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

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+0 Answers

#1
+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