Trinomial Factor

Summary: §144–§146: a real quadratic of the form $p^2-2pqz\cos \varphi +q^2{z}^2$ obtained as the product of a complex linear factor $qz-p(\cos \varphi +i\sin \varphi )$ and its conjugate. Every irreducible real quadratic factor of a polynomial can be written this way.

Sources: chapter9, chapter17

Last updated: 2026-05-11

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Why “trinomial”?

Chapter 2 established that every real polynomial decomposes into real linear factors and real quadratic factors (see factoring-polynomials, complex-conjugate-factors). But finding the complex linear factors is hard, and Euler proposes a different route: search directly for the real quadratic factor in a normalized form, then read off the two complex linear factors from it.

A real quadratic $p-qz+r{z}^2$ has complex linear factors precisely when $4pr>q^2$, i.e. when

$\frac{q}{2\sqrt{pr}}<1.$

Since this dimensionless quantity lies in $(-1,1)$, Euler sets

$\frac{q}{2\sqrt{pr}}=\cos \varphi ,q=2\sqrt{pr}\cos \varphi$

(source: chapter9, §145). To avoid the awkward $\sqrt{pr}$, he absorbs it: rename $p\to p^2$ and $r\to q^2$. The trinomial factor takes the canonical form

$\boxed{p^2-2pqz\cos \varphi +q^2{z}^2}$

and its two (complex) linear factors are

$qz-p(\cos \varphi +i\sin \varphi ),qz-p(\cos \varphi -i\sin \varphi ).$

When $\cos \varphi =\pm 1$, $\sin \varphi =0$ and both linear factors coincide and are real (source: chapter9, §145).

How to find $p$, $q$, $\varphi$ (§146–§149)

If $p^2-2pqz\cos \varphi +q^2{z}^2$ divides a polynomial $\alpha +\beta z+\gamma z^2+\delta z^3+\cdots$, then the polynomial vanishes at both $z=(p/q)(\cos \varphi +i\sin \varphi )$ and $z=(p/q)(\cos \varphi -i\sin \varphi )$. Substitute and use De Moivre: $z^n=(p/q{)}^n(\cos n\varphi \pm i\sin n\varphi )$. Writing $r=p/q$, the two substitutions give

$0=\alpha +\beta r\cos \varphi +\gamma r^2\cos 2\varphi +\delta r^3\cos 3\varphi +\cdots \pm i(\beta r\sin \varphi +\gamma r^2\sin 2\varphi +\delta r^3\sin 3\varphi +\cdots ).$

Adding and subtracting (and dividing the imaginary equation by $2i$) yields two real equations:

$0=\alpha +\beta r\cos \varphi +\gamma r^2\cos 2\varphi +\delta r^3\cos 3\varphi +\cdots ,$

$0=\beta r\sin \varphi +\gamma r^2\sin 2\varphi +\delta r^3\sin 3\varphi +\cdots .$

(source: chapter9, §148). Two equations in two unknowns $r,\varphi$. Each solution gives one trinomial factor; multiple solutions give multiple trinomial factors, and Euler claims they exhaust all of them (source: chapter9, §149).

The rule for any term: $z^n$ contributes $r^n\cos n\varphi$ to the first equation and $r^n\sin n\varphi$ to the second. Recall $\sin 0=0$, $\cos 0=1$, so a constant term $\alpha =\alpha z^0$ contributes $\alpha$ to the first equation and $0$ to the second.

§149 gives a useful generalization: multiplying the first equation by $\cos m\varphi$ and the second by $\sin m\varphi$ and combining produces

$0=\alpha \cos m\varphi +\beta r\cos (m-1)\varphi +\gamma r^2\cos (m-2)\varphi +\cdots ,$

$0=\alpha \cos m\varphi +\beta r\cos (m+1)\varphi +\gamma r^2\cos (m+2)\varphi +\cdots ,$

— any pair of these determines $r$ and $\varphi$.

Why this is a useful normalization

• The form makes the discriminant condition transparent: complex factors $⟺|\cos \varphi |<1$, real factors $⟺\cos \varphi =\pm 1$.
• The arc $\varphi$ is the argument of the complex root $(p/q)(\cos \varphi +i\sin \varphi )$ and $r=p/q$ is its modulus. The trinomial factor is therefore exactly $q^2(z-r{e}^{i\varphi })(z-r{e}^{-i\varphi })$ in modern notation, but Euler avoids invoking $e^{i\varphi }$ here even though eulers-formula is already in his toolbox.
• The form interlocks with de-moivre-formula and the trig power series cleanly: $z^n$ becomes a closed-form polynomial in $\cos \varphi$ and $\sin \varphi$.

This normalization is the engine of every result in Chapter 9 — from the closed-form factorization of $a^n\pm z^n$ (see factorization-of-an-plus-minus-zn) to the infinite product for $\sin z$.

Recovery from a recurrent series (Chapter 17)

Chapter 17 §348–§352 gives the reverse computation: when an algebraic equation has a dominant trinomial factor $1-2pz\cos \varphi +p^2{z}^2$ in its associated rational function, four consecutive terms $P,Q,R,S$ of a recurrent series with the equation’s scale determine both $p$ and $\cos \varphi$ in closed form:

$p=\sqrt{\frac{R^2-QS}{Q^2-PR}},\cos \varphi =\frac{QR-PS}{2\sqrt{(Q^2-PR)(R^2-QS)}}.$

This is the complex-roots analogue of Bernoulli’s method — see trinomial-factor-from-recurrent-series.

Related pages

• factoring-polynomials
• complex-conjugate-factors
• fundamental-theorem-of-algebra
• de-moivre-formula
• factorization-of-an-plus-minus-zn
• sine-infinite-product
• cosine-infinite-product
• chapter-9-on-trinomial-factors
• trinomial-factor-from-recurrent-series
• bernoullis-method-for-roots
• chapter-17-using-recurrent-series-to-find-roots-of-equations