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 $A$?
+15001 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.
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