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March 29, 2015

March 29, 2015

Posted by **Anonymous** on Monday, June 6, 2011 at 8:41pm.

EX:

1/(x-1)(x-2)^2 = A/(x-1) + B/(x-2) + C/(x-2)^2

Why do you have to repeat (x-2), instead of just putting B/(x-2)^2?

I know how to solve these types of problems, but I don't understand it conceptually.

- calculus -
**Reiny**, Monday, June 6, 2011 at 10:06pmSuppose you had to add

1/(x-1) + 1/(x-2) + 1/(x-2)^2

your common denominator would be (x-1)(x-2)^2

in the same way if you had

1/(x-1) + 1/(x-2)^2

you would still have that same common denominator of (x-1)(x-2)^2

So when you are trying to reverse the process, how do you know which way it was?

To allow for all possibilities we use the first version.

In the first case, if the 2nd term is missing, that will show up as B=0

- calculus -
**MathMate**, Monday, June 6, 2011 at 10:08pmYou could solve it as you put it:

1/(x-1)(x-2)^2 = A/(x-1) + B/(x-2) + C/(x-2)^2

or as

1/(x-1)(x-2)^2 = A/(x-1)+(Bx+C)/(x-2)^2

otherwise you'd be missing one parameter of type B/(x-2) in the most general form.

- calculus -
**Anonymous**, Monday, June 6, 2011 at 10:08pmWow that makes sense. Although, how come the C/(x-2)^2 doesn't follow the rule where the numerator must be one less power than the denominator?

Shouldn't it be Cx+D/(x-2)^2?

- calculus -
**MathMate**, Monday, June 6, 2011 at 11:55pmWhen you have B/(x-2), it takes care of the term Bx/(x-2)^2, so you use either one of the two, preferably the former.

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