Quartic Equation/Examples/6z^4 - 25z^3 + 32z^2 + 3z - 10 = 0

Example of Quartic Equations
The quartic equation:
 * $6 z^4 - 25 z^3 + 32 z^2 + 3 z - 10 = 0$

has solutions:
 * $-\dfrac 1 2, \dfrac 2 3, 2 + i, 2 - 1$

Proof
The integer divisors of $6$ and $-10$ are respectively:
 * $\pm 1, \pm 2, \pm 3, \pm 6$

and
 * $\pm 1, \pm 2, \pm 5, \pm 10$

Thus from Conditions on Rational Solution to Polynomial Equation, the possible rational solutions are:
 * $\pm 1, \pm \dfrac 1 2, \pm \dfrac 1 3, \pm \dfrac 1 6, \pm 2, \pm \dfrac 2 3, \pm 5, \pm \dfrac 5 2, \pm \dfrac 5 3, \pm \dfrac 5 6, \pm 10, \pm \dfrac {10} 3$

By trial, we find that $z = -\dfrac 1 2$ and $z = \dfrac 2 3$ are solutions.

Thus:
 * $\paren {2 z + 1} \paren {3 z - 2} = 6 z^2 - z - 2$

is a factor of $6 z^4 - 25 z^3 + 32 z^2 + 3 z - 10$.

The other factor is found to be:
 * $z^2 - 4 z + 5$

Hence:

Hence the result.