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## Algebra (all content)

### Course: Algebra (all content)>Unit 13

Lesson 13: Partial fraction expansion

# Partial fraction expansion: repeated factors

Sal performs partial fraction expansion upon (6x²-19x+15)/(x-1)(x-2)². Created by Sal Khan.

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• When he started the question and at the end, he said it was hairier. What's that supposed to mean?
(1 vote) • at I don't understand how is that if you ad B/(x-2) + C/(x-2)^2 you would get an expression something X + something over (x-2)^2 • B/(x-2) + C/(x-2)^2
Multiply B/(x-2) by (x-2)/(x-2) to get an equivalent fraction with the denominator (x-2)^2
B(x-2)/(x-2)^2 + C/(x-2)^2
Now we can add the numerators directly
( B(x-2) + C ) / (x-2)^2
Now expand the numerator
( Bx - 2B + C ) / (x-2)^2
There's your "something X + something over (x-2)^2 "
• I apologize in advance if this question is too generalized. I have been methodically advancing. Some topics are more difficult than others. I am bewildered by this topic. Completely. I passed over it in algebra but I am determined to master this topic. I've watched the videos (all of the ones leading up to this topic as well) multiple times. I think my problem is that I can't abstract this. The answer is more complex than the question. I don't want to move on. Any suggestions? • When you add two fractions with different denominators, you find the common denominator, do the calculations, and then add them. You end up with a single fraction.

Partial fractions are the reverse of this process. You start with a single fraction and split it into one or more fractions with different denominators.

The reason you would bother doing this is that some computations (especially once you get into calculus) can be greatly simplified by breaking up a fraction in this manner.
• when must I use the form (Bx + c ) /... or (Bx^2 +C) / ... ????
and how many forms are there? What's the rule of that? • (Bx + C) is for when you have an irreducible quadratic term (ax^2 + bx + c) in the denominator (possibly in the form: (x^2 - c)). In this problem the term is (x - a)^2, a subtle difference. In this case you need a fraction for each degree of the term.
So you get: B1 / (x - a) + B2 / (x - a)^2
If you have a quadratic that you can't reduce that is also repeated: (ax^2 + bx + c)^n
You will need to combine the irreducible quadratic approach with the repeated linear approach:
(B1 * x + C1)/(ax^2 + bx + c) + (B2 * x + C2)/(ax^2 + bx + c)^2 . . . (Bn * x + Cn)/(ax^2 + bx + c)^n
I hope that's clear.
• In the first Partial Fraction vid, you said that the degree in the nominator has to be less than the degree in the denominator (if that isn't the case, you'd have to do long division). Here clearly the degree in the nominator is greater, but it all worked out anyway. Why is that ? • Why is it C/(x-2)^2 and not (Cx+D)/(x-2)^2. In the previous video, didn't he say that the numerator had to be one degree less than the denominator. In the first scenario it is two degrees less • If the denominator is in the form ax^2+bx+c, then you are absolutely right and the numerator should be Cx+D. However, in this case, C / (x-2)^2 goes with the B / (x-2) just before it. When the two partial fractions are added, you get (B(x-2) + C)/(x-2)^2. As you can see, the numerator of this combined term is one degree less than the denominator. Sal explains how this works at .
• in what circumstance would this be applicable to a real world problem? • So this might help answer your question. Recently, I've been going through all the calculus modules on Khan Academy and I just reached a part of Differential Equations where I needed this skill (which is why I'm here). So far in my life, I have not needed to know how to do this because you're right - all on its own, this skill has very few real life uses. However, once you start doing some of the more advanced calculus problems, this is an essential technique for solving real world calculus problems.
• Sal,
The partial fraction videos were very helpful but i need to know how to solve for a_1,a_2 when I have (7s+12)/(s+2)^2.
thank You. • The partial fraction decomposition for a repeated linear factor is of the form:
(7s+12)/(s+2)^2 = A/(s+2) + B/(s+2)^2. Multiply through by (s+2)^2 and you get:
7s + 12 = A*(s+2) + B

There are various techniques from here... You can substitute in values of s to find A, B, or both. In this case s=-2 gives you -14 + 12 = A*0 + B or B = -2. But s=-1 gives you -7 + 12 = A + B or A + B = 5.

Alternatively you can expand the right hand side to get the polynomial Ax + (2A + B). Matching this against the terms on the left, we learn that A = 7 and 2A+B=12. Easily solved from there.
• If N(x)= x^2(x-1)(x^2+x+1) we get = A/x + [B/(x^2)] + C/(x-1) + (Dx+E)/(x^2+x+1).

My question is about the second term in []. • In the previous video, he said that your numerator of each part of the decomposed fraction must be one degree less that its denominator. E.g (Bx+C)/(x^2+2x+4)
Why when wrote the decomposed fraction out did he just write the constant C over the (x-2)^2 denominator. Shouldn't it have been Cx+D? 