Repeated linear factors in partial fractions

Question

I have a question about the following partial fraction:

$$\frac{x^4+2x^3+6x^2+20x+6}{x^3+2x^2+x}$$ After long division you get: $$x+\frac{5x^2+20x+6}{x^3+2x^2+x}$$ So the factored form of the denominator is $$x(x+1)^2$$ So $$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$$

Why is the denominator under $C$ not simply $x+1$? It is $x$ times $(x+1)^2$ and not $(x+1)^3$

Answer 1

Notice that the RHS after simplification must have an identical denominator as with the LHS.

The LHS denominator has a cubic term, hence the RHS must also be cubic.

So the choice for the term under $C$ has to be $(x+1)^2$, and not the ones you have suggested.

Answer 2

You need an $(x+1)^2$ in the denominator of at least one of the partial fractions, or when you sum them they would not have an $(x+1)^2$ term in the denominator of the sum.

But you might find it easier to solve:

$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{Bx + C}{(x+1)^2}$

And that is completely valid.

Answer 3

Explanation 1: If you have only $(x+1)$, your denominators are too weak:

$$\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+1} = \frac{A(x+1)}{x(x+1)}+\frac{Bx}{x(x+1)}+\frac{Cx}{x(x+1)}$$

Thus, you your denominator is quadratic whereas the desired denominator is $x(x+1)^2$, cubic. I mean, $\frac{C}{x+1}$ is redundant. Why? Just let $D=B+C$. Then we have $\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+1} = \frac{A}{x}+\frac{D}{x+1}$.

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Explanation 2: Would you do this?

$$\frac{1}{x^2} = \frac{A}{x} + \frac{B}{x \ \text{and not} \ x^2 ?}$$

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Explanation 3: Actually, some texts suggest

$$\frac{5x^2+20x+6}{x(x+1)^2} = \frac{\text{something}}{x}+\frac{\text{something}}{(x+1)^2}$$

where $\text{something}$ is an arbitrary polynomial with degree 1 lower than the denominator, i.e.

$$\frac{5x^2+20x+6}{x(x+1)^2} = \frac{D}{x}+\frac{Ex+F}{(x+1)^2}$$

This is actually equivalent to what your text suggests! How?

$$\frac{Ex+F}{(x+1)^2} = \frac{Ex+F+E-E}{(x+1)^2}$$

$$= \frac{E(x+1)+F-E}{(x+1)^2} = \frac{E}{x+1} + \frac{F-E}{(x+1)^2}$$

where $E=B$ and $F-E = C$

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If you're confused about equivalence or of why there's a squared, I think it's best to just play safe:

Partial fractions, if I understand right, is pretty much an ansatz in basic calculus, iirc (the word 'ansatz' is used here, but I guess different from how I use it).

Instead of trying to understand, I just go on the safe side and do:

$$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{(x+1)^2}$$

Similarly,

$$\frac{5x^2+20x+6}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{(x)^2}$$

$$\frac{5x^2+20x+6}{x(x+1)^3}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{(x+1)^2}+\frac{Ex^2+Dx+F}{(x+1)^3}$$

Maybe there's some rule about how something is redundant, but I don't care (if I start caring, I may end up with your question) because I'm covering everything: Whenever there's an exponent in the denominator, I do as many fractions at that exponent. The additional coefficients will end up as zero anyway.