6x^5+3x^4-2x^2+3x-8
I really need help with this one. I tried the Rational Zero Theorem but, either my understanding is flawed or it does not work with this polynomial. So, I figured it might be easiest to factor it.
But that came up with:
x(3x^3-1)(2x+1)(2x-3)-8 Which does not appear to be proper form to apply Factor Theorem: c is a zero if (x-c) is a factor of P(x). I definitely have binomial factors here but I think that the -8 negates the theorem.
Does anyone know if I'm wrong or have a solution?
Edit: I figured this out before it got approved (nice one mods).
Solution: The Rational Zero Theorem works on this polynomial. My confusion came from a misinterpretation of the theorem. So, remember: The Rational Zero Theorem will only give

*possible *zeros.
Using the theorem I got:
+/-1/6, +/-1/3, +/-1/2, +/-2/3, +/-1, +/-4/3, +/-2, +/-8/3, +/-4, +/-8
I got confused when I checked some of the values and found that most of them were

**not **zeros of the polynomial.
For anyone wondering, the Rational Zero Theorem basically says to find the factors of the constant term of the polynomial, 8 (-8 yeah but we're using |8| for the theorem) and then find the factors of the leading coefficient: 6.
So now we have the factors of 8: 1, 2, 4, 8 and the factors of 6: 1, 2, 3, 6
Divide the factors of 8 by those of 6 and remove the duplicates. Now the factors of 8 and (1, 2, 4,

/(1, 2, 3, 6) are the possible rational zeros of 6x^5+3x^4-2x^2+3x-8