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).

Guest Apr 14, 2019

#1**+1 **

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)

CPhill Apr 14, 2019