Partial Fractions: How to Factorize Fractions with Quadratic Denominators

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  • 1:40 Solving for Partial Fractions
  • 7:36 Solving for Partial…
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Lesson Transcript
Instructor: Kelly Sjol
Adding fractions with different denominators is something you probably learned how to do in algebra. In this lesson, learn how to do the opposite: take a complicated fraction and turn it into two simpler ones.

Example of a Complicated Fraction

Let's think about algebra for a minute. Let's say you're trying to add two fractions, like 1/(x+2) + 3/(x-1). You'd do this by multiplying the first term by (x-1)/(x-1) and the second term by (x+2)/(x+2). When you do that, you get x^2 + x-2 on the bottom, and you can just add up the two terms on the top so you get 4x+5. If I try to integrate these like in integral calculus, the two terms on the left-hand side look a lot more reasonable (the integral of 1/(x+2)dx=ln(x+2)+C) than the gigantic term on the right-hand side. In fact, if you gave me (4x+5)/(x^2 + x-2) and told me to integrate it, I'd probably look at you like you were nuts. And then I'd try substitution or I don't even know what else before giving it back to you and saying, 'Do it yourself.' Because the left-hand side is the same thing as the right-hand side, wouldn't it be nice if we could take this big, ugly, gigantic fraction and turn it into two smaller fractions that are easier to handle?

In this problem A and B are undetermined coefficients
Partial Fraction Undetermined Coefficients

Solving for Partial Fractions

This is what we call solving for partial fractions. So what I need to do is take my big, nasty fraction with a quadratic polynomial on the bottom and something less than quadratic on the top, and I'm going to factor the quadratic part. I've got x^2 + x-2. If I factor it out into two separate terms, I find that (x+2)(x-1) is the same thing as x^2 + x-2, so I can factor this quadratic as (x+2)(x-1). Once I've factored the bottom, I know that I can rewrite this entire equation as being equal to A/(x+2) + B/(x-1), because if I multiply the first term by (x-1)/(x-1) and the second term by (x+2)/(x+2), I get back something that looks like (4x+5)/(x+2)(x-1).

A and B are what we call undetermined coefficients. How do we find these 'undetermined' coefficients and make them 'determined'? Let's take a good look at our equation. We have (4x+5)/(x+2)(x-1) = A/(x+2) + B/(x-1). Like I said, if I want to combine these two fractions into one, I need to take the first fraction and multiply it by (x-1)/(x-1) and take the second fraction and multiply it by (x+2)/(x+2). When I do that on the right side, I get A(x-1)+B(x+2)/(x+2)(x-1). Because the bottoms of these equations are now the same, I can cancel them out. I can just multiply both sides of the equation by (x+2)(x-1), and I get 4x+5=A(x-1)+B(x+2). Now I'm going to gather the like terms. So in this case, I'm going to take every part of this right-hand side that has an x in it and collect them all together. So I get Ax-A+Bx+2B, and if I collect the two terms with x in them and factor out the x, I get 4x+5=x(A+B)+(2B-A).

With undetermined coefficients, the left-hand side must equal the right side for all x values
Partial Fractions Example 1

One of the most important things to realize about solving this partial fractions problem for these undetermined coefficients is that for the left-hand side to equal the right-hand side for all values of x - remember, we don't know what A and B are - the term with the x on the left-hand side has to equal the term with the x on the right-hand side. That is, 4x has to equal x(A+B). That means A+B=4. Similarly, the term without the x on the left-hand side has to equal the term without the x on the right-hand side. So I can write another equation that is 5=2B-A. So now I have two equations and two unknowns - A and B are still my undetermined coefficients. You can solve these two equations for A and B using whatever method you prefer. I like substitution, so what I would do is say A=4 - B, and I would plug 4 - B into my second equation, 5=2B - 4 - B, just to get rid of the A. If I solve that for B, I get B=3. Then I can plug B into either one of these equations and solve for A, A=4 - 3, and I get A=1.

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