STRICTLY REAL FUNDAMENTAL THEOREM OF ALGEBRA USING POLYNOMIAL INTERLACING

Abstract Without resorting to complex numbers or any advanced topological arguments, we show that any real polynomial of degree greater than two always has a real quadratic polynomial factor, which is equivalent to the fundamental theorem of algebra. The proof uses interlacing of bivariate polynomials similar to Gauss’s first proof of the fundamental theorem of algebra using complex numbers, but in a different context of division residues of strictly real polynomials. This shows the sufficiency of basic real analysis as the minimal platform to prove the fundamental theorem of algebra.


Introduction
Let us consider a univariate polynomial of degree N with its independent variable as well as fixed coefficients belonging to the field of real numbers R, the smallest field containing all rational numbers and satisfying the least upper bound property [7]: c n x n ∈ R, x ∈ R, c n ∈ R for 0 ≤ n ≤ N. (1.1) The fundamental theorem of algebra (FTA), stated solely in terms of real numbers, asserts the factorisability of any such polynomial into a product of linear (degree one) and quadratic (degree two) polynomials with real coefficients [3]. Since c 0 = 0 forces x to be a trivial factor, we restrict our consideration to f (x) with c 0 0. Ignoring trivial factorisations for N ≤ 2, the FTA can be rephrased in a succinct version for N ≥ 3. S. Basu [2] Even though the minimal algebraic elements for proving the FTA have been elucidated [1,4,8], the minimal analysis platform to prove the FTA has not been characterised. We will show that the basic properties of the real number field suffice to prove Theorem 1.1 by considering the linear residue of the division of N n=0 c n x n by ( which we will derive shortly. Both h N (a, b) and bh N−1 (a, b) + c 0 are bivariate polynomials in a and b. By studying their interlacing [2], we will show that there are A, B ∈ R which satisfy h N (A, B) = 0 and Bh N−1 (A, B) + c 0 = 0, thereby proving Theorem 1.1.
The crucial steps will be to prove the following statements.
(1) There is b < 0 such that, for any fixed b < b, h N (a, b) and bh N−1 (a, b) + c 0 have the maximum possible number of (real) roots in the variable a which alternate with each other. For brevity, this property is abbreviated as interlacing.
(2) For some b > b, interlacing fails either due to the failure in the alternating of the roots, or because h N (a, b) or bh N−1 (a, b) + c 0 has a smaller number of (real) roots than its degree in a.
Other strictly real proofs of the fundamental theorem of algebra (see [5,6]) rely on advanced topological arguments.

Establishing interlacing
Directly checking for interlacing is intractable, so we resort to checking the interlacing of Sturm-like pairs h m−1 (a, b), h m (a, b) (see [9]), where h m (a, b) is the coefficient of x N−m+1 at the mth step of polynomial division of f (x) by x 2 − ax − b, as described below. Without initiating division, The first division step is to replace x 2 with ax + b in the term with the largest power, c N x N , leading to Next we divide the remaining term with highest power, (c N a + c N−1 )x N−1 , to obtain the next term in the remainder: The fundamental theorem of algebra 251 For larger m, we continue to show that with the following recursion formulae, which can be proved by induction: .2)    252 S. Basu [4] PROOF. Suppose that |a| ≥ C √ |b|. Then which completes the proof of the lemma.
LEMMA 2.2. Chooseb ≤ −1 such that, for 0 ≤ n ≤ N, which completes the inductive proof of (2.9). However, for m ′ = 3, we have |c N | = |h 1 (a, b)| < |c N |, which is a contradiction. Therefore, Lemma 2.2 must hold. [5] The fundamental theorem of algebra 253 THEOREM 2.3. For all b <b ≤ −1 and all integers m such that 2 ≤ m ≤ N, the pairs h m −1 (a, b), h m (a, b) are interlacing in a. To recapitulate, for clarity: (1) for fixed b <b, h m (a, b) has m − 1 distinct (real) roots in the variable a, which we designate by α (m) pairs h m−1 (a, b), h m (a, b) for 2 ≤ m ≤ N have interlacing roots, that is, PROOF. Fix b <b ≤ −1. We proceed by induction over m. Interlacing holds vacuously for the base case of m = 2 as h 1 (a, b) has no roots. For the induction step, suppose that 3 ≤ m ≤ N and assume that the result holds for m − 1, that is, 1 (a, b) are interlacing. The number of roots of h m−2 (a, b) is equal to its degree in a and h m−2 (a, b) changes sign at these roots. Therefore, the signs of h m−2 (α (m−1) If all the inequalities are strict, then B belongs to the interlacing set. Then there exists t > 0 such that (α i − t, α i + t) and (β j − t, β j + t) are disjoint. This implies that h N (α i − t, B)h N (α i + t, B) < 0 and (Bh N−1 (α j − t, B) + c 0 )(Bh N−1 (α j + t, B) + c 0 ) < 0. From the continuity of bivariate polynomials, there exists δ with 0 < δ < −B/2 such that h N (a, b) and bh N−1 (a, b) + c 0 do not change sign inside circles of radius δ centred around (α i ± t, B) and (β j ± t, B), respectively. Since h N (α i − t, b) h N (α i + t, b) < 0 for B ≤ b < B + δ, h N (a, b) has at least one root in (α i − t, α i + t) for B ≤ b < B + δ and, since (bh N−1 (β j − t, b) + c 0 bh N−1 (β j + t, b) + c 0 < 0 for B ≤ b < B + δ, bh N−1 (a, b) + c 0 has at least one root in (β j − t, β j + t) for B ≤ b < B + δ. But then interlacing holds for B ≤ b < B + δ, which contradicts the definition of B as the infimum of the noninterlacing set. Therefore, at least two neighbouring quantities in (3.1) must be equal. The value of A is given by these equal quantities at b = B, thereby proving Theorem 1.1.
Additionally, we have proven that we can always find B < 0 satisfying Theorem 1.1. To obtain the complete factorisation of f (x), Theorem 1.1 can be applied onward from N−2 n=0 d n x n until it halts due to exhausting all powers of x.