MTH61: PARTIAL FRACTIONS

Partial Fractions

Partial fractions are the fractions used for the decomposition of a rational expression. When an algebraic expression is split into a sum of two or more rational expressions, then each part is called a partial fraction. Hence, basically, it is the reverse of the addition of rational expressions. Similar to fractions, a partial fraction will have a numerator and denominator, where the denominator represents the decomposed part of a rational function.

In mathematics, we can see many complex rational expressions. If we try to solve the problems in a complex form, it will take a lot of time to find the solution.

To avoid this complexity, we have to continue the problem by reducing the complex form of the rational expression into the simpler form. Partial fraction decomposition is one of the methods, which is used to decompose rational expressions into simpler partial fractions. This process is more useful in the integration process. In this article, you will learn the definition of the partial fraction, partial fraction decomposition, partial fractions of an improper fraction with solved examples in detail.

A way of “breaking apart” fractions with polynomials in them.

What are Partial Fractions?

We can do this directly:

That is what we are going to discover:

How to find the “parts” that make the single fraction
(the “partial fractions“).

Why Do We Want Them?

First of all … why do we want them?

Because the partial fractions are each simpler.

This can help solve the more complicated fraction. For example it is very useful in Integral Calculus.

Partial Fraction Decomposition

So let me show you how to do it.

The method is called “Partial Fraction Decomposition”, and goes like this:

Step 1: Factor the bottom

Step 2: Write one partial fraction for each of those factors

Step 3: Multiply through by the bottom so we no longer have fractions

Step 4: Now find the constants A₁ and A₂

Substituting the roots, or “zeros”, of (x−2)(x+1) can help:

And we have our answer:

That was easy! … almost too easy …

… because it can be a lot harder!

Now we go into detail on each step.

Proper Rational Expressions

Firstly, this only works for Proper Rational Expressions, where the degree of the top is less than the bottom.

• Proper: the degree of the top is less than the degree of the bottom.
  

  Proper:

  degree of top is 1 degree of bottom is 3

• Improper: the degree of the top is greater than, or equal to, the degree of the bottom.
  

  Improper:

  degree of top is 2 degree of bottom is 1

If your expression is Improper, then do polynomial long division first.

Factoring the Bottom

It is up to you to factor the bottom polynomial. See Factoring in Algebra.

But don’t factor them into complex numbers … you may need to stop some factors at quadratic (called irreducible quadratics because any further factoring leads to complex numbers):

So the factors could be a combination of

• linear factors
• irreducible quadratic factors

When you have a quadratic factor you need to include this partial fraction:

B₁x + C₁(Your Quadratic)

Factors with Exponents

Sometimes you may get a factor with an exponent, like (x−2)³ …

Like this:

The same thing can also happen to quadratics:

Sometimes Using Roots Does Not Solve It

Even after using the roots (zeros) of the bottom you can end up with unknown constants.

So the next thing to do is:

Oh my gosh! That is a lot to handle! So, on with an example to help you understand:

A Big Example Bringing It All Together

Here is a nice big example for you!

x²+15(x+3)² (x²+3) 

x²+15(x+3)²(x²+3)  =  A₁x+3 + A₂(x+3)² + Bx + Cx²+3

Now multiply through by (x+3)²(x²+3):

x²+15 = (x+3)(x²+3)A₁ + (x²+3)A₂ + (x+3)²(Bx + C)

There is a zero at x = −3 (because x+3=0), so let us try that:

(−3)²+15 = 0 + ((−3)²+3)A₂ + 0

And simplify it to:

24 = 12A₂

so A₂=2

Let us replace A₂ with 2:

x²+15 = (x+3)(x²+3)A₁ + 2x²+6 + (x+3)²(Bx + C)

Now expand the whole thing:

x²+15 = (x³+3x+3x²+9)A₁ + 2x²+6 + (x³+6x²+9x)B + (x²+6x+9)C

Gather powers of x together:

x²+15 = x³(A₁+B)+x²(3A₁+6B+C+2)+x(3A₁+9B+6C)+(9A₁+6+9C)

Separate the powers and write as a Systems of Linear Equations:

Simplify, and arrange neatly:

Now solve.

You can choose your own way to solve this … I decided to subtract the 4th equation from the 2nd to begin with:

Then subtract 2 times the 1st equation from the 2nd:

Now I know that B = −(1/2).

We are getting somewhere!

And from the 1st equation I can figure that A₁ = +(1/2).

And from the 4th equation I can figure that C = +(1/2).

Final Result:

And we can now write our partial fractions:

x²+15(x+3)²(x²+3)  =  12(x+3) + 2(x+3)² + −x + 12(x²+3)

Phew! Lots of work. But it can be done.

(Side note: It took me nearly
an hour to do this, because I had to
fix 2 silly mistakes along the way!)

Summary

• Start with a Proper Rational Expressions (if not, do division first)
• Factor the bottom into:
  • linear factors
  • or “irreducible” quadratic factors
• Write out a partial fraction for each factor (and every exponent of each)
• Multiply the whole equation by the bottom
• Solve for the coefficients by
  • substituting zeros of the bottom
  • making a system of linear equations (of each power) and solving
• Write out your answer!

REPEATED QUADRATIC ROOTS

IMPROPER PARTIAL FRACTIONS