+1207 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 \((A,B)\).
+129852 We can solve this by the method of partial fractions
Note that the first denominator can be factored as ( x -2) (x+ 1)
Multiply everything through by (x-2) (x + 1) and we get that
x + 7 = A( x + 1) + B( x - 2) simplify
x + 7 = (A + B)x + (A - 2B) equating terms, we have that
A + B =1 (1)
A - 2B = 7 → -A + 2B = -7 (2)
Add (1) and (2) and we have that
3B = -6
B = -2
So A -2 = 1
A = 3
(A,B) = ( 3, -2)
