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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).

 Apr 14, 2019
 #1
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+2

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)

 

 

cool cool cool

 Apr 14, 2019

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