You mean rational zeroes. You applied the Rational Roots theorem to find all possible rational roots correctly, but you may have made a mistake when checking if P(x) is zero. But then, it may be that there are no rational zeroes. So, the fact that none of the candidades yields zero is not proof that a mistake has been made.
There are a few ways you can simplify the problem. The reason why you got so many possibilities is because 24 has so many divoisors. So, you could do a change of variables, e.g.:
x = t + 1
Then the constant term of the polynomial as a function of t will be the value at t = 0, which corresponds to x = 1, which is -2. So, that's a huge simplification.
If we define Q(t) = P(t+1)
Then to coefficient of t^4 will be 2, the constant term will be P(1) = -2
The Rational Roots theorem then yields the possible roots:
t = 2, -2, 1, -1, 1/2, -1/2
and therefore:
x = t+1 = 3, -1, 2, 0, 3/2, 1/2
We know that x = 0 is not possible, so the list has shrunk to:
x = 3, -1, 2, 3/2, 1/2
I checked them all (quickly, I could have made mistakes), and none of them worked, so it seems that there are no rational solutions.
I need a little help with this problem.
P(x)=2x^4-17x^3+47x^2-58x+24
(A) Find all possible candidates for real zeroes. I got +-1, +-2, +-3, +4, +-6,+-8, +-12, +-1.5, +-.5, and +-24.
Then I have to find all real zeroes. I tried all of the possible candidates and none of them worked. Could you please tell me where I may have gone wrong and if there is an easier way to find real zeros?
Thank you!
2 answers
Thank you so much for your help!