algebra-precalculuspolynomialsroots
According to many sources, the fundamental theorem of algebra states that every polynomial of degree $n$ has exactly n roots. But where's the other root when $b^2-4ac = 0$?
What's the other root of $4x^2 – 32x + 64$, for example? (the real root is 4).
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.
The fundamental theorem of algebra is stated in the complex domain ($\mathbb C$). This reminded us that maybe we could have an answer with complex ($\mathbb C$) viewpoint.
Let $f(z)$ be a polynomial over $\mathbb C$, and $u\in\mathbb C$ be a root of $f(z)$ of multiplicity $m$. This is expressed by an equation: $$f(z)=(z-u)^m g(z)$$ where $g(z)$ is a polynomial that does not have $u$ as its root.
Clearly $f$ brings $u$ to $0$, from one complex plane to another. This characterizes the notion of root geometrically:
Since $f$ is continuous, points near $u$ are mapped to some points near $0$. By complex analysis, a small loop around $u$ should be mapped to another loop winding $m$ times around the origin. The following picture shows the case when $m=2$:
For every complex number $a$ we can assign a color for it according to its angle $\operatorname{arg} a$. The brighter color means that the angle is closer to $0^\circ$:
Let $f(z)=(z+1)^3(z-2)$. For every point $z$ on the complex plane we paint the corresponding color for $\operatorname{arg}f(z)$:
The horizontal axis is real and the vertical one is imaginary. The small green squares indicate the unit length.
When we travel along a path $\gamma$ around $z=-1$, which is the root of multiplicity $3$, the color changes (white to black) three times. This shows that the corresponding path $f(\gamma)$ loops three times around the origin of the codomain. The discussion of the other root $z=2$ is similar.
This characterizes the notion of multiplicity geometrically, so I will try to prove this property (or, at least, give an idea of how it happens, because I'm not really familiar with complex analysis). But so far you can just think that the multiplicity is (or coincides with) how many times the color changes. That's the viewpoint that I wanted to provide.
(In fact, you can count the change of the color around a circle that contains all roots. The result coincides with the degree of $f$. This is related to a proof of the fundamental theorem of algebra.)
I will work on this statement:
Let $f$ has a root $u$ of multiplicity $m$, and $f$ is described as $f(z)=(z-u)^mg(z)$ where $g(z)$ is nonzero at $u$.
Let $\gamma:[0,2\pi]\to\mathbb C$ defined by $\theta\mapsto u+e^{i\theta}$ be a small circle that loops once around $u$ and does not contain other roots inside it. Then $\gamma$ is mapped to a curve $\gamma_2$ by $f$: $$\begin{matrix}\gamma_2:=f\circ\gamma:&[0,2\pi]&\to&\mathbb C\\ & \theta &\mapsto & f(u+re^{i\theta}).\end{matrix}$$ which loops $m$ times around the origin, just as the second picture of this answer.
Since $\gamma$ does not contain other root of $f$, $\gamma$ contains no root of $g$. This means if we let the radius $r$ tend to zero, then $\gamma$ shrinks to $u$ gradually without passing through any root of $g$. Thus the corresponding curve $g\circ \gamma$ can shrink to a point without passing through zero.
(I implicitly used the continuity of $g$.)
Now, let's simplify $\gamma_2$: $$\begin{aligned} \gamma_2(\theta)=f(u+re^{i\theta}) &= ((u+re^{i\theta})-u)^mg(u+re^{i\theta})\\[0.7em] &= re^{im\theta}g(u+re^{i\theta}). \end{aligned}$$
Since $g(u+re^{i\theta})$ can shrink continuously to a point without passing through the origin, the net change of angle of the loop $g(u+re^{i\theta})$ w.r.t the origin is zero.
(I implicitly use some property of the homotopy)
So the net change of angle of $\gamma_2$ is only caused by $re^{im\theta}$, which winds $m$ times around the origin.
Best Answer
The fundamental theorem of algebra requires roots to be counted with multiplicity. In conjunction with the factor theorem it implies that every univariate polynomial with complex coefficients can be broken into linear factors (of the form $x-a$, corresponding to root $a$); the multiple roots are the ones appearing more than once.
Here $4x^2-32x+64=4(x-4)^2$ and the root $4$ appears twice; it is a double root.