Simplify. - 14x^3 / x^3-5x^4 A) - 14/5x-1; where x is not equal to 1/5,0 B) - 14x/1-5x; where x is not equal to 1/5 C) 1-5x/-14x; where x is not equal to 0 D) - 14/1-5x; where x is not equal to 1/5, 0

It's D. Factor x^3 out of the denominator leaving 1 - 5x, then cancel out the x^3 in the numerator with the one you factored out of the denominator. That leaves you with 1 - 5x cannot equal 0. Solve that for x and you get 1/5. However, since we are talking vertical asymptotes and the x^3 is a removable discontinuity, we have to count it as a "problem". Therefore, the other value of x that is not allowed is x = 0.