Question

Find the illegal values of b in the fraction 2b² + 3b - 10 over b² - 2b - 8 .

A. b = −2 and −4
B. b = −2 and 4
C. b = −5, −2, 2, and 4
D. b = −5 and 2

Answer 1

(2b² +3b-10) / (b²-2b-8).

This fraction is invalid for all numbers rendering the denominator nil.
We notice that b²-2b-8 is a quadratic equation, so to factorize it let's find its roots:
b'= 4 and x" = -2. FOR THESE VALUES b= - 2 and b = 4 , render the fraction impossible