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# Plz help!

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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
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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?

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.

!

Dec 21, 2020
#2
+32020
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asinus has it right as far as he went, so:

Alan  Dec 21, 2020