ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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A quartic solved through its resolvent cubic

Example

For

$f (x) = x^{4} - 10 x^{2} + 1,$

the resolvent cubic is

$R_{f} (y) = y^{3} + 10 y^{2} - 4 y - 40 = (y - 2) (y + 2) (y + 10) .$

Taking the root $y = 2$ leads to

$x^{2} = 5 \pm 2 \sqrt{6},$

so the four roots are

$\pm (\sqrt{2} + \sqrt{3}), \pm (\sqrt{2} - \sqrt{3}) .$

Facts & Assumptions

Given: The quartic $f (x) = x^{4} - 10 x^{2} + 1$ .

[L1]

For $x^{4} + a x^{3} + b x^{2} + c x + d$ , the resolvent cubic is $y^{3} - b y^{2} + (a c - 4 d) y - (a^{2} d + c^{2} - 4 b d)$ (The coefficient formula and discriminant of the quartic resolvent).

[L2]

The roots of the resolvent are the three sums of products obtained from the three pairings of the four quartic roots (The resolvent cubic of a monic quartic).

Verification

technique · direct

1.1L1algebra

Here $a = 0$ , $b = - 10$ , $c = 0$ , and $d = 1$ , so [L1] gives $R_{f} (y) = y^{3} + 10 y^{2} - 4 y - 40.$ Direct substitution shows that $y = 2$ is a root, and polynomial division yields the displayed factorization.

2.1L2step 1.1algebra

Use the resolvent root $y = 2$ to choose the pairing in [L2]. Seek a factorisation $f (x) = (x^{2} + u x + v) (x^{2} - u x + w) .$ The paired products give $v + w = y = 2$ , while the constant term gives $v w = 1$ , so $v = w = 1$ . Comparing the $x^{2}$ coefficient gives $v + w - u^{2} = - 10$ , hence $u^{2} = 12$ . Taking $u = 2 \sqrt{3}$ yields $f (x) = (x^{2} + 2 \sqrt{3} x + 1) (x^{2} - 2 \sqrt{3} x + 1) .$

3.1step 2.1algebra∎

Solving the two quadratic factors in step 2.1 gives $- \sqrt{3} \pm \sqrt{2}$ and $\sqrt{3} \pm \sqrt{2}$ , which are exactly $\pm (\sqrt{2} + \sqrt{3}), \pm (\sqrt{2} - \sqrt{3}) .$ Thus the resolvent root leads to the four displayed quartic roots.

Depends on

Used by

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Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Ash, Basic Abstract Algebra, quartic examples (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, quartic resolvent examples (standard reference, not scraped)