How does depressing a polynomial help you factor it?

Question

I've noticed that a lot of the derivations for the cubic and quartic polynomial solutions require you to depress them, but what about in a case by case scenario?

How does depressing this cubic, $x^3+x^2+1$, into this cubic, $u^3-\frac13u+\frac{31}{27}$, actually help you factor it? It only appears to make the problem more difficult to me.

It's more that putting it into a standard form makes it more amenable to general algorithms. Anyway, you've cherry-picked a somewhat unusual example. $x^3+x^2+1$ is "simpler" to look at than its depressed version; $x^3+10x^2+\frac{31}{1998}x+\frac{5997}{166009}$ isn't really any simpler than its depressed version. — Patrick Stevens, Feb 28 '16 at 13:45
Are you familiar with the cos and cosh substitutions to find a zero of a cubic of the form $x^3+a x+b$? — DanielWainfleet, Feb 28 '16 at 14:55

score 2 · Accepted Answer · answered Mar 07 '16 at 04:15

The reason is a generic one that holds for many problems in general. Whenever we try to solve a problem, it is helpful to try to identify special cases and see if we can reduce every other case to those. This is especially useful when:

There are symmetries. Such as when you assume without loss of generality that $a < b$.
The number of parameters can be decreased. Such as here (see below).
There is a unique canonical form. Such as prenex normal form for first-order formulae.

In this case, originally we want to factorize an arbitrary cubic $a x^3 + b x^2 + c x + d$. There are $4$ natural parameters here, one for each coefficient.

It should immediately occur to you that we can reduce the problem by one parameter by factoring out $a$, getting $a ( x^3 + \frac{b}{a} x^2 + \frac{c}{a} x + \frac{d}{a} )$. You might object that if we started with an integer polynomial now we might have a non-integer polynomial, so it only appears to make the problem more difficult. Whether that is true depends on whether we can solve the resulting 'reduced' problem, which is now to factor a monic cubic (leading coefficient $1$).

We now have $3$ parameters, and it turns out that we can perform a trick to reduce by yet one more. If you view the graph of the cubic function, it is nothing more than a translation along the $x$ axis, which translates the roots and hence transforms the factors in a simple way. Is there a special position we can translate it to that makes it easier to factor? It happens that one particular translation works, namely the one that eliminates the second coefficient. You can see that letting $y = x + \frac{b}{3a}$ causes the equivalent polynomial in $y$ to have zero second coefficient.

We now only need to factor a cubic of the form $y^3 + p y + q$, which has only $2$ parameters. From this point it's probably by trial and error that Cardano discovered that one could make a further substitution $z = y + \frac{p}{3y}$ and somehow the powers of $z$ become multiples of $3$ and we essentially have a quadratic in $z^3$.

The reason that depressing the cubic and quartic work in the stronger sense of producing closed radical formulae for their roots is kind of not explainable, because it crucially involves the non-trivial fact that the general quintic does not have any such closed formula, which in turn comes down to some group theoretic facts about permutations (of roots in this case).

There is a more straightforward method to find the general formula for the cubic and quartic using Lagrange resolvents, that directly exposes the group theoretic structure. In some sense the methods using 'depressing' are ad-hoc, even though they fall under the usual classes of methods we should try, as explained earlier.

But there is a way to show why that z substitution works. Otherwise Cardano must have been groping in the dark for a long time. — Nεo Pλατo, Dec 22 '19 at 11:35
@NeoPlato: There is no need to show it. If you followed the line of thinking in this post, you would see that it is very natural to try what Cardano did, because it is in some sense exactly the same as the previous step, where the common goal is to cancel the next highest power. — user21820, Dec 22 '19 at 16:33
@NeoPlato: You're welcome! In fact, that's precisely how I remember the trick. =) — user21820, Dec 22 '19 at 17:58

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

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Put $x=u-\frac{1}{3}$, $x^3+x^2+1= (u-\frac{1}{3})^{3}+(u-\frac{1}{3})^{2}+1= u^3-\frac{u}{3}+\frac{29}{27}$

For reduced cubic $y^{3}+3py-2q=0$,

$y=\omega^{k}\sqrt[3]{q+\sqrt{q^{2}-p^{3}}}+\omega^{2k}\sqrt[3]{q-\sqrt{q^{2}-p^{3}}}$ for $k=0,1,2$ and $\omega=e^{\frac{2\pi i}{3}}$

Also see

edited Apr 13 '17 at 12:20

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answered Feb 28 '16 at 18:51

Ng Chung Tak

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How does depressing a polynomial help you factor it?

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