When you are decomposing fractions into constants

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.

Suppose 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

You 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.

Wow 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?

When 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.

When decomposing a fraction into constants using partial fraction decomposition, the goal is to express the given rational function as a sum of simpler fractions. Each simpler fraction should have a denominator that is a power of a linear factor from the original denominator.

In the example you provided: 1/((x-1)(x-2)^2) = A/(x-1) + B/(x-2) + C/(x-2)^2

Here's the concept behind why we need to repeat (x-2) instead of just using B/(x-2)^2:

When decomposing a fraction into partial fractions, we need to consider all the possible factors in the denominator that give rise to distinct fractions. In this case, the denominator has two linear factors: (x-1) and (x-2). The factor (x-2) appears twice, so we need to account for each power of it when decomposing the fraction.

So, we start with:

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

Now, we need to determine the expression for the third fraction. Since the factor (x-2) appears twice in the denominator, we know that it will contribute two terms to the decomposition. Let's assume the expression for the third fraction is D/(x-2)^2.

Using a common denominator, we can rewrite the right-hand side as a single fraction:

A/(x-1) + B/(x-2) + D/(x-2)^2

To find the value of D, we can multiply both sides of the equation by the denominator on the left-hand side, which is (x-1)(x-2)^2:

1 = A(x-2)^2 + B(x-1)(x-2) + D(x-1)

Now, we can solve for D. We can choose a convenient value of x to simplify the equation. For example, if we choose x=2, the term A(x-2)^2 will become zero since (x-2) is being squared:

1 = D(2-1)

1 = D

So, we found that D equals 1. This means that the third term is 1/(x-2)^2.

Now we can rewrite the original equation as:

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

And we have successfully decomposed the fraction into constants.

In summary, by repeating the denominator (x-2) in two separate terms, we account for the presence of the linear factor being raised to different powers, allowing us to express the original fraction as a sum of simpler fractions.