 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.

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

For example: 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. 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: Here is a fraction with two linear factors: Remember the importance of structure in the house of cards? Our structure here is: 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: Since the left-hand side must equal the right-hand side, the numerators must equal each other, and we can simplify our equation: 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: Quadratic factors have x raised to the second power: Let’s decompose this equation: 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: As with the linear equations, we use common denominator and then equate numerators: Substituting x = -2 (to wipe out A and B), gives C = 3.

Expanding the right-hand side multiplications and grouping terms: What if we compare the left-hand side with the right-hand side? To be clear, we are looking at: 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: Using C = 3 gives us A = 2 and B = 1. The result is: Linear Factors Raised to a Power

What about a linear factor raised to the second power? The structure here is:

[Image_Link]/cimages/multimages/16/eparfra7b.

png” alt=”the structure of a PFD with a linear factor rasied to a second power” />

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

[Image_Link]/cimages/multimages/16/eparfra7c.

png” alt=”equating numerators for the PFD of linear factors raised to a second power” />

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: Quadratic Factors Raised to a Power

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

[Image_Link]/cimages/multimages/16/eparfra8a.

png” alt=”example of a PFD of 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: As before, we equate the numerators: Expanding the right-hand side and grouping terms gives us:

[Image_Link]/cimages/multimages/16/eparfra8c1.

png” alt=”RHS expansion and term grouping for the PFD example of a quadratic factor raised to a second power” />

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: Guidelines

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.

[Image_Link]/cimages/multimages/16/eparfra9a.

png” alt=”example which violates the first rule of PFD” />

Do a 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: 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 .

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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 