Quintic Equation/Examples/z^5 - 2z^4 - z^3 + 6z - 4 = 0

Example of Quintic Equations
The quintic equation:
 * $z^5 - 2 z^4 - z^3 + 6 z - 4 = 0$

has solutions:
 * $1, 1, 2, -1 \pm i$

Proof
The integer divisors of $4$ are respectively:
 * $\pm 1, \pm 2, \pm 4$

Thus from Conditions on Rational Solution to Polynomial Equation, the possible rational solutions are:
 * $\pm 1, \pm 2, \pm 4$

Performing some trial divisions:

z^4  - z^3 - 2 z^2 - 2 z + 4 --- z - 1 ) z^5 - 2 z^4 -  z^3 + 0 z^2 + 6 z - 4        z^5 -   z^4        ---               -z^4 -   z^3               -z^4 +   z^3              -                     -2 z^3 + 0 z^2                     -2 z^3 + 2 z^2                      -                             -2 z^2 + 6 z                             -2 z^2 + 2 z                              ---                                      4 z - 4                                      4 z - 4                                      ---

Hence we have that $z - 1$ is a factor of $z^5 - 2 z^4 - z^3 + 6 z - 4$, and so $z = 1$ is a solution.

Trying the same thing again:

z^3        - 2 z - 4 - z - 1 ) z^4 - z^3 - 2 z^2 - 2 z + 4       z^4 - z^3        -              0    -2 z^2 - 2 z                   -2 z^2 + 2 z                           -4 z + 4                           -4 z + 4

Hence we have that $z - 1$ is another factor of $z^5 - 2 z^4 - z^3 + 6 z - 4$, and so $z = 1$ is another solution.

Trying another trial division:

z^2 + 2 z + 2 - z - 2 ) z^3 + 0 z^2 - 2 z - 4       z^3 - 2 z^2        ---              2 z^2 - 2 z              2 z^2 - 4 z              ---                      2 z - 4                      2 z - 4                      ---

Hence we have that $z - 2$ is another factor of $z^5 - 2 z^4 - z^3 + 6 z - 4$, and so $z = 2$ is another solution.

We are left with:

Hence the result.