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24 April, 03:13

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

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Answers (1)
  1. 24 April, 03:36
    0
    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
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