Question 174804
what is the illegal values of b in the fraction 2b^2+3b-10 over b^2-2b-8?
:
The value for b which results in 0 in the denominator, to find that solve for b:
b^2 - 2b - 8 = 0
Factor 
(b-4)(b+2) = 0
Two solutions
b = 4
b = -2
These are the illegal values for b, substitute these values for b in the denominator to see that we have division by 0 for these values