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:

\(\ds z^2 - 4 z + 5\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds z\)

\(=\)

\(\ds \dfrac {4 \pm \sqrt {16 - 4 \times 5} } 2\)

Quadratic Formula: $a = 1$, $b = -4$, $c = 5$

\(\ds \)

\(=\)

\(\ds 2 \pm i\)

Hence the result.

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: Polynomial Equations: $34$