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For, 4x^2-3x+5/(x-1)^2(x+2) we need: A/(x-1)^2+B/(x-1)+C/(x+2)

It is also correct to write Ax+B/(x-1)^2 + C/(x+2) but the fractions are not then reduced to the simplest form.

How do the 2nd fractions simplify to give the 1st set of fractions?

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- Thread starter Rach123
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- #1

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For, 4x^2-3x+5/(x-1)^2(x+2) we need: A/(x-1)^2+B/(x-1)+C/(x+2)

It is also correct to write Ax+B/(x-1)^2 + C/(x+2) but the fractions are not then reduced to the simplest form.

How do the 2nd fractions simplify to give the 1st set of fractions?

- #2

dynamicsolo

Homework Helper

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For, 4x^2-3x+5/(x-1)^2(x+2) we need: A/(x-1)^2+B/(x-1)+C/(x+2)

It is also correct to write Ax+B/(x-1)^2 + C/(x+2) but the fractions are not then reduced to the simplest form.

How do the 2nd fractions simplify to give the 1st set of fractions?

They don't exactly: the constants in the second set will be a little different. They are just saying that the two sums are equivalent, but the first sum is preferred for certain purposes. (I don't agree that the first set is

What you would actually find for the first two terms in the first set would be

[A/(x-1)^2] + [B/(x-1)] = [A/(x-1)^2] + [{B(x-1)}/(x-1)^2] =

[{A+Bx-B}/(x-1)^2] ;

they then consolidated the A-B in the numerator into a single constant and relabeled the coefficients. But the A and B in the second set will not be then same as they are in the first set (which is a usage I find a bit sloppy)...

You are correct in saying that you can't just rearrange the second set to get the first one.

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