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3 votes
Find the illegal values of b in the fraction: 2b^2 + 3b - 10 over b^2 - 2b -8

2 Answers

6 votes
The illegal values of b in the fraction are:

-2 , 4
User Peterhurford
by
6.8k points
2 votes
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.

denominator = b² - 2b -8 ⇒⇒⇒ by factoring
b² - 2b -8 = 0
(b-4)(b+2) = 0
b - 4 = 0 or b + 2 = 0
∴ b = 4 or -2


the illegal values of b in the fraction = -2 , 4


User Dodgrile
by
7.2k points
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