Partial Fraction Decomposition with Linear Factors

1. What Are Partial Fractions and Why Do We Need Them?

In algebra, you often combine fractions by finding a common denominator. For example:

Partial fraction decomposition is the reverse of this process. We start with the complicated fraction, $\frac{3x-1}{x^2-2x-3}$, and figure out the simpler fractions, $\frac{2}{x-3}$ and $\frac{1}{x+1}$, that add up to it.

As mentioned, the primary reason you'll do this in the IB curriculum is for calculus. Trying to integrate $\frac{3x-1}{x^2-2x-3}$ directly is very difficult. However, integrating the sum $\frac{2}{x-3}+\frac{1}{x+1}$ is straightforward, as it's just the sum of two simple integrals.

The video below gives a quick overview of this idea and its importance in the AA HL course.

A crucial first step is to check if the fraction is proper or improper. A rational expression is a proper fraction if the degree of the numerator's polynomial is less than the degree of the denominator's polynomial.

• Proper: $\frac{2x+1}{x^2+x-2}$ (degree 1 over degree 2)
• Improper: $\frac{3x^2+1}{x^2+x-2}$ (degree 2 over degree 2)
• Improper: $\frac{x^3}{x^2+x-2}$ (degree 3 over degree 2)

The techniques in this lesson apply directly to proper fractions, which is the focus of your syllabus for this topic.

2. Case 1: Distinct Linear Factors

This is the most straightforward case. If the denominator of a proper rational fraction can be factored into distinct linear terms, like $(x-a)(x-b)(x-c)$, we can decompose it.

For a fraction like $\frac{P(x)}{(x-a)(x-b)}$, the decomposition will have the form:

Our job is to find the values of the constants $A$ and $B$. The general method is:

1. Write the equation with the unknown constants $A$, $B$, etc.
2. Multiply both sides by the original denominator to clear all fractions. This creates an identity that is true for all values of $x$.
3. Solve for the constants.

The most efficient way to solve for the constants in this case is the substitution method (sometimes called the Heaviside cover-up method). By substituting the roots of the denominator (e.g., $x=a$ and $x=b$) into the identity, we can make terms disappear and solve for the constants one by one.

The following video provides an excellent, step-by-step demonstration of this method.

An alternative method is equating coefficients. After clearing the denominator, you expand the right side, collect like terms (e.g., all $x^2$ terms, all $x$ terms, all constant terms), and equate their coefficients with the corresponding coefficients on the left side. This creates a system of linear equations to solve for $A$ and $B$. While always valid, it can be slower than substitution for this case.

3. Case 2: Repeated Linear Factors

What happens if the denominator has a factor raised to a power, like $(x-a{)}^2$ or $(x-b{)}^3$? This is a repeated linear factor.

If a proper fraction has a denominator with a factor $(x-a{)}^n$, the partial fraction decomposition must include a term for every power of that factor from 1 up to $n$.

For a denominator like $(x-a{)}^2(x-b)$, the form of the decomposition is:

A common mistake is to forget the $\frac{A}{x-a}$ term. You must include it!

To solve for the constants $A$, $B$, and $C$, the most effective strategy is a hybrid approach:

1. Use the substitution method for any non-repeated factors (like substituting $x=b$ to find $C$) and for the highest power of the repeated factor (substituting $x=a$ to find $B$).
2. You will still have at least one unknown constant left ($A$ in this case). To find it, you can either:
   • Substitute another convenient value for $x$ (like $x=0$).
   • Use the equating coefficients method for a single coefficient (e.g., the coefficient of the highest power of $x$).

This video from Maths Genie demonstrates this hybrid method perfectly.

Sometimes, one of the coefficients in the decomposition might turn out to be zero. This is perfectly fine and happens in exam questions. It just means that particular partial fraction term was not part of the sum. The following clip shows just such an example.

4. Putting It All Together: An IB-Style Problem

Let's walk through the structure of a typical IB question. You will need to show your work clearly to get full method marks.

Problem: Express $\frac{x^2-3x-2}{(x+3)(x-1{)}^2}$ in the form $\frac{A}{x+3}+\frac{B}{x-1}+\frac{C}{(x-1{)}^2}$.

Step 1: Set up the identity.
First, write the identity by multiplying by the denominator:
$x^2-3x-2=A(x-1{)}^2+B(x+3)(x-1)+C(x+3)$
This line is crucial for method marks.

Step 2: Use substitution for the 'easy' constants.

• To find $A$, let $x=-3$. This makes the $B$ and $C$ terms zero.
  $(-3{)}^2-3(-3)-2=A(-3-1{)}^2+0+0$
  $9+9-2=A(-4{)}^2$
  $16=16A⟹A=1$
• To find $C$, let $x=1$. This makes the $A$ and $B$ terms zero.
  $(1{)}^2-3(1)-2=0+0+C(1+3)$
  $1-3-2=4C$
  $-4=4C⟹C=-1$

Step 3: Find the remaining constant.
We still need to find $B$. We can use any convenient method. Let's substitute $x=0$, using our known values of $A=1$ and $C=-1$.
$(0{)}^2-3(0)-2=A(0-1{)}^2+B(0+3)(0-1)+C(0+3)$
$-2=A(1)+B(-3)+C(3)$
$-2=(1)(1)-3B+(-1)(3)$
$-2=1-3B-3$
$-2=-2-3B⟹0=-3B⟹B=0$

Step 4: Write the final answer.
Always finish by writing out the final decomposition clearly.
$\frac{x^2-3x-2}{(x+3)(x-1{)}^2}=\frac{1}{x+3}-\frac{1}{(x-1{)}^2}$

This exact problem is solved in the video we've been using. You can watch the solution to confirm your understanding and see the steps laid out.

Conclusion

Today we have learned a crucial algebraic technique for rewriting complex rational functions as sums of simpler ones. This skill is not an end in itself but a powerful tool you will need for integration in calculus.

Key Takeaways:

• Partial fraction decomposition only applies directly to proper fractions, where the numerator's degree is less than the denominator's.
• The form of the decomposition depends on the denominator's factors. For distinct linear factors like $(x-a)(x-b)$, the form is $\frac{A}{x-a}+\frac{B}{x-b}$.
• For repeated linear factors like $(x-a{)}^2$, you must include a term for each power: $\frac{A}{x-a}+\frac{B}{(x-a{)}^2}$.
• A hybrid strategy of substitution (for easy constants) followed by substituting a convenient value (like $x=0$) or equating coefficients is often the most efficient way to solve for the unknown numerators.

In our next lesson, we will shift gears slightly to focus on another key aspect of IB-style communication: rounding numerical answers to an appropriate degree of accuracy and understanding the difference between exact answers and approximations. This is a fundamental skill for presenting your work correctly on Papers 1 and 2.