Ask Question
26 August, 02:30

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

+1
Answers (1)
  1. 26 August, 03:43
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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, ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers