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Find constants $A$ and $B$ such that \[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\] for all $x$ such that $x\neq -1$ and $x\neq 2$. Give your answer as the ordered pair $(A,B)$.

 May 26, 2019
 #1
avatar+4330 
+2

Partial Fraction Decomposition does the job. 

 May 26, 2019
 #2
avatar+23786 
+3

Find constants $A$ and $B$ such that \[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\] for all $x$ such that $x\neq -1$ and $x\neq 2$. Give your answer as the ordered pair $(A,B)$.

 

\(\begin{array}{|lrcll|} \hline & \dfrac{x + 7}{x^2 - x - 2} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad x^2 - x - 2 = (x+1)(x-2) \\\\ & \dfrac{x + 7}{(x+1)(x-2)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad \cdot (x+1)(x-2) \\\\ & x + 7 &=& \dfrac{A(x+1)(x-2)}{(x - 2)} + \dfrac{B(x+1)(x-2)}{(x + 1)} \\\\ & \mathbf{x + 7} &=& \mathbf{A(x+1) + B(x-2)} \\ \hline x=2 : & 2 + 7 &=& A(2+1) + B(2-2) \\ & 9 &=& 3 A \\ & A &=& \dfrac{9}{3} \\ & \mathbf{A} &=& \mathbf{3} \\ \hline x=-1 : & -1 + 7 &=& A(-1+1) + B(-1-2) \\ & 6 &=& -3B \\ & B &=& -\dfrac{6}{3} \\ & \mathbf{B} &=& \mathbf{-2} \\ \hline \end{array}\)

 

\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{3}{x - 2} - \dfrac{2}{x + 1}\)

 

 

laugh

 May 27, 2019

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