Find the illegal values of b in the fraction 2b² + 3b - 10 over b² - 2b - 8 .

1 answer:

6 0

(2b² +3b-10) / (b²-2b-8).

This fraction is invalid for all numbers rendering the denominator nil.
We notice that b²-2b-8 is a quadratic equation, so to factorize it let's find its roots:
b'= 4 and x" = -2. FOR THESE VALUES b= - 2 and b = 4 , render the fraction impossible

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