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).
+526 \({x+2 \over x^2-x-2} = {x+2 \over (x-2)(x+1)} = {A\over x-2}+{B\over x+1}\)
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 \(({4\over 3},{-1\over 3})\).
+526 \({x+2 \over x^2-x-2} = {x+2 \over (x-2)(x+1)} = {A\over x-2}+{B\over x+1}\)
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 \(({4\over 3},{-1\over 3})\).
