Solve a polynomial through its factors rather than by its degree (#272)

[Rafael-SOWNet]

Closes #272.

What was wrong

The solver read a polynomial equation only as a polynomial: gather the monomials, take the degree, apply the formula for that degree — Cardano at three, Ferrari at four, nothing at all above four. A factorization was never looked for, and one handed over outright was expanded away before the degree was taken.

solve  (x - 1) * (x^2 - 3) = 0    →  { 1, -(-1 + (26 + 18i)^(1/3) + 10/(26 + 18i)^(1/3))/3, … }
solve  x^2 - 3 = 0                →  { sqrt(3), -sqrt(3) }

The second factor is answered exactly when asked on its own. Handed to the solver as part of a product, it went through a cubic instead.

The half the issue does not mention: roots went missing

A sweep over 572 polynomials built from factors whose roots are known in advance (every product and every expansion of two or three factors drawn from a pool of linear and irreducible quadratics) found that ten of them came back short of roots:

x^5 - 2x^3 - x^2 + 2 = 0      is    { 1, sqrt(2), -sqrt(2) }
     = (x-1)(x^2-2)(x^2+x+1)  now   all five, including (-1 ± i·sqrt(3))/2

Above degree four the equation went to the numeric solver whole, and that searches the real axis — so the complex conjugate pair was simply absent. A root set two roots short does not look any different from a complete one.

After the fix, 0 of 572 are short of roots, and no case in either direction returns a value that is not a root.

What changed

• A product is solved factor by factor, since a product is zero exactly where one of its factors is.
• Where the equation is a sum, its rational roots are divided out first, leaving a factor of low enough degree for the existing formulas to answer in one piece.

PolynomialFactoring.TryFactor could not be reused as it stands: it offers a factorization only when the polynomial splits completely into whole linear factors, which is right for the simplifier (a partly factored sum is not a nicer way to write the same thing) and exactly wrong here.

Soundness

Zeroing one factor makes the product zero only where the other factors are defined, so each root is checked against the whole product before it is kept. x * (1/x) is undefined at 0, not zero there. The verification every solver shares cannot catch this — it drops a root only on positive evidence, and a NaN residual is deliberately not evidence (SolveStatement.IsSpurious). Pinned by a test.

Deliberately left alone

a*x^n + b is answered whole by inverting x^n = -b/a, which gives the roots as a + bi. Splitting it at its rational roots leaves the rest to be dug out of a quotient:

	inversion (kept)	splitting (rejected)
x^3 - 8 = 0	2, (-1/2 ± i·sqrt(3)/2)·2	2, (-2 ∓ sqrt(-12))/2

Nothing is gained, so two-term polynomials are excluded, and a test pins that decision. The remaining ugliness there — x^6 - 1 answers a root as 1/2 + i·sin(5/3·π) — is filed separately as #743 rather than folded into this.

Measured

Against 40ce1093:

equation	was	now
x^3 - x^2 - 3x + 3 = 0	-(-1 + (26+18i)^(1/3) + 10/(26+18i)^(1/3))/3, …	{ 1, sqrt(3), -sqrt(3) }	160 ms → 9 ms
x^5 - x^4 - 7x^3 + 7x^2 + 12x - 12 = 0	{ -2, -1.7320508075688774… - 2.3e-15i, 1, … }	{ 1, 2, -2, sqrt(3), -sqrt(3) }	153 ms → 12 ms
x^5 - 2x^3 - x^2 + 2 = 0	{ 1, sqrt(2), -sqrt(2) } — two roots missing	all five	303 ms → 250 ms
(x - 1)(x^2 - 3) = 0	Cardano	{ 1, sqrt(3), -sqrt(3) }	138 ms → 8 ms
x^3 - 6x^2 + 11x - 6 = 0	{ 1, 3, 2 }	{ 1, 2, 3 }	160 ms → 1 ms

4931 unit tests and 130 F# tests pass. The 117-problem corpus stays at 112 solved with 0 wrong, 0 error, 0 timeout, and gets slightly faster overall (12.08 s → 11.26 s), so the extra factoring probe costs nothing measurable on non-polynomial equations.

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