Find the illegal values of b in the fraction 2b^2+3b-10/b^2-2b-8

A. b = - 5 and 2

B. b = - 2 and 4

C. b = - 2 and - 4

D. b = - 5, - 2, 2, and 4

Question

24 April, 03:13

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Answers (1)

Know the Answer?

Arielle Clayton · Answer 1

That is the ones that make the denomenator 0

b^2-2b-8=0

solve

factor

(b-4) (b+2) = 0

b=4 and - 2

answer is B

Find the illegal values of b in the fraction 2b^2+3b-10/b^2-2b-8A. b = - 5 and 2B. b = - 2 and 4C. b = - 2 and - 4D. b = - 5, - 2, 2, and 4