + 1\,$ is an irreducible quadratic:
• write $\,x^2 + 1\,$ as $\,1x^2 + 0x + 1\,$ to see that $\,a = 1\,$, $\,b = 0\,$ and $\,c = 1\,$
• the discriminant of $\,x^2 + 1\,$ is $\,b^2 - 4ac = 0^2 - 4(1)(1) = -4\,$, which is negative
• $\,x^2 + 1\,$ cannot be factored using real numbers
$\,x^2 + x + 2\,$ is an irreducible quadratic, since $\,b^2 - 4ac = 1^2 - 4(1)(2) = -7\,$ is negative.
Therefore, $\,x^2 + x + 2\,$ can't be factored into linear factors using only real numbers.
• FACTORS:
In a product (things multiplied), the things being multiplied are called factors.
• LINEAR FACTORS:
A linear factor is a factor of the form $\,ax + b\,$, where $\,a\,$ and $\,b\,$ are real numbers with $\,a \ne 0\,$.
For example: $\,x\,$, $\,x - 1\,$, and $\,\sqrt 2x + \pi\,$ are linear factors.
• FACTORIZATION OF POLYNOMIALS WITH REAL NUMBER COEFFICIENTS:
Every polynomial with real number coefficients can be factored into linear factors and irreducible quadratics.
Note: Even though it can be done doesn't mean that it's easy to do!
For an arbitrary cubic polynomial, it can already be difficult.
• ZERO OF A FUNCTION:
A zero of a function is an input, whose corresponding output is zero.
That is:   $\,c\,$ is a zero of $\,f\,$   if and only if   $\,f(c) = 0$
• RELATIONSHIP BETWEEN ZEROS AND FACTORS OF POLYNOMIALS:
Let $\,P\,$ be a polynomial.
The following are equivalent:
• $\,c\,$ is a zero of $\,P\,$
• $\,x - c\,$ is a factor of $\,P(x)\,$
For example, consider $\,P(x) = x^2 + x -2\,$.
The number $\,1\,$ is a zero of $\,P\,$, since $\,P(1) = 1^2 + 1 - 2 = 0\,$. Thus, $\,x-1\,$ is a factor.
The number $\,-2\,$ is a zero of $\,P\,$, since $\,P(-2) = (-2)^2 + (-2) - 2 = 0\,$. Thus, $\,x - (-2) = x + 2\,$ is a factor.
So, $\,P(x) = (x-1)(x + 2)\,$.
If we know the zeros of a polynomial, then we know the (non-constant) factors!
• FACTORING QUADRATICS:
Every quadratic can be readily factored (or determined to be irreducible).
You may be able to factor simple quadratics using standard, well-rehearsed methods.
In a pinch, though, here's a ‘Foolproof Quadratic Factorization Method’:

Let $\,P(x) := ax^2 + bx + c\,$ with $\,a\ne 0\,$.
• Use the quadratic formula to find the zeros: $$\left(ax^2 + bx + c = 0 \ \text{ and }\ a\ne 0\right) \qquad \text{ if and only if } \qquad x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
• Use the zeros to get the factors.
• Be careful—you may need to supply a constant factor yourself.
The only factor the zeros don't give us is a possible constant factor.
• For example, suppose a quadratic polynomial has zeroes $\,1\,$ and $\,-2\,$.
The most we can say is that $\,P(x) = K(x-1)(x+2)\,$ for some nonzero constant $\,K\,$.
If we additionally know that (say) the leading coefficient of the polynomial is $\,5\,$, then $\,P(x) = 5(x-1)(x+2)\,$.
• DISTINCT:
Depending on context, the word distinct is often used in mathematics to mean different; or, it can refer to exactly one.
• A QUADRATIC WITH DISTINCT LINEAR FACTORS:
The quadratic $\,2x^2 - x - 1\,$ can be factored into distinct linear factors: $$\,2x^2 - x - 1 = (2x+1)(x-1)\,$$ Here, the linear factors $\,2x + 1\,$ and $\,x-1\,$ are distinct (different)—they correspond to different zeros.
The linear factor $\,2x+1\,$ is distinct (there is exactly one).
The linear factor $\,x-1\,$ is distinct (there is exactly one).
• MULTIPLICITY:
The quadratic $\,x^2 - 2x + 1 = (x-1)^2\,$ does not have distinct (different) linear factors.
In other words, $\,x-1\,$ is not a distinct linear factor (there is not exactly one).
Here, the factor $\,x - 1\,$ corresponds to the zero $\,x = 1\,$, which has a multiplicity of $\,2\,$ (meaning there are exactly two factors of $\,x-1\,$).
• DEGREE OF A POLYNOMIAL:
The degree of a polynomial is the highest power to which $\,x\,$ is raised.
For example, $\,2x + 3 = 2x^1 + 3\,$ has degree $\,1\,$. All linears factors have degree $\,1\,$.
Also, $\,x^2 + 4x - 5\,$ has degree $\,2\,$. All quadratic factors have degree $\,2\,$.

## A Simple Example: PFE with Distinct Linear Factors

This simple example shows the basic process of Partial Fraction Expansion.
The way to handle situations other than distinct linear factors is discussed in subsequent sections.

Find the partial fraction expansion of $\,\displaystyle\frac{5x-2}{2x^2 - x - 1}\,$
• STEP 1:
Check that the degree of the numerator is strictly less than the degree of the denominator.
The numerator, $\,5x-2\,$, has degree $\,1\,$.
The denominator, $\,2x^2 -x - 1\,$, has degree $\,2\,$.
$\,1\,$ is strictly less than $\,2\,$.
Check!
• STEP 2:
Completely factor the denominator into linear factors and irreducible quadratics.
$2x^2 - x - 1 = (2x+1)(x-1)$
(This denominator has distinct linear factors.)

Foolproof Factorization Method’:
If you don't quickly see the factorization, use the quadratic formula to find that $-\frac 12\,$ and $\,1\,$ are the zeros of $\,2x^2 - x - 1\,$.
So, $\,x+\frac 12\,$ and $\,x-1\,$ are factors.
Supply the constant factor (the leading coefficient) yourself.

Thus, $\,2x^2 - x - 1 = \underbrace{2(x+\frac 12)}^{\text{multiply together}}(x-1) = (2x+1)(x-1)\,$.
• STEP 3:
Different types of factors in the denominator give rise to different term(s) in the partial fraction expansion.
Each distinct linear factor $\,ax+b\,$ gives rise to a term of the form $\,\displaystyle\frac{A}{ax+b}\,$ in the PFE:
• The factor $\,2x+1\,$ gives rise to a term of the form $\,\displaystyle\frac{A}{2x+1}\,$.
• The factor $\,x-1\,$ gives rise to a term of the form $\,\displaystyle\frac{B}{x-1}\,$.
• Together: $$\frac{5x-2}{2x^2 - x - 1} \ =\ \frac{5x-2}{(2x+1)(x-1)} \ =\ \frac{A}{2x+1} + \frac{B}{x-1}$$
• To complete the process, we must find the unknown constants $\,A\,$ and $\,B\,$.
• STEP 4:
Solve for the unknown constants.
 $$\begin{gather} \frac{5x-2}{(2x+1)(x-1)}\cdot \color{red}{(2x+1)(x-1)} = \left[\frac{A}{2x+1} + \frac{B}{x-1}\right]\color{red}{(2x+1)(x-1)}\cr\cr 5x - 2 = A(x-1) + B(2x+1)\ \ \ \ (*) \end{gather}$$ Clear fractions: multiply both sides of the equation by the original denominator (in factored form) cancel When there are only distinct linear factors, you don't need to solve a system of equations.Hooray! Choose $\,x = 1\,$. This makes $\,x-1\,$ equal zero, and hence $\,A\,$ disappears: $$\begin{gather} 5(1) - 2 = A(1-1) + B(2\cdot 1+1)\cr 3 = 3B\cr B = 1 \end{gather}$$ Choose $\,x = -\frac 12\,$. This makes $\,2x+1\,$ equal zero, and hence $\,B\,$ disappears: $$\begin{gather} 5(-{\textstyle\frac 12}) - 2 = A(-{\textstyle\frac12}-1) + B(2\cdot -{\textstyle\frac 12}+1)\cr -\frac 92 = -\frac 32A\cr A = 3 \end{gather}$$ Equation (*) must be true for all real numbers $\,x\,$. Therefore, equation (*) must be true for any particular value of $\,x\,$. Choose values that make one of the unknowns disappear!
• STEP 5:
Report the answer. Spot-check to gain confidence in your result.

Substitute in the values of $\,A\,$ and $\,B\,$ found in the prior step: $$\frac{5x-2}{2x^2 - x - 1} = \frac 3{2x+1} + \frac 1{x-1}$$ A quick check for an easy value of $\,x\,$ doesn't take much time, and catches many mistakes.
When $\,x = 0\,$ is available, it's often easiest to use this value for your spot-check.
Don't drop the question mark over the equals sign until you're sure that both sides are equal! $$\begin{gather} \frac{5\cdot 0-2}{2\cdot 0^2 - 0 - 1} \overset{?}{=} \frac 3{2\cdot 0+1} + \frac 1{0 - 1}\cr\cr \frac{-2}{-1} \overset{?}{=} 3 - 1\cr\cr 2 = 2\cr \text{Check!} \end{gather}$$
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Partial Fraction Expansion: Linear Factors

On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
AVAILABLE MASTERED
IN PROGRESS
 (MAX is 26; there are 26 different problem types.) Want textboxes to type in your answers? Check here: