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One version of the fundamental theorem of algebra states that all non-constant polynomials over $\Bbb{C}$ with complex coefficients have a zero.

The other version states that all non-constant polynomial with real coefficients factors as a product of deg 1 and deg 2 polynomials with negative discriminant.

I understand how to go from the first one to the second. How does the second imply the first? I tried to break up the coefficient into the real and imaginary part and got stuck. How should I proceed?

William Ambrose
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  • Once a non-constant polynomial $f$ is factored into a product of degree $1$ and $2$ polynomials, it must have a zero. If $f$ has at least one degree $1$ factor, say $(x-\alpha)$, then $\alpha$ is a zero of $f$. Otherwise, if all of its factors are degree $2$ polynomials, you could just apply the quadratic formula to obtain a (complex) root. – ilovebulbasaur Apr 28 '20 at 03:55
  • but 2) only applies to polynomials with real coefficient. I need the result for polynomials with complex coefficient. – William Ambrose Apr 28 '20 at 04:03

2 Answers2

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If $P(x)=\sum_{j=0}^n A_jx^j,$ with $n>0\ne A_n,$ let $Q(x)=\sum_{i=0}^n\bar A_jx^j.$

Observe that $P(x)Q(x)$ is a polynomial of positive degree, with real co-efficients.

The "other version" of FTA implies that $P(x)Q(x)$ factors over $\Bbb C$ as a product of deg-$1$ polynomials, so there exists $z\in \Bbb C$ with $P(z)Q(z)=0.$ Now if $P(z)=0,$ then we are done. And if $Q(z)=0$ then $P(\bar z)=\overline {Q(z)}=\bar 0=0.$

DanielWainfleet
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Consider the complex conjugate. Multiply the polynomial with its conjugate to get a polynomial with real coefficient, then apply 2).

William Ambrose
  • 442