SOLUTION: what is the illegal values of b in the fraction 2b^2+3b-10 over b^2-2b-8?

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Question 174804: what is the illegal values of b in the fraction 2b^2+3b-10 over b^2-2b-8?
Found 2 solutions by Earlsdon, ankor@dixie-net.com:
Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
What are the "illegal" values of b in the fraction:
%282b%5E2%2B3b-10%29%2F%28b%5E2-2b-8%29
The "illegal" (sometimes called "excluded") values of b are those values of b that would make the denominator equal zero. So, we set the denominator equal to zero and solve for b.
b%5E2-2b-8+=+0 Solve by factoring.
%28b%2B2%29%28b-4%29+=+0 Apply the zero product rule.
b%2B2+=+0 or b-4+=+0, so that...
highlight%28b+=+-2%29 or highlight%28b+=+4%29 These are the "illegal" values of b.

Answer by ankor@dixie-net.com(12692) About Me  (Show Source):
You can put this solution on YOUR website!
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