List the possible rational roots of 2x^3 + 17x^2 + 23x - 42 = 0
My answer: the possible rational roots are {±1, ±2, ±3, ±6, ±7, ±14, ±21, ±42. ±1/2, ±3/2, ± 7/2, ±21/2}.
Correct?
yes, and guess what?
x = 1 works!
after a quick synthetic division I got
2x^3 + 17x^2 + 23x - 42 = 0
(x-1)(x^2 + 19x + 42) = 0
and that quadratic factors very nicely again.
Yes, this is what you get when you apply the Rational Roots theorem.
But your list is very long, so it is of little use. It is a bit like answering the question by saying that all possible rational roots are the members of Q (as the problem didn't specifically say to list all the roots that the Rational Roots theorem yields).
Incorrect
Yes, your answer is correct. To find the possible rational roots of the polynomial equation 2x^3 + 17x^2 + 23x - 42 = 0, you can apply the Rational Root Theorem. According to the theorem, the possible rational roots are all the divisors of the constant term (42) divided by the divisors of the leading coefficient (2).
In this case, the divisors of 42 are ±1, ±2, ±3, ±6, ±7, ±14, ±21, and ±42, while the divisors of 2 are ±1 and ±2. So, combining all possible combinations of these divisors, you get:
±1, ±2, ±3, ±6, ±7, ±14, ±21, ±42
Additionally, according to the theorem, since the leading coefficient is 2, you need to consider the reciprocals of the divisors of 2 as well:
±1/2, ±3/2, ±7/2, ±21/2
Therefore, the possible rational roots of the given equation are {±1, ±2, ±3, ±6, ±7, ±14, ±21, ±42, ±1/2, ±3/2, ±7/2, ±21/2}, just as you stated.