Site Loader

What if you had a way to expand certain large math expressions into smaller pieces? This would make some calculus integrals easier to solve.

In this lesson, we explore such a method: partial fraction decomposition.

Best services for writing your paper according to Trustpilot

Premium Partner
From $18.00 per page
4,8 / 5
Writers Experience
Recommended Service
From $13.90 per page
4,6 / 5
Writers Experience
From $20.00 per page
4,5 / 5
Writers Experience
* All Partners were chosen among 50+ writing services by our Customer Satisfaction Team

Partial Fraction Decomposition

Building a house of cards is getting complicated results from something simpler, which is the opposite of partial fraction decomposition (PFD), where you get simpler results from something complicated. But just like a house of cards, partial fraction decomposition needs structure and rules.

Partial Fraction Decomposition is just like a house of cards.
A House of Cards

I’m sure you’ve seen fractions before, but what about polynomial fractions? These have polynomials for both the numerator and denominator.

For example:

polynomial fraction

Both polynomials are in standard form: the terms are ordered from the highest exponent to the lowest. Here’s another key idea: the order of a polynomial, which is the highest numbered exponent. Our standard form numerator polynomial has a first term of 2x1. The numerator order is 1.

The order of the denominator polynomial is 2, since x2 is the highest exponent. The first rule of PFD is the denominator order must be greater than the numerator order.

Setting up the Work

Please don’t panic with the following equation. This is just four cases of PFD combined into a single expression.

four cases of PFD

The (x + 2) factor is linear, while (x2 – 2x + 2) is quadratic. The (x – 1)2 is a linear factor raised to a power, while (x2 + 4)2 is a quadratic factor raised to a power.

The arrows point to the PFD.

Linear Factors

Linear factors have x raised to the first power:

examples of linear factors

Here is a fraction with two linear factors:

polynomial fraction with 2 linear factors

Remember the importance of structure in the house of cards? Our structure here is:

structure of the decomposition for linear factors

Like a house of cards, isn’t it? Okay, maybe not, but we have a rule for linear factors: for each linear factor, write a new fraction of a capital letter over the linear factor. Then add these new fractions together.In this example, the right-hand side of the equation becomes the sum of two new fractions.

This is how we find a common denominator:

common denominator for linear factors

Since the left-hand side must equal the right-hand side, the numerators must equal each other, and we can simplify our equation:

equating numerators

Any value may be substituted for x, although some values will simplify better than others. Letting x = 1 wipes out the A since 1 – 1 = 0, and with some algebra gives us B = 1. Letting x = -2, we can wipe out the B with -2 + 2 = 0, gives us A = 3.

We now have our decomposition:

the resulting decomposition for the linear factors example

Quadratic factors have x raised to the second power:

examples of quadratic factors

Let’s decompose this equation:

example with a quadratic factor

Quadratic Factors

The first factor in the denominator is quadratic. Here’s the rule for quadratic factors: write Ax + B in the numerator of a new fraction and the quadratic factor as the denominator.The second factor in the denominator is linear. In our new fraction, it gets a single letter over the linear factor.

Our structure here is:

structure for PFD when dealing with a quadratic factor

As with the linear equations, we use common denominator and then equate numerators:

equating numerators for the quadratic factor example

Substituting x = -2 (to wipe out A and B), gives C = 3.

Expanding the right-hand side multiplications and grouping terms:

expansion of the RHS and grouping terms

What if we compare the left-hand side with the right-hand side? To be clear, we are looking at:

the LHS and RHS of the quadratic factor example

On the right-hand side, look at what is multiplying the x2 – it is A + C. Now we look at the left-hand side and see the x 2 term being multiplied by 5. Our conclusion? A + C = 5.The 8 all by itself on the left-hand side compares to what on the right-hand side? Here’s a hint: on the right-hand side, find the terms not being multiplied by an x or an x2. You are correct! 2B + 2C = 8.

Having made these comparisons, we can write:

conclusions after comparing RHS with LHS

Using C = 3 gives us A = 2 and B = 1. The result is:

final result for the quadratic factor example

Linear Factors Raised to a Power

What about a linear factor raised to the second power?

a PDF with a linear factor raised to a second power

The structure here is:

We know about linear factors. The new fraction has a single letter over the linear factor. But here we have (x – 1)2.

The power of 2 requires not one, but two new fractions. So, if it were (x – 1)3, we would have three fractions before the (x + 2) fraction (which now has a numerator of D) and they would have as denominators (x – 1), (x – 1)2, and (x -1)3, with corresponding numerators of A, B, and C. The rule for factors raised to the power n is: the right-hand side is the sum of n new fractions where the numerator of each fraction is the typical numerator for this type of factor in PFD, but the denominators are the factor raised to the first power, the second power, and all the way up to the power n.Getting back to our example, the common denominator on the right-hand side is (x – 1)2. Writing each right-hand side fraction over this common denominator and adding gives us:

the RHS of the example of linear factor raised to a second power

Since the denominators of the left-hand and the right-hand sides are the same, we equate the numerators:

Letting x = 1, which wipes out both A and C, gives us B = 2. For x = -2 (to wipe out B), we get C = 3. Substituting x = 0 (to leave 5 by itself on the left-hand side) and using the values for B and C gives us A = 1.The final answer is:

final result of the PDF example of a linear factor raised to a second power

Quadratic Factors Raised to a Power

Let’s do one more example. Our last case considers a quadratic factor raised to a second power:

You already know how to do this. It’s the same as having a linear factor raised to the second power, except the numerators have a letter times x + a letter. Here’s our structure:

the structure of a PFD with a quadratic factor raised to a second power

As before, we equate the numerators:

equating numerators in the PFD example of a quadratic factor raised to a second power

Expanding the right-hand side and grouping terms gives us:

Comparing each right-hand side group to the left-hand side gives A = 1 (1x3 = Ax3) and B = -1 (-1x2 = Bx2). From 4A + C = 5 (5x = (4A + C)x) and knowing A + 1, we get C = 1. From 4B + D = -6 and knowing B = -1, we get D = -2.Our resulting decomposition is:

final result of the PFD example of a quadratic factor raised to a second power


In a house of cards, structure is very important.

The bottom must be greater than the top. This will help you remember the first rule of partial fraction decomposition: the denominator order must be greater than the numerator order. What if that’s not the case? The answer is polynomial long division. For example, here’s a case where the numerator and denominator are both of order 2. Warning: the first rule of partial fraction decomposition is not satisfied.

Do a polynomial long division:

polynomial long division

In polynomial long division, we do the subtraction step by changing the signs of 4x2 + 4x – 8 to -4x2 – 4x + 8 and adding this to 4x2 – 1. The remainder is -4x + 7.

The constant 4 plus the remainder written over the divisor x2 + x – 2 now has a polynomial fraction where partial fraction decomposition is allowed:

result of the polynomial long division and PFD

Another useful guide predicts the number of unknown letters: the number of unknown letters is the same as the order of the denominator. For example, decomposing an order 4 denominator requires 4 unknown letters.Enough partial fraction decompositions. Now we can return to another challenging pastime – adding floors to a house of cards .

. .

Lesson Summary

Partial fraction decomposition allows complicated polynomial fractions to be written as the sum of simpler fractions.

In this lesson, we used examples to showcase the rules for four cases of partial fraction decompositions:

  1. Linear factors
  2. Quadratic factors
  3. Linear factors raised to a power
  4. Quadratic factors raised to a power

Post Author: admin


I'm Eric!

Would you like to get a custom essay? How about receiving a customized one?

Check it out