# Concept Decomposing of Partial Fractions

• babby
In summary: Therefore, equating coefficients allows us to solve for the unknowns A, B, and C, which in turn helps us decompose the original fraction into smaller, more manageable parts. In summary, equating coefficients works because it allows us to solve for unknowns and manipulate equations while still maintaining the equivalence of both sides.
babby
I have a few questions about decomposing partial fractions. I know how to solve these problems, but I just don't understand why I'm doing some of the things.

1. Why does equating coefficients work? I don't understand the idea behind it.

2. 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?

Thanks.

$$\frac{A}{x-1} + \frac{Bx+C}{(x-2)^2}$$.
The point is that the numerator can be anything with degree less than the degree of the denominator.

g_edgar said:
$$\frac{A}{x-1} + \frac{Bx+C}{(x-2)^2}$$.
The point is that the numerator can be anything with degree less than the degree of the denominator.

Ah, that was what I was originally thinking, but forgot the variable. But is there a reason why you have to repeat the function when doing it the other way? I don't understand how it works. Is it because the first part of the function:

B/(x-2) eliminates the need for a numerator with one degree less than the denominator seen in

C/(x-2)^2?

OK, I understand that part now, but I have another question:

Can somebody explain to me why equating coefficients work?

Example:
8x^3+13x = Ax^3 + 2Ax + Bx^2 + 2B + Cx + D

expanded into:

8x^3 + 13x = Ax^3 + Bx^2 + (2A+C)x + 2(B+D)

where A,B,C,D are constants.

Why does 8 = A; 0 = B; 13 = 2A + C; etc.

I know they have same power variables, but why does this actually work? Thanks!

The technique involves the assumption that for the correct values of unknowns such as A, B, and C, both sides of the equation are equivalent for all x. This assmption would still hold after manipulation to put both sides in polynomial form. It should be obvious that two polynomial functions of x are equal for all x if the corresponding coefficients are equal.

## 1. What is concept decomposing of partial fractions?

Concept decomposing of partial fractions is a method used to simplify and rewrite a complex fraction into smaller, simpler fractions. It involves breaking down the numerator into smaller fractions with different denominators and then adding them together.

## 2. Why do we need to decompose partial fractions?

Decomposing partial fractions allows us to solve integrals and differential equations more easily. It also helps us to simplify complex algebraic expressions and make them more manageable.

## 3. How do you decompose partial fractions?

To decompose partial fractions, you need to follow a few steps:

• Factor the denominator of the original fraction into its linear and irreducible quadratic factors.
• Write the original fraction as a sum of smaller fractions with the same denominators as the factors in the previous step.
• Find the unknown coefficients of each smaller fraction by equating the numerators of the original fraction and the decomposed fractions.
• Simplify the smaller fractions, if possible.
• Combine the smaller fractions back into one fraction to get the final solution.

## 4. What are some common mistakes when decomposing partial fractions?

Some common mistakes when decomposing partial fractions include:

• Not factoring the denominator correctly.
• Not setting up the equations to find the unknown coefficients correctly.
• Forgetting to include all the factors in the denominator when writing the smaller fractions.
• Not simplifying the smaller fractions before combining them back into one fraction.

## 5. Can you provide an example of concept decomposing of partial fractions?

Sure, consider the fraction 3x + 2 divided by x2 + 5x + 6. To decompose this fraction, we first factor the denominator into (x + 2)(x + 3). Then, we can write the original fraction as A/(x + 2) + B/(x + 3), where A and B are the unknown coefficients we need to find. Next, we equate the numerators of the original fraction and the decomposed fractions to get the equations 3x + 2 = A(x + 3) + B(x + 2). Solving for A and B, we get A = 1 and B = 2. Therefore, the decomposed fraction is 1/(x + 2) + 2/(x + 3).

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