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 = −5, −2, 2, and 4
D. b = −2 and 4

Answer 1

1) It seems that the fraction is:

2b^2 + 3b - 10
---------------------
b^2 - 2b - 8

2) Given that the division by 0 is not defined, the illegal values of b are those that make b^2 - 2 b - 8 = 0

3) Solve b^2 - 2b - 8 = 0

You can factor: b^2 - 2b - 8 = (b - 4 ) (b + 2)

=> b = 4 and b =- 2

4) Answer: option D. b = - 2 and 4.