Question

Find the zeroes of 6x^4+x^3+2x^2-4x+1

Answer 1

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 $x=-\dfrac{1\pm\sqrt3i}2$.