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Suppose with with $A, B, C \in \mathbb{R}$. What is $A$?

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

 Sep 1, 2019
edited by Guest  Sep 1, 2019
 #1
avatar+109719 
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Please edit this and write it properly.

 Sep 1, 2019
 #3
avatar+109719 
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Thanks for doing as I requested.

Melody  Sep 2, 2019
 #2
avatar+25225 
+2

Suppose with with \(A, B, C \in \mathbb{R}\).

What is \(A\)?
\(\dfrac{1}{x^3-x^2-21x+45}=\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x - 3)^2}\)

 

\(x^3-x^2-21x+45=(x+5)(x-3)^2\)

 

\(\begin{array}{|lrcll|} \hline & \mathbf{\dfrac{1}{x^3-x^2-21x+45} } &=&\mathbf{\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x - 3)^2} } \\\\ & \dfrac{1}{(x+5)(x-3)^2} &=&\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x-3)^2} \quad | \quad \times (x+5)(x-3)^2 \\\\ & 1 &=& A(x-3)^2 + B(x+5)(x-3) + C(x+5) \\ \hline x=-5: & 1 &=& A(-5-3)^2 + B(-5+5)(-5-3) + C(-5+5) \\ & 1 &=& A(-8)^2 + B(0)(-8) + C(0) \\ & 1 &=& 64A \\ & \mathbf{A} &=& \mathbf{ \dfrac{1}{64} } \\ \hline \end{array}\)

 

 

laugh

 Sep 1, 2019

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