Suppose with with $A, B, C \in \mathbb{R}$. What is $A$?
\(\frac{1}{x^3-x^2-21x+45}=\frac{A}{x+5}+\frac{B}{x-3} + \frac{C}{(x - 3)^2}\)
+26367 Suppose with with \(A, B, C \in \mathbb{R}\).
What is \(A\)?
\(\dfrac{1}{x^3-x^2-21x+45}=\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x - 3)^2}\)
\(x^3-x^2-21x+45=(x+5)(x-3)^2\)
\(\begin{array}{|lrcll|} \hline & \mathbf{\dfrac{1}{x^3-x^2-21x+45} } &=&\mathbf{\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x - 3)^2} } \\\\ & \dfrac{1}{(x+5)(x-3)^2} &=&\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x-3)^2} \quad | \quad \times (x+5)(x-3)^2 \\\\ & 1 &=& A(x-3)^2 + B(x+5)(x-3) + C(x+5) \\ \hline x=-5: & 1 &=& A(-5-3)^2 + B(-5+5)(-5-3) + C(-5+5) \\ & 1 &=& A(-8)^2 + B(0)(-8) + C(0) \\ & 1 &=& 64A \\ & \mathbf{A} &=& \mathbf{ \dfrac{1}{64} } \\ \hline \end{array}\)
