I was watching the Khan Academy video on the Fundamental Theorem of Algebra when I got confused by something that Sal Khan states. From what I understand, the Theorem says that the complex zeros of a polynomial function always come in pairs because they are conjugates of each other. But exactly what would happen with a 3rd degree polynomial with no real zeros? Why can you not have 3 complex zeros with a 3rd degree polynomial but it is possible with a 4th degree polynomial?
Finally, what then would the zeros of a 3rd degree polynomial with no real zeros be?
This is the video:
https://www.khanacademy.org/math/algebra2/polynomial_and_rational/fundamental-theorem-of-algebra/v/fundamental-theorem-of-algebra-intro
Best Answer
If you have a polynomial with real coefficients, then complex roots always come in conjugate pairs. It is however altogether possible that you could a construct a cubic polynomial with three complex roots -- just take $(x-z_1)(x-z_2)(x-z_3)$ for any complex $z_1,z_2,z_3$. However you will find that when you expand this polynomial out, the coefficients will not be real, unless you picked 3 real roots, or two complex conjugate root and one real root.
The reason why any polynomial with real coefficients of odd degree (including cubics) must have at least one real root is because the highest power term dominates when the variable $x$ gets large. Assuming the coefficient of the highest term $x^n$ is 1, since $(-x)^n=-x^n$ when $n$ is odd, then when $x\to\infty$ and $-\infty$, then your polynomial will tend to $\infty$ and $-\infty$ respectively. Somewhere in between it must cross the $x$ axis by the intermediate value theorem, giving us a real root.
The reason why complex roots come in conjugate pairs, is because complex conjugation respects addition and multiplication. More precisely, if $z=x+iy$ and $w=a+bi$, and $\overline{z}$ represents the complex conjugate of $z$, then:
\begin{align*}\overline{z+w}&=\overline{x+iy+a+bi} \\ &=\overline{(x+a)+(y+b)i} \\ &=(x+a)-(y+b)i \\ &=(x-yi)+(a-bi) \\ &=\overline{z}+\overline{w} \end{align*}
and \begin{align*} \overline{zw}&=\overline{(x+iy)(a+bi)}\\ &=\overline{(xa-yb)+(xb+ya)i} \\ &=(xa-yb)-(xb+ya)i \\ &=(x-yi)(a-bi) \\ &=\overline{z}\cdot\overline{w} \end{align*}
Repeated application of the second identity gives $\overline{z^n}=(\overline{z})^n$. So suppose we know that $x=w$ is a root of a polynomial
$$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$
where $a_1,\ldots,a_n$ are real numbers. Then
\begin{align*} p(\overline{w})&=a_n\overline{w}^n+a_{n-1}\overline{w}^{n-1}+\cdots+a_1\overline{w}+a_0 \\ &= a_n\overline{w^n}+a_{n-1}\overline{w^{n-1}}+\cdots+a_1\overline{w}+a_0 \\ &= \overline{a_nw^n}+\overline{a_{n-1}w^{n-1}}+\cdots+\overline{a_1w}+\overline{a_0} \\ &= \overline{a_nw^n+a_{n-1}w^{n-1}+\cdots+a_1w+a_0} \\ &= \overline{p(w)} \\ &= 0 \end{align*}
meaning that $\overline{w}$, the complex conjugate was also a root, and therefore complex roots come in conjugate pairs. Note that it was crucial that the $a_i$ were all real, because then they are equal to their complex conjugate.