3x^3 -4x^2-17x+6=0
I'm just told to solve it, it's in a chapter about polynomials. I know the answers, 3,1/3, and -2. I just don't know how to get them.
trial and error doesn't make sense when you are talking on infinitely many numbers to try on.. i mean, it is a waste of time.. try decreasing the number of numbers you are going to try,,
the rational root test tells you that if you have a polynomial of degree $\displaystyle n$ whose leading coefficient is $\displaystyle a_n$ and whose constant term is $\displaystyle a_0$, then the possible roots of the polynomial is given by $\displaystyle \frac{p}{q}$ where $\displaystyle p$ is a factor of $\displaystyle a_0$ and $\displaystyle q$ is a factor of $\displaystyle a_n$
in your example, $\displaystyle a_0 = 6$ and $\displaystyle a_n = 3$
so $\displaystyle p$ can be equal to $\displaystyle \pm 1, \pm 2, \pm 3, \pm 6$
and $\displaystyle q$ can be equal to $\displaystyle \pm 1, \pm 3$
from here, you may start your trial and error by choosing any combination of $\displaystyle \frac{p}{q}$
after getting one, you may do previous poster's suggestion by taking out and factoring the quadratic factor..