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).
x^2 - x - 2 factors as (x -2) ( x + 1)
Multiply through by this factorization and we have that
1x + 7 = A( x + 1) + B( x - 2) simplify
1x + 7 = Ax + A + Bx - 2B equate coefficients
1x + 7 = (A + B) x + (A - 2B)
This implies that
A + B = 1 ⇒ - A - B = - 1 (1)
A - 2B = 7 (2)
Add (1) and (2) and we have that
-3B = 6
B = -2
And
A + B = 1
A - 2 = 1
A = 3
So (A, B) = ( 3, -2)