ch1.4.14-bombelli-rule

Ch 1.4.14 — Of the Rule of Bombelli

Summary: Bombelli’s method reduces any quartic to a cubic by writing the quartic as a difference of two squares. One root of the auxiliary cubic then yields the four roots of the quartic via two quadratics.

Sources: chapter-1.4.14

Last updated: 2026-05-03

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Motivation (§765)

Euler observes that once one root of a quartic is found, the others satisfy a cubic — so solving quartics requires solving cubics. Bombelli (16th-century Italian) devised a systematic reduction. See bombelli-rule for a conceptual summary.

Setup (§766)

The general quartic $x^4+a{x}^3+b{x}^2+cx+d=0$ is assumed equal to:

${\left(x^2+\frac{a}{2}x+p\right)}^2-(qx+r{)}^2=0$

Expanding and matching coefficients with the given quartic yields three conditions:

1. $\frac{a^2}{4}+2p-q^2=b$, i.e., $q^2=\frac{a^2}{4}+2p-b$
2. $ap-2qr=c$, i.e., $2qr=ap-c$
3. $p^2-r^2=d$, i.e., $r^2=p^2-d$

These three equations determine $p$, $q$, $r$.

Reduction to a cubic (§767)

Multiplying the first equation (taken four times: $4q^2=a^2+8p-4b$) by the third ($r^2=p^2-d$) gives $4q^2{r}^2$. Squaring the second equation also gives $4q^2{r}^2=(ap-c{)}^2$. Equating and simplifying:

$\boxed{8p^3-4b{p}^2+(2ac-8d)p-a^{2}d+4bd-c^2=0}$

This is the Bombelli cubic in $p$ (source: §767). It always has at least one real root (since it’s a cubic). Any one value of $p$ suffices.

Recovering $q$ and $r$ (§768)

With $p$ determined:

$q=\sqrt{\frac{a^2}{4}+2p-b},r=\frac{ap-c}{2q}$

(If $q=0$ from the formula, use $r^2=p^2-d$ directly.)

Extracting the four roots (§768)

The equation $(x^2+\frac{a}{2}x+p{)}^2=(qx+r{)}^2$ splits into two quadratics:

$x^2+\frac{a}{2}x+p=qx+r⟹x^2=\left(q-\frac{a}{2}\right)x-p+r$

$x^2+\frac{a}{2}x+p=-qx-r⟹x^2=-\left(q+\frac{a}{2}\right)x-p-r$

Each quadratic yields two roots via the quadratic formula; together they give all four roots.

Worked example 1 (§769–770)

$x^4-10x^3+35x^2-50x+24=0$

Here $a=-10$, $b=35$, $c=-50$, $d=24$. The Bombelli cubic becomes $2p^3-35p^2+202p-385=0$, with roots $p=5,7,\frac{11}{2}$. All three give the same four quartic roots: $x=1,2,3,4$.

Worked example 2 (§771)

$x^4-16x-12=0(a=b=0,c=-16,d=-12)$

The Bombelli cubic reduces to $p^3+12p-32=0$. Substituting $p=2t$ gives $t^3+3t-4=0$, with root $t=1$, so $p=2$. Then $q=2$, $r=4$, and the two quadratics give:

$x=1\pm \sqrt{3}\text{and}x=-1\pm \sqrt{-5}$

Worked example 3 (§772)

$x^4-6x^3+12x^2-12x+4=0$

The Bombelli cubic is $p^3-6p^2+14p-12=0$, root $p=2$. Then $q=1$, $r=0$, giving:

$x=2\pm \sqrt{2}\text{and}x=1\pm \sqrt{-1}$

Related pages

• bombelli-rule
• quartic-equations
• ch1.4.13-quartic-equations
• ch1.4.15-new-method-quartic
• cubic-equations
• cardanos-rule
• quadratic-equations
• equations