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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?

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.

Derek 朕會功夫
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4 Answers4

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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

Warren Hill
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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.

lsp
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  • I don't understand what you mean by "multiplicity of the zero x = -1 is just 1". Can you explain it more please? – Derek 朕會功夫 Feb 06 '14 at 07:03
  • The explanation of lsp is the right one. – kmitov Feb 06 '14 at 07:03
  • In your example, you added $\frac{B}{x+1}+\frac{C}{x+1}$ and the denominator remains the same. But no one says that I have to do it like this though; I can do $\frac{B(x+1)+C(x+1)}{(x+1)^2}$ and everything is still valid. – Derek 朕會功夫 Feb 06 '14 at 07:10
  • @Derek朕會功夫 But the fraction you just mentioned in you comment is reducible ! You can cancel out $(x+1)$ from numerator and denominator. – lsp Feb 06 '14 at 07:12
  • @lsp - But no one says I have to cancel them out. – Derek 朕會功夫 Feb 06 '14 at 07:13
  • Ok. I'll explain you. But first explain me why you are writing $\frac{x}{(x+2)(x+1)} = \frac{A}{x+2} + \frac{B}{x+1}$ and not $\frac{x}{(x+2)(x+1)} = A + B$ – lsp Feb 06 '14 at 07:15
  • Well it's because if you multiply the denominators together, the denominator in the left will be the same as the right. Since they are the same, I will be left with $x=A(x+1)+B(x+2)$. – Derek 朕會功夫 Feb 06 '14 at 07:17
  • Why not write it as just $A+B$ ? As we can re write $A+B$ as $\frac{A(x+1)}{x+1}+\frac{B(x+2)}{x+2}$. This will still give the denominator as $(x+1)(x+2)$. What do you comment on this ? – lsp Feb 06 '14 at 07:19
  • @lsp - Writing it as $A+B$ works too, but doesn't it mean that the fraction is equal to $0$? Okay something doesn't sound right here. – Derek 朕會功夫 Feb 06 '14 at 07:23
  • @Derek朕會功夫 So did you get it now ? What it will lead you to when you consider reducible fractions ? – lsp Feb 06 '14 at 07:25
  • Yea I get it now, we add the square to prevent getting 0's. But every step I did is correct, and if every step is true then the statement $\frac{x}{(x+2)(x+1)}=0$ should be true as well. Why isn't it the case here? – Derek 朕會功夫 Feb 06 '14 at 07:27
  • @Derek朕會功夫 Just go through the link that I have provided. That will make things very clear for you. – lsp Feb 06 '14 at 07:30
  • @lsp - The Wikipedia article doesn't explain why $\frac{x}{(x+2)(x+1)}=0$ is false... – Derek 朕會功夫 Feb 06 '14 at 07:34
  • @Derek朕會功夫 Not only $\frac{x}{(x+2)(x+1)} = \frac{A}{x+2} + \frac{B}{x+1}$, according to your logic all fractions in such form are $0$ which is absolutely false. – lsp Feb 06 '14 at 07:36
  • @lsp - I know it is definitely false, and I understand now why the square is needed, but I want to know where in between the process went wrong since each step is true. – Derek 朕會功夫 Feb 06 '14 at 07:38
  • Your steps went wrong from where you started using reducible fractions, i.e., fractions of the form $\frac{A(x+1)}{(x+1)}$. In your case in the first step itself where you have used $\frac{C}{(x+1)}$ assuming it to be $\frac{C(x+1)}{(x+1)^2}$ – lsp Feb 06 '14 at 07:40
  • @lsp - But $\frac{C}{x+1}$ is the same as $\frac{C(x+1)}{(x+1)^2}$... isn't it? – Derek 朕會功夫 Feb 06 '14 at 07:45
  • @Derek朕會功夫 Instead of just bluntly trying to disprove everything just take a moment and think as to what happens when you do that here. – lsp Feb 06 '14 at 07:48
  • @lsp - Okay I will take a look at it again later. Anyway thank you so much for explaining the reason why the square is needed. – Derek 朕會功夫 Feb 06 '14 at 07:55
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$\frac{A}{x+2} + \frac{B}{x+1}+\frac{C}{x+1}=\frac{A(x+1)^2+B(x+1)(x+2) + C(x+1)(x+2)}{(x+2((x+1)^2}$

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.

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

kmitov
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  • If I multiply $(x+2)(x+1)(x+1)$ I still get $(x+2)(x+1)^2$ back though. – Derek 朕會功夫 Feb 06 '14 at 06:55
  • @Derek Yes, but it's not the lowest common multiple of the denominators. – Daryl Feb 06 '14 at 06:56
  • The proper cmmon denominator in the second case is $(x+2)(x+1)$ – kmitov Feb 06 '14 at 06:57
  • @kmitov - Indeed the common denominator is $(x+2)(x+1)$, but why would you care about common denominator? I mean you can still multiply everything like they are different and solve the equation at the end. – Derek 朕會功夫 Feb 06 '14 at 07:01
  • But you can try and see the result. – kmitov Feb 06 '14 at 07:08
  • @kmitov - Indeed the common denominator is (x+2)(x+1), but why would you care about common denominator? I mean you can still multiply everything like they are different and solve the equation at the end. – Derek 朕會功夫 16 mins ago You have to care. – kmitov Feb 06 '14 at 07:19