By the rational root theorem, you have the following candidates for roots:
\(\pm\dfrac16,\pm\dfrac13,\pm\dfrac12\)
Plugging in each of these will tell you which one is actually a zero. You'll find that both \(x=\dfrac12\) and \(x=\dfrac13\) both work, which means \(x-\dfrac12\) and \(x-\dfrac13\) are linear factors to the quartic.
To find the remaining factor(s), divide the quartic by the known factors:
\(\dfrac{6x^4+x^3+2x^2-4x+1}{x-\dfrac12}=6x^3+4x^2+4x-2\)
\(\dfrac{6x^3+4x^2+4x-2}{x-\dfrac13}=6x^2+6x+6\)
Since \(6x^2+6x+6=6(x^2+x+1)=0\) has no real roots, you are left with
\(6x^4+x^3+2x^2-4x+1=\left(x-\dfrac12\right)\left(x-\dfrac13\right)(6x^2+6x+6)=(2x-1)(3x-1)(x^2+x+1)=0\)
which has two real zeros at \(x=\dfrac12\), \(x=\dfrac13\). It also has two complex roots at \(x=-\dfrac{1\pm\sqrt3i}2\).
