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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≠-1 and x≠2.

 Jul 8, 2020
 #1
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Notice that x-2 and x+1 are factors of x^2-x-2.

 

\(\frac{x+7}{x^2-x-2}=\frac{a}{x-2}+\frac{b}{x+1}\\ \frac{x+7}{x^2-x-2}=\frac{a(x+1)}{x^2-x-2}+\frac{b(x-2)}{x^2-x-2}\\ x-7=a(x+1)+b(x-2)\\ x-7=ax+a+bx-2b\\\)

ax+bx must add up to the x term on the left side, which is x. a-2b must subtract to the constant on the left side, -7. We then have a system of equations.

 

a+b=1

a-2b=-7

 

We find that a= -5/3 and b=8/3.

 Jul 8, 2020
 #2
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+2

Thanks for the help!! I believe a-2b should equal 7, though, not -7.

mathmathj28  Jul 8, 2020
edited by mathmathj28  Jul 10, 2020

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