Use the Rational Zero Theorem to list all possible rational zeros for the given function.?

Question

f(x) = 6x4 + 2x3 -3x2 +2

Can someone please explain how to solve this one?

Answer 1

Hi,
Well, the rational zero theorem says that the POSSIBLE rational zeros are given by +- the factors of 2/6. So, that would be:
+-(1/1, 1/2, 1/3,1/6, 2/1, 2/2, 2/3, 2/6)
=+-(1, 1/2, 1/3, 1/6, 2, 2/3) which doesn't give a very promising situtation.
I'm guessing that you were only asked to find the possible rational roots, because, as a matter of fact, all of the roots are complex.
Persuing the rational zeros a bit further, If we go to Descartes rule of signs we can test for the number of positive rational roots. There are two sign changes in the function, so we either have two positive rational roots or zero.
The same goes for possible negative roots. If we substitute -x in the equation we get this:
f(-x) = 6(-x)^4 +2(-x)^3 -3(-x)^2 +2 which gives us this:
f(-x) 6x^4-2x^3-3x^2 +2 which also has two sign changes. So, you either have two or zero negative roots.
If you plug this eqution into a calculator such as a TI-89 or a computer program, you'll find the following roots:
x1and x2 = .733732+-.344055i
x3 and x4 = .567066+-.431273i
Which, of course, shows that there are no real roots. Also, if you graph it, you'll see that it never crosses or touches the x-axis.


Hope this helps some.
FE

Answer 2

Using The Rational Zero Theorem

Answer 3

Potential rational roots of this function will be: ±1, 3, 1/11, 3/11

Answer 4

Formeng is right on the substance, even if the word "factors" when applied to a quotient is technically a bit sloppy.