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

 Jun 14, 2021

Best Answer 

 #1
avatar+524 
+2

\({x+2 \over x^2-x-2} = {x+2 \over (x-2)(x+1)} = {A\over x-2}+{B\over x+1}\)

 

Multiplying by (x - 2)(x + 1) both sides, 

\(A(x+1)+B(x-2)=x+2\)

 

Put x = -1  ⇒ -3B = 1

                      B = -1/3

Put x = 2  ⇒  3A = 4

                     A = 4/3 

 

Thus the answer is \(({4\over 3},{-1\over 3})\)

 Jun 14, 2021
 #1
avatar+524 
+2
Best Answer

\({x+2 \over x^2-x-2} = {x+2 \over (x-2)(x+1)} = {A\over x-2}+{B\over x+1}\)

 

Multiplying by (x - 2)(x + 1) both sides, 

\(A(x+1)+B(x-2)=x+2\)

 

Put x = -1  ⇒ -3B = 1

                      B = -1/3

Put x = 2  ⇒  3A = 4

                     A = 4/3 

 

Thus the answer is \(({4\over 3},{-1\over 3})\)

amygdaleon305 Jun 14, 2021

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