How to depress a cubic

The title of this post sounds like the opening to a bad joke: How do you depress a cubic? … Insert your own punch line.

A depressed cubic is a simplified form of a cubic equation. The odd-sounding terminology suggests that this is a very old idea, older than the current connotation of the word depressed. That is indeed the case. According to Etymonline the term depress originally meant “put down by force, conquer” and the psychological use of the of the word came later. To this day you’ll occasionally hear of a button being depressed.

A depressed cubic equation is depressed in the sense that the quadratic term has been removed. Such an equation has the form

x^3 + cx + d = 0

Once you’ve put the equation in depressed form you’ve conquered the quadratic term.

So how do you put a cubic equation in depressed form? First, divide by the leading coefficient, then use the change of variables [1]

x = t - \frac{b}{3}.

For example, suppose we start with the equation

11x^3 + 19x^2 + 20x + 22 = 0.

We first turn this into

x^3 + \frac{19}{11} x^2 + \frac{20}{11}x + 2 = 0

Then we set x = t – 19/33. This gives us

t^3 + \frac{299}{363} t + \frac{47972}{35937} = 0

which has no quadratic term.

We can use Mathematica to show that this works in general:

    Simplify[x^3 + b x^2 + c x + d /. x -> (t - b/3)]

This returns

t^3 + \left(c-\frac{t b^2}{3}\right) t + \left(d + \frac{2 b^3}{27}-\frac{c b}{3}\right)
This post shows that an analogous change of variables works for higher-order polynomials as well.

Application to elliptic curves

If you look into elliptic curves, you’ll often see them defined as a set of points satisfying

y^2 = x^3 + a x + b

Why no quadratic term? Because you can always remove it using the process above. Well, not quite always. The depression trick doesn’t work for elliptic curves over finite fields of characteristic 2 or 3. If you’d like to read more about this exception, see this post.

How to solve a depressed cubic equation

That is the subject of the next post.


[1] You could do the change of variables first, using x = tb/a. This removes the quadratic term, but leaves the leading coefficient a.

One thought on “How to depress a cubic

  1. Ambrose Bierce used “press”, but he could as easily used “depress” to get both meanings in a single phrase: Piano – A parlor utensil for subduing the impenitent visitor. It is operated by pressing the keys of the machine and the spirits of the audience.

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