Linear Factor Rule

**The Anomaly** · Jul3-09, 04:43 AM

I'm studying the Partial Fraction method of integration, and I believe I understand the fundamental idea of it. However, much of it is based on a rule that the book calls the Linear Factor Rule. It is the following:

For each factor of the form (ax+b)^m the partial fraction decomposition contains the following sum of m partial fractions:

$LaTeX Code: \\frac{A_1}{(ax+b)}$ + $LaTeX Code: \\frac{A_2}{(ax+b)^2}$ + ... + $LaTeX Code: \\frac{A_m}{(ax+b)^m}$

I'm assuming that the proof of this is either assumed, or was done in a Precalculus course or something. But could you help me out with proving it? It just doesn't make much sense at this point.

**g_edgar** · Jul3-09, 05:53 AM

Originally Posted by The Anomaly

I'm studying the Partial Fraction method of integration, and I believe I understand the fundamental idea of it. However, much of it is based on a rule that the book calls the Linear Factor Rule. It is the following:

For each factor of the form (ax+b)^m the partial fraction decomposition contains the following sum of m partial fractions:

$LaTeX Code: \\frac{A_1}{(ax+b)}$ + $LaTeX Code: \\frac{A_2}{(ax+b)^2}$ + ... + $LaTeX Code: \\frac{A_m}{(ax+b)^m}$

I'm assuming that the proof of this is either assumed, or was done in a Precalculus course or something. But could you help me out with proving it? It just doesn't make much sense at this point.

No, the proof that such numerators exist is found only in more advanced courses. But to proceed in integration you do not need that proof. Even if you have not proved that such coefficients exist in all cases, if such coefficients exist (as they always do) in the case you are doing, then you can go ahead with the integration.

**The Anomaly** · Jul3-09, 05:55 AM

Originally Posted by g_edgar

No, the proof that such numerators exist is found only in more advanced courses. But to proceed in integration you do not need that proof. Even if you have not proved that such coefficients exist in all cases, if such coefficients exist (as they always do) in the case you are doing, then you can go ahead with the integration.

Alright, that makes sense. I just wanted to make sure that I wasn't missing some obvious proof from precalculus.

Thanks for the reply!

EDIT: And for curiosity, when my Calculus book refers to advanced algebra (Not exactly for this example, but for others) are they referring to Precalculus algebra? Or way more advanced stuff that I have not approached yet?