Thread by @MathPrinceps

@jamestanton God wants us to write our polynomials like this:

1ax^0
1ax^1 + 1bx^0
1ax^2 + 2bx^1 + 1cx^0
1ax^3 + 3bx^2 + 3cx^1 + 1dx^0
1ax^4 + 4bx^3 + 6cx^2 + 4dx^1 + 1ex^0,

with their coefficients multiplied by entries drawn from Pascal's triangle. Makes everything nicer.

In particular, when the coefficients a, b, c, ... are in geometric progression (i.e., b = ar, c = ar^2, etc.) then so are these polynomials. Indeed, the nth polynomial is then just

a(x + r)^n.

So in this case. all the roots coincide: all n of them are equal to -r = -b/a.

In the general case, something similar is true. If we define a "shift operation," s, which takes a to b, and b to c, and c to d, and so forth, then we have for the nth polynomial the form

[(x + s)^n]a,

where s^2 means "do s twice," and s^3 means "do s three times," etc.

When the sequence a, b, c, ... is geometric, the shift operation s is exactly the same as multiplying by r, so we get back to the previous result: the nth polynomial is then a(x + r)^n. Otherwise, though, and in general, we have the [(x + s)^n]a form, which is almost as nice.

In particular, one consequence of the [(x + s)^n]a interpretation of God's preferred polynomials is that

ax^2 + 2bx + c = (ax + b)x + (bx + c),

which makes it particularly easy to compute

a(ax^2 + 2bx + c) - (ax + b)^2,

which is the the polynomial counterpart of ac - b^2.

When the sequence a, b, c is geometric, ac - b^2 = 0. In this case, the same is true of the polynomial sequence, as we've seen. But what happens in general?

Well,

a(ax^2 + 2bx + c) = (ax + b)(ax) + a(bx + c),

so

a(ax^2 + 2bx + c) - (ax + b)^2 =

a(bx + c) - b(ax + b).

And of course

a(bx + c) - b(ax + b) = ac - b^2.

That is, we conclude that, in general,

a(ax^2 + 2bx + c) - (ax + b)^2 = ac - b^2.

This is what is known as "completing the square." Since

a(ax^2 + 2bx + c) = (ax + b)^2 + ac - b^2.

If ac - b^2 > 0, then both sides are > 0.

It's hard to imagine making the algebra here any prettier, or more natural. And the result is very striking:

a(ax^2 + 2bx + c) - (ax + b)^2 = ac - b^2

means that the polynomial analog of ac - b^2 turns out to have no x dependence -- and, indeed, is just ac - b^2 itself.

But what's most interesting about all this is that the same ideas apply all the way along the sequence of God's preferred polynomials.

In particular, the next step is to compute

a(ax^3 + 3bx^2 + 3cx + d) - (ax + b)(ax^2 + 2bx + c),

which, it turns out, is equally lovely.

This computation leads to the so-called "depressed cubic" which is a cubic polynomial in ax + b with no quadratic term. And one can continue in this same way to arrive at "depressed" quartics, quintics, etc. They're all polynomials in ax + b, with no subleading term.

And since all these "depressed" polynomials are polynomials in ax + b, we know that when the coefficient sequence a, b, c, ... is geometric, they must take the form (ax + b)^n = 0 -- which means that their coefficients must all vanish in this case. Like ac - b^2, ad - bc, etc.

Indeed, one easily finds, for example, that

a^2(ax^3 + 3bx^2 + 3cx + d) =

(ax + b)^3 +

3(ac - b^2)(ax + b) +

a(ad - bc) - 2b(ac - b^2).

And this "depressed" cubic in ax + b has for its coefficients polynomials in ac - b^2 and ad - bc -- both zero in the geometric case.

I thought you might enjoy this string of ideas, which one can easily make into a "general theory of depression." It goes farther than this, too, as I imagine will not surprise you. Really quite lovely stuff, which puts "completing the square" into its proper algebraic context.

A final remark: in this context, the cubic formula takes on a peculiarly elegant and memorable form: if ax^3 + 3bx^2 + 3cx + d = 0, then

ax + b = cbrt[(ac - b^2)(ar + b)] + cbrt[(ac - b^2)(as + b)],

where r, s are the roots of

(ac - b^2)x^2 + (ad - bc)x + (bd - c^2) = 0.