Extra Square in Partial Fraction

Question

So I understand why this is true: $$ \frac{x}{(x+2)(x+1)} = \frac{A}{x+2} + \frac{B}{x+1} $$

But there's a special rule in partial fraction that I just couldn't get it. When you have a term that is squared, I must add another fraction with the term squared in the denominator: $$ \frac{x}{(x+2)(x+1)^2} = \frac{A}{x+2} + \frac{B}{x+1} + \frac{C}{(x+1)^2} $$

Why? Shouldn't $(x+1)^2$ be treated like $(x+1)(x+1)$ and do this instead?: $$ \frac{x}{(x+2)(x+1)^2} = \frac{A}{x+2} + \frac{B}{x+1} + \frac{C}{x+1} $$

Why is the extra square needed?

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To visualize it easier, I have tried to substitute terms with letters.

$$ \frac{x}{(x+2)(x+1)}\\ \text{Let x+2 be X, x+1 be Y:}\\ = \frac{A}{X} + \frac{B}{Y}\\ = \frac{AY+BX}{XY} $$

And I can do the same thing for this:

$$ \frac{x}{(x+2)(x+1)^2} = \frac{x}{(x+2)(x+1)(x+1)}\\ \text{Let x+2 be X, x+1 be Y, x+1 be Z:}\\ = \frac{A}{X} + \frac{B}{Y} + \frac{C}{Z}\\ = \frac{AYZ + BXZ + CXY}{XYZ}\\ \\ \frac{x}{(x+2)(x+1)(x+1)} = \frac{AYZ + BXZ + CXY}{XYZ}\\ x = AYZ + BXZ + CXY $$

I don't see any problem in this.

Answer 1

Because the zero $x=-1$ of the denominator has a multiplicity of $2$, the partial fractions must contain the $2$nd power of $(x+1)$. Though there is no necessity that the partial fraction must contain $1$st power of $(x+1)$.

In your explanation, $$\frac{B}{x+1} + \frac{C}{x+1} = \frac{B+C}{x+1}$$ $$\frac{A}{x+2} + \frac{B+C}{x+1} = \frac{A(x+1)+(x+2)(B+C)}{(x+2)(x+1)}$$ Check the denominator here. What is the power of $(x+1)$ here ? It is $1$. But it should have been $2$ right? Where did it vanish ? It vanished because of the mistake that you have done in neglecting the repeated root in denominator. Here the multiplicity of the zero $x=-1$ is just $1$, where it should have been $2$.

You can refer this if still in doubt.

EDIT: Since you are not fully aware on how partial fraction decomposition works. Read this.

Answer 2

The common denominator must be $(x+2)(x+1)^2$. You cannot obtain it from your second representation.

Answer 3

$\frac{A}{x+2} + \frac{B}{x+1}+\frac{C}{x+1}=\frac{A(x+1)^2+B(x+1)(x+2) + C(x+1)(x+2)}{(x+2((x+1)^2}$

Then

$x=Ax^2+2Ax+A+Bx^2+3Bx+2B+Cx^2+3Cx+2C$

and

$A+B+C=0$

$2A+3B+3C=1$

$A+2B+2C=0$

You can try to solve this system.

Answer 4

Consider

$$ \frac{A}{x+2}+\frac{B}{x+1}+\frac{C}{x+1} = \frac{A}{x+2}+\frac{B+C}{x+1}=\frac{A(x+1)+(B+C)(x+1)}{(x+1)(x+2)} $$

This does not contain the $(x+1)^2$ term in the denominator so does not cover all possible solutions. Which is why it does not work.

While

$$ \frac{A}{x+2}+\frac{B}{x+1}+\frac{C}{(x+1)^2} = \frac{A(x+1)^2+B(x+1)(x+2)+C(x+2)}{(x+1)^2(x+2)} $$

Does.

We can expand the numerator

$$ \begin{align} A(x+1)^2+B(x+1)(x+2)+C(x+2) &= A(x^2+2x+1)+B(x^2+3x+2)+C(x+2)\\ &= (A+B)x^2+(2A+3B+C)x+(A+2B+2C) \end{align}$$

Which we can use to find A, B and C