Quintic Equation/Examples/z^5 - 2z^4 - z^3 + 6z - 4 = 0
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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:
\(\ds z^2 + 2 z + 2\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \dfrac {-2 \pm \sqrt {2^2 - 4 \times 2} } 2\) | Quadratic Formula: $a = 1$, $b = 2$, $c = 2$ | ||||||||||
\(\ds \) | \(=\) | \(\ds -1 \pm \sqrt {1 - 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1 \pm i\) |
Hence the result.
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Polynomial Equations: $101$