Why partial fraction decomposition of $ \frac {1}{(x^2-3)^2 (x+7)}$ is $\frac{A}{(x+7)} + \frac {Ax+B}{x^2-3}+\frac {Cx+D}{(x^2-3)^2}$

Asked May 25 '23 at 16:34

Active May 25 '23 at 16:57

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I am curious about its reason. I read something that are saying that this comes from Bezout Identity and idea is nothing but if we divide something by lets say degree $3$ then remainder at most can be degree $2$ like usual division.

But for here, we divide something by degree $4$ which is $(x^2-3)^2$ so why can't we directly write $ \frac {1}{(x^2-3)^2 (x+7)} = \frac{A}{(x+7)} + \frac {Bx^3+Cx^2+Dx+E}{(x^2-3)^2}$

edited May 25 '23 at 16:48

asked May 25 '23 at 16:34

Fuat Ray

1

The numerator over $x-3$ should be a constant. Otherwise, the fraction is not simple. – ajotatxe May 25 '23 at 16:36
Sorry I mess up, that should be $(x^2-3)$ not $x-3$ – Fuat Ray May 25 '23 at 16:46
I want to ask this questions for this $\frac {1}{(x^2-3)^2 (x+7)}$. Okay, if things that I wrote for its decompositions then what should it be? – Fuat Ray May 25 '23 at 16:49
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Review the theorem https://en.wikipedia.org/wiki/Partial_fraction_decomposition#Statement and beware that $x^2-3$ is not irreductible in $\Bbb R[x].$ – Anne Bauval May 25 '23 at 16:50
Okay, roughly theorem says that you can not take square of non irreducible polynomial, right? – Fuat Ray May 25 '23 at 16:54
See – tryst with freedom May 25 '23 at 16:54
and – tryst with freedom May 25 '23 at 16:55
Right. A simple fraction is of the form $\frac{a(x)}{(p(x))^k}$ where $p$ is irreducible and $\deg a<\deg p.$ – Anne Bauval May 25 '23 at 16:57
What if we encounter with this situation. What should we do? It seems to me that there should be method @AnneBauval – nozalp10 May 25 '23 at 17:30
Note: the first letter $A$ should be a different letter, call it $H$ for example. Meanwhile, the final version in the question is also correct: $ \frac {1}{(x^2-3)^2 (x+7)} = \frac{P}{(x+7)} + \frac {Qx^3+Rx^2+Sx+T}{(x^2-3)^2}$ See if you can compare the letters – Will Jagy May 25 '23 at 17:32
To begin with, apply the statement of the theorem (linked above), i.e. write the decomposition with indeterminate coefficients (like you did, but correctly). – Anne Bauval May 25 '23 at 17:34
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Does this answer your question? The numerators in partial fraction decomposition involving repeated factors Do that for each of the three (real) roots (one simple and two double) of your denominator. – Anne Bauval May 25 '23 at 17:36

Why partial fraction decomposition of $ \frac {1}{(x^2-3)^2 (x+7)}$ is $\frac{A}{(x+7)} + \frac {Ax+B}{x^2-3}+\frac {Cx+D}{(x^2-3)^2}$

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