Finding all real zeros of the polynomial? Okay, so I need to find all real zeros in this polynomial...

$$f(x) = 2x^3 + x^2 - 13x + 6$$

I know that the first step is to find the factors of 6 and 2, then see which when multiplied by the other coefficients have them add up to equal zero, but none of the factors I tried came out to zero. Is there an easier way to go about doing this???
Your method will in general not find all real, but only all rational zeroes.
If the leading coefficient were 1 instead of 2, all rational zeroes would have to be divisors of 6 (i.e. $\{\pm1, \pm2, \pm3, \pm6\}$).
However with a leading coefficient of 2, one should also check halves of these values (i.e. also {$\pm\frac12, \pm\frac32\}$).
Plugging in $x=2$, you will find that it is in fact a root. By polynomial division you thus obtain a quadratic for the other roots, which you can solve (or you will happen to find the remaining roots also by trying the above candidates).
$$2x^{3}+x^{2}-13x+6=(x-2)(2x^{2}+5x-3)=(x-2)(x+3)(2x-1)$$ 
so the zeros are $-3$, $\frac{1}{2}$ and $2$.