SOLUTION: help find the illegal values of b in the fraction {{{(2b^2+3b-10)/(b^2-2b-8)}}}

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Question 40079: help
find the illegal values of b in the fraction %282b%5E2%2B3b-10%29%2F%28b%5E2-2b-8%29
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
"Illegal" values are those values of b which make the denominator of the fraction equal zero...it is forbidden to divide by zero, ya know...
so let's see what those values are...
b^2 - 2b - 8 = 0
(b - 4)(b + 2) = 0
b = 4 or b = -2
Thus we're not allowed to have b = 4 or b = -2...we say they are not in the domain of this expression...