Back To CourseMath 104: Calculus
13 chapters | 105 lessons
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?
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).
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.
So I have A=1 and B=3, I can plug those into my original fractions and I get (4x+5)/(x+2)(x-1) = 1/(x+2) + 3/(x-1). Because I'm unsure of myself, I want to simplify the right-hand side and make it one fraction to make sure it equals the original fraction on the left-hand side. I'm going to multiply the first term by (x-1)/(x-1) and multiply the second term by (x+2)/(x+2). On the top of the fraction I get (x-1)+(3x+6), which simplifies to (4x+5)/(x+2)(x-1). That matches up with what was on the left-hand side. So indeed, I can rewrite this complicated equation as the sum of two much easier equations.
Let's try another example. Let's say we have (2x+16)/(x^2 + x-6). This one's a little bit harder because we don't know the answer going in, unlike the previous question. Our first step, because we're going to write this as the sum of two fractions, is to find out the factors that go into x^2 + x-6. That looks like it's going to be x minus something times x plus something, and 2 and 3 work, so I have (x-2)(x+3). Now that we've factored the bottom of this equation, I know that the two fractions we're going to add up to get (2x+16)/(x^2 + x-6) are going to be A/(x-2) + B/(x+3). So A and B are my undetermined coefficients. To determine what they are, I'm going to multiply the first term of this equation by (x+3)/(x+3), and I'm going to multiply the second term by (x-2)/(x-2). So I can write this out as A(x+3)+ B(x-2), which equals Ax+3A+Bx-2B. And let's again group the terms that have x in them. Let's match the tops of the two equations, so I have 2x+16=(A+B)x+(3A-2B). To make this equation work for all values of x, I get A+B=2 as one equation, and I get 3A-2B=16 as my second equation, so I've got one for the x term and one for the non-x term.
To solve this, because I like substitution, I'm going to solve A+B=2 for A, so I get A=2 -B. I'm going to plug that in for A in this second equation: 3(2-B) - 2B=16. If I simplify that I get B= -2. If B= -2, I can plug that in to either equation, A=2-(-2), and I find that A=4. Okay, so I've determined my two coefficients and I can put those back into my two fractions. If I add 4/(x-2) to -2/(x+3), I get (2x+16)/(x^2 + x-6). Just to make sure, I'm going to do that addition, and I get 4(x+3)/(x-2)(x+3) - 2(x-2)/(x-2)(x+3). If I simplify the top and bottom, I get (2x+16)/(x^2 + x-6). That equals my original fraction, so it looks good. I just took a really complicated fraction and made it look a lot nicer by using two smaller fractions.
The key with partial fractions is to split up a single, nasty fraction by factoring and determining what the undetermined coefficients A and B are. You do that by grouping all of the x terms and all the non-x terms and setting them equal on the left side and on the right side. Once you've done this, you'll end up with, perhaps, two equations for A and B. You'll have two equations and two unknowns so you can solve for your coefficients, and you have simplified your complicated fraction into a couple smaller, easier-to-deal-with fractions.
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Back To CourseMath 104: Calculus
13 chapters | 105 lessons