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Suppose \[\frac{1}{x^3-x^2-21x+45}=\frac{A}{x+5}+\frac{B}{x-3} + \frac{C}{(x - 3)^2}\]where $A$, $B$, and $C$ are real constants. What is $A$?

 Dec 21, 2020
 #1
avatar+10824 
+1

Suppose

\(\frac{1}{x^3-x^2-21x+45}=\frac{A}{x+5}+\frac{B}{x-3} + \frac{C}{(x - 3)^2} \)

where A, B and C are real constants. What is A?

 

Hello Guest!

 

\(\frac{1}{x^3-x^2-21x+45}=\frac{A}{x+5}+\frac{B}{x-3} + \frac{C}{(x - 3)^2} \)

\(\frac{1}{x^3-x^2-21x+45}=\frac{A(x-3)^2+B(x-3)(x-5)+C(x+5) }{x^3-x^2-21x+45}\\ A(x-3)^2+B(x-3)(x-5)+C(x+5)=1 \)

 

How does it go on?  I don't have a solution.

frown !

 Dec 21, 2020
 #2
avatar+31512 
+1

 

asinus has it right as far as he went, so:

Alan  Dec 21, 2020

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