The Fundamental Theorem of Algebra

The gap in your toolkit

Here is everything you can currently do with the roots of a polynomial. The Factor Theorem says that $p(r) = 0$ if and only if $(x - r)$ divides $p(x)$, so a root and a linear factor are two views of the same thing. The Rational Root Theorem narrows the search for a rational root to a finite list of candidates built from the constant term and the leading coefficient. Synthetic division tests a candidate quickly and, when it works, hands you the quotient.

Notice what none of that does: it never promises that anything is there. The Rational Root Theorem tests candidates and can reject every one of them. Take $p(x) = x^2 - 2$. Its candidate list is $\pm 1, \pm 2$, and not one of them is a root. The Rational Root Theorem has done its job perfectly and told you there is no rational root, but $p$ does have roots, $\sqrt{2}$ and $-\sqrt{2}$. That theorem was simply not built to see them.

A root can escape you in a second way as well. The polynomial $x^2 + 1$ has no real root at all, because $x^2 \ge 0$ for every real $x$, so $x^2 + 1$ is never smaller than $1$. Its roots are $i$ and $-i$, which the real line has no room for. Each time the search failed, the fix was to widen the number system, from the rationals to the reals, then from the reals to the complex numbers. So the honest question is this: does it ever end? Is there some polynomial, perhaps with complex coefficients already, whose roots escape the complex numbers too and force us to invent a fourth kind of number?

The Fundamental Theorem of Algebra

The answer is no, and the reason is the theorem this lesson is named for.

The Fundamental Theorem of Algebra. Every polynomial of degree at least $1$ whose coefficients are complex numbers has at least one complex root.

Read that statement slowly, because four of its words are load-bearing.

“Degree at least $1$.” The hypothesis is not decoration. The constant polynomial $p(x) = 7$ has no root, since $7 = 0$ is never true, and a theorem that claimed otherwise would be false on the first line. Non-constant is exactly the condition.

“Complex coefficients.” The coefficients themselves are allowed to be complex, not merely real. This looks like generosity and is actually a necessity, as the proof below will show.

“At least one complex root.” A real number is a complex number whose imaginary part is $0$, so a real root is one of the possible outcomes. What the theorem refuses to promise is that the root is real, or rational, or anything you could guess. It promises only that some complex number kills the polynomial.

“Has.” The theorem is an existence statement and nothing more. It does not tell you what the root is, and it does not hand you a formula. Finding the root is still your job, with the Rational Root Theorem, division, and the quadratic formula. What changes is that you now know the search is not in vain.

Why algebra alone cannot prove it

The name is misleading, and pleasantly so. The Fundamental Theorem of Algebra is a theorem about algebra that cannot be proved by algebra, and there is a sharp reason why.

Imagine someone shows you a proof that uses nothing but the ordinary rules of arithmetic: adding, multiplying, dividing coefficients, rearranging, comparing degrees. An argument like that never asks what kind of numbers the coefficients are, so it would run word for word inside the rational numbers. But the theorem is false there. The polynomial $x^2 - 2$ has rational coefficients and no rational root. A proof built only from algebraic rules would therefore prove something false, so no such proof can exist. Any correct proof has to use a property that the real numbers have and the rationals do not, and that property is completeness, the absence of gaps. Completeness is what makes continuity work, and continuity is the ingredient every proof of this theorem quietly imports. There are proofs called “algebraic” that squeeze the analysis down to two small facts. Those facts are that every odd-degree real polynomial has a real root and that every complex number has a square root, but both are themselves proved with continuity. The analysis never leaves. It only hides.

So let us see how far algebra does get, and look honestly at the step where it stops. Write a monic polynomial of degree $n \ge 1$, which costs nothing since dividing by the leading coefficient changes no roots:

Set $S = |a_{n-1}| + \cdots + |a_1| + |a_0|$, and let $z$ be any complex number on the circle $|z| = R$, where $R$ is chosen larger than both $1$ and $S$. The modulus of a sum is never more than the sum of the moduli. That is because laying the arrows end to end can never carry you further from the start than the total of their lengths, and usually carries you less far. So the tail of the polynomial obeys

The middle step uses $R > 1$, which makes every power $R^k$ with $k \le n-1$ no bigger than $R^{n-1}$; the last step uses $R > S$. And $R^n$ is precisely $|z^n|$. On a large enough circle, then, the leading term $z^n$ is strictly bigger in size than the whole rest of the polynomial put together. The tail can nudge $p(z)$ away from $z^n$, but it can never nudge it as far as the origin.

Now watch where $p$ sends that circle. As $z$ travels once around $|z| = R$, the point $z^n$ travels around the origin $n$ times. You can see the mechanism in the powers of $i$: multiplying by $i$ is a quarter turn. Sure enough, $i^2 = -1$ is a half turn, $i^3 = -i$ is three quarters of a turn, and $i^4 = 1$ is a full turn. Raising to the $n$th power multiplies the turning by $n$. So the image of the big circle is a closed loop that wraps $n$ times around the origin. By the size bound above, the tail is far too small to drag that loop across the origin and cancel a lap.

Now shrink $R$ toward $0$. The loop deforms continuously as it goes. When $R$ finally reaches $0$ the circle has collapsed to the single point $z = 0$, so its image has collapsed to the single point $p(0) = a_0$. If $a_0 = 0$, then $0$ is a root and there is nothing left to prove. Otherwise the image is now a point sitting away from the origin, enclosing nothing. So the loop begins wrapped around the origin and ends clear of it, and a loop cannot stop enclosing a point without crossing it. At the radius where the crossing happens, some $z$ satisfies $p(z) = 0$. That $z$ is the root the theorem promises.

Every word of that story is believable, and not one word of it is algebra. “Deforms continuously”, “wraps around”, “cannot stop enclosing without crossing”: these are claims about continuity and about the plane, and turning them into proofs is the work of analysis and topology. That is the honest edge of the tools you hold right now. The theorem is true, the picture is the right picture, and the proof belongs to a later course. What you can do today is take the theorem and squeeze everything out of it, which turns out to be a great deal.

From one root to all of them

One root does not sound like much. It is in fact everything, because the Factor Theorem lets you cash it in for a smaller problem.

Feed the Fundamental Theorem a polynomial and it returns a root $r$. Feed that root to the Factor Theorem and it returns a factor $(x - r)$. Divide the factor out and you hold a polynomial of degree one lower, which the Fundamental Theorem applies to just as well. Repeat until nothing is left. Induction is the tool that turns “repeat” into a proof.

That proof is where the phrase “complex coefficients” in the theorem earns its keep. Even when $p$ starts out with real coefficients, the quotient $q$ usually does not. Divide the perfectly real $x^2 + 1$ by the factor $(x - i)$ and you get $x + i$, whose constant term is not real. If the Fundamental Theorem only promised roots for polynomials with real coefficients, the induction would stall on its very first quotient. Stating the theorem for complex coefficients is what lets its conclusion survive being fed back into its own hypothesis.

The factorization also settles a question the theorem never mentioned: are there any other roots hiding somewhere?

Multiplicity, and the honest way to count

The factorization is allowed to repeat a factor, and that repetition is the single most useful piece of bookkeeping in this chapter. It deserves a name and a careful definition.

Definition. Let $p$ be a polynomial and $r$ a complex number. The multiplicity of $r$ in $p$ is the number $m$ for which

for some polynomial $q$. Concretely, $m$ counts how many times you can divide $p$ by $(x - r)$ before the division stops coming out even. A root of multiplicity $1$ is called simple, and a root of multiplicity $2$ or more is called repeated.

Notice that this definition never mentions a factorization, so it cannot depend on which factorization you happened to write down. It does agree with the factorization, though, and here is why. Collect all the copies of $(x - r)$ in the complete factorization, say there are $m$ of them, and let $q(x)$ be $a$ times the product of the other $n - m$ linear factors. Then $p(x) = (x - r)^m q(x)$, and evaluating the leftovers at $r$ gives

a product of the leading coefficient with numbers $r - s_j$ in which every $s_j$ is a root different from $r$. None of those numbers is zero, and $a$ is not zero, so $q(r) \ne 0$. Counting copies of the factor and running the division test give the same $m$ every time.

Now the counting can be stated without a lie in it. Let the distinct roots of $p$ be $s_1, s_2, \ldots, s_k$, with multiplicities $m_1, m_2, \ldots, m_k$. Every one of the $n$ linear factors carries exactly one of those roots, so

Root Counting Theorem. A polynomial of degree $n \ge 1$ with complex coefficients has exactly $n$ roots counted with multiplicity, and at most $n$ distinct roots. It has exactly $n$ distinct roots if and only if every root is simple.

Both halves of that last sentence are worth checking. If there are $n$ distinct roots, then $n$ multiplicities, each at least $1$, add up to $n$, which forces every one of them to equal $1$. And if every multiplicity is $1$, the sum of the $k$ multiplicities is just $k$, so $k = n$.

The three ways a degree-$2$ polynomial can behave are already all of the behaviour there is, only smaller:

You met these three cases in the quadratic chapter, filed under the discriminant $\Delta = b^2 - 4ac$. Their discriminants are $1$, $0$, and $-4$: positive gives two distinct real roots, zero gives one repeated root, negative gives two non-real roots. The middle row is the one that breaks the careless slogan. The polynomial $x^2 - 2x + 1$ has degree $2$ and exactly one distinct root, so “a degree-$n$ polynomial has $n$ roots” is simply false as a sentence about distinct numbers. What every row does share is a right-hand column of length $2$. Counted with multiplicity, over the complex numbers, the count is always the degree.

Complete factorization in practice

Nothing about the theorem changes your method. You still hunt for a root with the Rational Root Theorem, divide it out, and keep going until what remains is small enough to finish with the quadratic formula. What has changed is that you now know the hunt terminates with a complete factorization every time.

Common mistakes