Editing Fundamental theorem of algebra (section)

===Real-analytic proofs===
Even without using complex numbers, it is possible to show that a real-valued polynomial ''p''(''x''): ''p''(0) ≠ 0 of degree ''n'' > 2 can always be divided by some quadratic polynomial with real coefficients.<ref>{{Cite journal |last=Basu |first=Soham |date=October 2021|title=Strictly real fundamental theorem of algebra using polynomial interlacing|journal=[[Australian Mathematical Society|Bulletin of the Australian Mathematical Society]] |language=en |volume=104 |issue=2 |pages=249–255 |doi=10.1017/S0004972720001434 |mr=4308140|doi-access=free }}</ref> In other words, for some real-valued ''a'' and ''b'', the coefficients of the linear remainder on dividing ''p''(''x'') by ''x''<sup>2</sup> − ''ax'' − ''b'' simultaneously become zero.

: <math>p(x) = (x^2 - ax - b) q(x) + x\,R_{p(x)}(a, b) + S_{p(x)}(a, b),</math>

where ''q''(''x'') is a polynomial of degree ''n'' − 2. The coefficients ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') are independent of ''x'' and completely defined by the coefficients of ''p''(''x''). In terms of representation, ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') are bivariate polynomials in ''a'' and ''b''. In the flavor of Gauss's first (incomplete) proof of this theorem from 1799, the key is to show that for any sufficiently large negative value of ''b'', all the roots of both ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') in the variable ''a'' are real-valued and alternating each other (interlacing property). Utilizing a [[Sturm's theorem|Sturm-like]] chain that contain ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') as consecutive terms, interlacing in the variable ''a'' can be shown for all consecutive pairs in the chain whenever ''b'' has sufficiently large negative value. As ''S''<sub>''p''</sub>(''a'', ''b'' = 0) = ''p''(0) has no roots, interlacing of ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') in the variable ''a'' fails at ''b'' = 0. Topological arguments can be applied on the interlacing property to show that the locus of the roots of ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') must intersect for some real-valued ''a'' and ''b'' < 0.