Why do repeated linear factors have to be dealt with in this way?

Asked Dec 18 '14 at 00:18

Active Dec 18 '14 at 00:59

Viewed 348 times

When dealing with partial fractions, and your denominator has a repeated linear factor, the way to solve is this:

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

$2x+3=A+B(x-2)$

and so on.

Now, even though I know the following method doesn't work, I don't understand why it doesn't work.

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

$2x+3=A(x-2)+B(x-2)$

Why can't I treat repeated linear factors like other linear factors? I'm sure the answer is very obvious, but I can't see it right now. I mean, I can see that I can't set $x=2$ here. But yeah. I feel like I'm missing something.

edited Dec 18 '14 at 00:59

Git Gud

31,356

asked Dec 18 '14 at 00:18

Korgan Rivera

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1

you say yourself that you know it doesn't work, and then you ask why you can't do that? Well, you can't do it because it does not work. – Ittay Weiss Dec 18 '14 at 00:20
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Because $\frac{A}{x-2}+\frac{B}{x-2}=\frac{A+B}{x-2}$. Is that really possible? – Thomas Andrews Dec 18 '14 at 00:24
In particular, can you solve: $2x+3=(A+B)(x-2)$? Is $2x+3$ a multiple of $x-2$? – Thomas Andrews Dec 18 '14 at 00:27
@IttayWeiss The question is, why doesn't it work. – Korgan Rivera Dec 18 '14 at 00:42
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is it a philosophical question? like why is $2+3\ne 4$? – Ittay Weiss Dec 18 '14 at 00:44
@KorganRivera didn't say what? – Thomas Andrews Dec 18 '14 at 00:46
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Reference question for partial fractions. – Git Gud Dec 18 '14 at 01:05
@ThomasAndrews You answered my question. I must have messed up the arithmetic to not have seen that before. Thanks. – Korgan Rivera Dec 18 '14 at 01:11

Why do repeated linear factors have to be dealt with in this way?

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