2
$\begingroup$

This question already has an answer here:

I was wondering why, not how, partial fractions work the way we are normally taught to do. To be specific:

We are told that, when we have a second degree expression on the bottom that can't be factored out (such as the expression $x^2+5x+7$), we are to have a first degree expression on the top (and not just a constant). For example,

$$ \frac{5x+1}{(x-1)(x^2+3x+10)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+3x+10}, $$ where the capital constants($A,B$, and $C$) are to be determined. I watched Khan Academy, and it was explained that in the partial fractions, the leading degree of the numerator must be exactly one less than the leading degree of the denominator. Why is this a rule?

Second question: why does a denominator expression that has multiplicity at least two has to have multiple partial fractions? For example, we are told that

$$ \frac{3x+2}{(x+2)(x-5)^2} $$ must be decomposed into the form $$ \frac{A}{x+2}+\frac{B}{x-5}+\frac{C}{(x-5)^2}. $$ Khan Academy says that this is because have just the $(x-5)^2$ denominator would leave the original problem still being able to be decomposed further. But I am not entirely convinced by this. Can anyone give me a better explanation why there has to be $\frac{C}{(x-5)^2}$ AND $\frac{B}{x-5}$ (not just $\frac{C}{(x-5)^2}$)?

Please don't explain how I can do partial fractions. I already know how to decompose a fraction into partial fractions; I want to know the answers to the above two questions.

$\endgroup$

marked as duplicate by Git Gud, user91500, Watson, Daniel W. Farlow, Hans Lundmark May 21 '16 at 17:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Good luck: math.stackexchange.com/a/368790 $\endgroup$ – Git Gud May 21 '16 at 9:16
  • 1
    $\begingroup$ I think the reason for these rules is "because we can't do any better". For example, in the latter, how would you express $x/(x-5)^2$ if you were only allowed to use $C/(x-5)^2$ (for constant $C$)? Similar reasoning behind the former one. $\endgroup$ – Wojowu May 21 '16 at 9:21
0
$\begingroup$

In the example

$$\frac{3x+2}{(x+2)(x-5)^2}=\frac{A}{x+2}+\frac{B}{x-5}+\frac{C}{(x-5)^2}$$

You will find that the degree of the LHS is $1$ in the numerator and $3$ in the denominator. This implies that the degree of the entire fraction is $-2$.

On the RHS, you will find that the degree for the first fraction is $-1$, the second is $-1$, and the last is $-2$.

And if you've been taught right, the larger degree usually takes over the degree of a polynomial, right?

For example, $x^3+x^2$ has a degree of $3$ since the degree of the second term is smaller than the first.

To get a degree of $-2$ on the RHS, we must have something along the following lines:

$$\frac{A}{x+2}+\frac{B}{x-2}=Dx^{-2}+Ex^{-3}+\dots$$

In other words, you'd want the $x^{-1}$ term to cancel out, which is not doable unless you have two fractions so that they may cancel each other.

This leaves us with a degree of $-2$, which corresponds to the LHS.

For examples like

$$\frac{x+1}{x-1}=1+\frac2{x-1}$$

You see the "$1$" is actually an $x^0$, and the LHS has a degree of $0$ as well. So this works out well.

$\endgroup$