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What does the fundamental theorem of algebra guarantee about a non-constant polynomial?
It has at least one complex root
The theorem says that every non-constant polynomial with complex coefficients has at least one complex root. Apply it repeatedly by factoring out each root as you find it, and you get the stronger version everyone actually uses: a polynomial of degree n has exactly n roots in the complex numbers, counting repeats. This is why the complex numbers feel complete in a way the reals do not. Over the reals a polynomial can have no roots at all, as x squared plus one does, and its graph simply never touches the axis. Allowing i, the square root of minus one, fills every gap, and no further extension of the number system is needed to solve polynomial equations. That property is called being algebraically closed. Note that the theorem is about existence, not method: it promises the roots are there but gives no formula for finding them, and for degree five and above no general formula in radicals exists.
Real polynomials with complex roots always have them in conjugate pairs, which is why an odd-degree real polynomial must cross the axis somewhere.
Why the other answers are wrong
- It has only real roots. Plenty of polynomials have no real roots at all, such as x squared plus one.
- It can always be factorised over the integers. Many polynomials are irreducible over the integers, and roots are often irrational or complex.
- It has an even number of roots. The number of roots equals the degree, which can be odd, as with a cubic.
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