The Fundamental Theorem of Algebra

Learning goals

State the Fundamental Theorem of Algebra and why its proof needs ideas beyond algebra
Explain how removing one root at a time factors a polynomial into $n$ linear factors
Define multiplicity and find it by dividing by $(x - r)$ repeatedly
Count roots with multiplicity, and distinguish that count from the number of distinct roots
Argue why the complex numbers need no further enlargement

The gap in your toolkit

The Factor Theorem says $p(r) = 0$ 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 then narrows the search for a rational root to a finite candidate list. Synthetic division tests each candidate and hands you the quotient when one works. None of that promises a root exists. It only promises that if a rational root is there, it is on the list.

A root can escape that search in two different ways. Take $p(x) = x^2 - 2$ . Its rational candidates are $\pm 1, \pm 2$ , and not one of them is a root, yet $p$ does have roots, $\sqrt{2}$ and $-\sqrt{2}$ . The Rational Root Theorem was simply never built to see an irrational one. Or take $x^2 + 1$ : since $x^2 \ge 0$ for every real $x$ , the value $x^2 + 1$ is never smaller than $1$ , so no real number is a root at all. Its roots are $i$ and $-i$ . 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 whose roots escape the complex numbers too and force a fourth kind of number into existence?

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.

Four parts of that statement are each doing real work.

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. The proof further down needs that room.

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 satisfies the polynomial.

Has. The theorem only asserts that a root exists. It does not hand you a formula. It does not even guarantee that the Rational Root Theorem, synthetic division, and the quadratic formula will find that root. Those tools succeed only on polynomials built to have accessible roots. What changes is that you now know a root is genuinely there, whether or not those particular tools happen to find it.

Check your understanding

Exactly one of these statements is true. Which one?

A Every polynomial with real coefficients has at least one real root. B The constant polynomial

p(x) = 7

has a complex root. C Every polynomial of degree at least

1

with complex coefficients has at least one complex root. D The theorem tells you the value of the root it promises.

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 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 too. But the theorem is false there. The polynomial $x^2 - 2$ has rational coefficients and no rational root. A proof built only from those rules would therefore prove something false, so no such proof can exist. Every standard proof of this theorem instead leans on continuity, the fact that the real and complex numbers have no gaps to fall through, which the rationals do not have.

That is as far as this course goes. The theorem is true, and turning that picture into a full proof is the work of a later course in analysis. What you can do today is take the theorem as given and see how much it buys you, which turns out to be a great deal.

From one root to all of them

One root does not sound like much, but the Factor Theorem turns it into a smaller problem of the same kind.

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 that factor out and you hold a polynomial of degree one lower, which the Fundamental Theorem applies to again, just as well as it did the first time. Repeat, and each step removes exactly one more root and lowers the degree by exactly one. Starting from degree $n$ , that process stops after $n$ steps, when the last quotient is just the leading coefficient $a$ left over. Collecting every factor pulled out along the way gives the complete factorization

p(x) = a(x - r_1)(x - r_2)\cdots(x - r_n),

where $r_1, \ldots, r_n$ are the $n$ roots, repeated if the same root came up more than once.

That repeated argument is where “complex coefficients” in the theorem earns its keep. Even when $p$ starts out with real coefficients, the quotient after one division 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, this argument would stall on its very first quotient. Stating the theorem for complex coefficients is what lets it apply to its own leftovers, step after step.

The factorization also settles a question the theorem never mentioned: are there any other roots hiding somewhere? Suppose $c$ is any complex number with $p(c) = 0$ . Substituting into the factorization gives $a(c - r_1)(c - r_2)\cdots(c - r_n) = 0$ , and a product of complex numbers is zero only when one of its factors is. The leading coefficient $a$ is never zero, so one of the remaining factors must vanish: $c - r_i = 0$ for some $i$ , that is, $c = r_i$ . Every root of $p$ is therefore one of the $n$ numbers $r_1, \ldots, r_n$ , and by the factorization each of those numbers really is a root. The list is complete: no root of $p$ can hide anywhere else.

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.

A degree-4 polynomial written as four linear factors. The factor (x - 1) appears twice, so the root 1 has multiplicity 2. There are 4 factors and therefore 4 roots counted with multiplicity, but only 3 distinct roots: 1, 2, and -3.

Look at what repeats. The factor $(x - 1)$ appears twice in that factorization, while $(x - 2)$ and $(x + 3)$ each appear once. That count of copies is called the root’s multiplicity: the root $1$ has multiplicity $2$ , and the roots $2$ and $-3$ each have multiplicity $1$ . A root of multiplicity $1$ is called simple, and a root of multiplicity $2$ or more is called repeated.

Definition. Let $p$ be a polynomial that is not the zero polynomial, and let $r$ be a root of $p$ . The multiplicity of $r$ in $p$ is the positive integer $m$ for which

p(x) = (x - r)^m\,q(x), \qquad q(r) \ne 0

for some polynomial $q$ . Concretely, $m$ counts how many times you can divide $p$ by $(x - r)$ before the division stops coming out even. That matches counting copies of $(x - r)$ in the complete factorization: however you count, the same $m$ comes out.

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

m_1 + m_2 + \cdots + m_k = n.

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 $k = n$ .

A degree- $2$ polynomial can behave in exactly three ways, and each one is a pattern that reappears at every higher degree too, just combined with others:

Polynomial	Complete factorization	Distinct roots	Roots with multiplicity
$x^2 - 3x + 2$	$(x-1)(x-2)$	$1$ and $2$ (two)	$1, 2$ (two)
$x^2 - 2x + 1$	$(x-1)^2$	$1$ (one)	$1, 1$ (two)
$x^2 + 1$	$(x-i)(x+i)$	$i$ and $-i$ (two)	$i, -i$ (two)

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.

Check your understanding

How many distinct complex roots does $p(x) = (x - 4)^2(x^2 + 1)$ have?

2

3

4

5

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 a complete factorization genuinely exists, even for a polynomial whose roots these particular tools cannot reach.

Worked example 1 Factor $x^3 - 4x^2 + 9x - 10$ completely

This one works out cleanly because a rational root is on the candidate list. The leading coefficient is $1$ and the constant term is $-10$ , so the Rational Root Theorem offers the candidates $\pm 1, \pm 2, \pm 5, \pm 10$ . Test the small ones:

p(1) = 1 - 4 + 9 - 10 = -4, \qquad p(2) = 8 - 16 + 18 - 10 = 0.

So $2$ is a root, and by the Factor Theorem $(x - 2)$ divides $p$ . Synthetic division with $2$ on the coefficients $1, -4, 9, -10$ brings down $1$ , then produces $-4 + 2 = -2$ , then $9 - 4 = 5$ , then $-10 + 10 = 0$ . The zero remainder confirms the root and the quotient is

x^2 - 2x + 5.

The quotient is a quadratic, so finish it with the quadratic formula rather than hunting further:

x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(5)}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i.

The complete factorization is therefore

p(x) = (x - 2)\bigl(x - (1 + 2i)\bigr)\bigl(x - (1 - 2i)\bigr).

Degree $3$ , three linear factors, three distinct roots, every one of them simple.

Worked example 2 Find the multiplicity of the root $2$ in $x^4 - 5x^3 + 6x^2 + 4x - 8$

Start as usual. The candidates are $\pm 1, \pm 2, \pm 4, \pm 8$ , and $p(1) = 1 - 5 + 6 + 4 - 8 = -2$ while

p(2) = 16 - 40 + 24 + 8 - 8 = 0,

so $2$ is a root. That fact alone tells you the multiplicity is at least $1$ , and nothing more. To learn the real multiplicity you have to keep dividing.

Divide by $(x - 2)$ using synthetic division on $1, -5, 6, 4, -8$ . The remainder is $0$ and the quotient is $x^3 - 3x^2 + 0x + 4$ . Now test the quotient at $2$ , because the question is whether $(x-2)$ goes in again:

2^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0.

It does. Dividing $1, -3, 0, 4$ by $2$ leaves remainder $0$ and quotient $x^2 - x - 2$ . Test once more:

2^2 - 2 - 2 = 0,

so $(x - 2)$ divides a third time, leaving the quotient $x + 1$ . Test that quotient at $2$ as well, and this time $2 + 1 = 3$ , which is not zero, so the divisions stop here. Assembling everything,

p(x) = (x - 2)^3(x + 1).

The root $2$ has multiplicity $3$ and the root $-1$ has multiplicity $1$ . Their multiplicities add to $3 + 1 = 4$ , which is the degree, exactly as the Root Counting Theorem demands. The polynomial has four roots counted with multiplicity and only two distinct ones.

Check your understanding

A polynomial $p$ has degree $5$ and complex coefficients, and its only distinct roots are $-1$ and $3$ . If $-1$ has multiplicity $2$ , what is the multiplicity of $3$ ?

1

2

3

5

Common mistakes

Saying “a degree $n$ polynomial has $n$ roots” and stopping there. Over the reals the claim is false, since $x^2 + 1$ has no real root, and counting distinct roots it is false too, since $x^2 - 2x + 1 = (x-1)^2$ has only the root $1$ . The true sentence is “exactly $n$ roots, counted with multiplicity, over the complex numbers.”
Expecting the theorem to hand you the roots. It is an existence statement. It never produces a root. It never even guarantees that the Rational Root Theorem and the quadratic formula will land on one, since those tools only reach polynomials built to have accessible roots.
Guessing a multiplicity instead of testing it. One successful division by $(x - r)$ proves the multiplicity is at least $1$ and nothing more. Keep dividing, and stop only when a remainder comes out nonzero.
Applying the theorem to a constant. Degree at least $1$ is a hypothesis, not decoration. The polynomial $p(x) = 7$ has no roots whatsoever, and the theorem never claimed it did.

Key takeaways

The Fundamental Theorem of Algebra. Every polynomial of degree at least $1$ with complex coefficients has at least one complex root.
Complete factorization. Removing one root and one factor at a time gives $p(x) = a(x - r_1)(x - r_2)\cdots(x - r_n)$ , where $a$ is the leading coefficient, and those $n$ numbers are all of the roots there are.
Multiplicity. The root $r$ has multiplicity $m$ when $p(x) = (x - r)^m q(x)$ with $q(r) \ne 0$ . It counts how many times $(x - r)$ appears as a factor, and you find it by dividing repeatedly until the remainder is nonzero.
Counting. The multiplicities of the distinct roots sum to $n$ . A degree- $n$ polynomial has exactly $n$ roots counted with multiplicity and at most $n$ distinct roots, with equality exactly when every root is simple.
The complex numbers are enough. No polynomial equation with complex coefficients, however unpleasant, can force you to invent a new kind of number.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Polynomial factorization (Khan Academy)
Fundamental Theorem of Algebra (Math is Fun)
Zeros of polynomial functions (free textbook) (OpenStax)
The fundamental theorem of algebra (history) (MacTutor)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

A sketch of what the proof actually uses

The core lesson shows that no proof built only from algebra’s ordinary rules can work: such a proof would run inside the rational numbers too, and the theorem is false there. Here is a sketch of what a real proof leans on instead. It is not a full proof, since it depends on ideas about continuous change that belong to a later course in analysis, but it shows where those ideas do their work.

Write a monic polynomial of degree $n \ge 1$ , which costs nothing since dividing by the leading coefficient changes no roots. Call the variable $z$ rather than $x$ , as a reminder that $z$ now ranges over the whole complex plane, not just the real line:

p(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0.

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, because laying the arrows end to end can never carry you further from the start than the total of their lengths. So the tail of the polynomial obeys

\begin{aligned} |a_{n-1}z^{n-1} + \cdots + a_0| &\;\le\; |a_{n-1}|R^{n-1} + \cdots + |a_1|R + |a_0| \\ &\;\le\; S\,R^{n-1} \;<\; R \cdot R^{n-1} = R^n. \end{aligned}

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. Multiplying by $i$ is a quarter turn, $i^2 = -1$ is a half turn, and 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.

A schematic of what p does to circles, not the exact shape for any one polynomial. On a large circle the leading term dominates, so the image loops around the origin. As the radius shrinks to 0 the image shrinks to the single point p(0), which does not enclose the origin. Somewhere in between, an image loop must pass through the origin, and there sits a root.

Every word of that sketch depends on ideas about continuous change: that a loop can deform gradually, and that it cannot stop enclosing a point without sweeping across it. Turning those words into a proof is the work of analysis and topology, courses still ahead of you.

The complete factorization proof, written out by induction

The core lesson explains the complete factorization by repeating “find a root, divide it out” until nothing is left. Turning “repeat” into a proof takes mathematical induction, proving a claim first for the smallest case and then showing it survives one more step, forever.

A polynomial of degree $n \ge 1$ splits into exactly $n$ linear factors over the complex numbers#

Let $p$ have degree $n \ge 1$ , complex coefficients, and leading coefficient $a$ . We prove by induction on $n$ that there are complex numbers $r_1, r_2, \ldots, r_n$ , not necessarily different from each other, with $p(x) = a(x - r_1)(x - r_2)\cdots(x - r_n)$ .

For the base case, take $n = 1$ , so $p(x) = ax + b$ with $a \ne 0$ . Factoring the leading coefficient out gives $p(x) = a\left(x + \frac{b}{a}\right)$ , which is $a\bigl(x - (-\tfrac{b}{a})\bigr)$ . That is the required shape, with the single root $r_1 = -\frac{b}{a}$ .

Now suppose the claim holds for every polynomial of degree $n - 1$ with complex coefficients, and let $p$ have degree $n \ge 2$ and leading coefficient $a$ . By the Fundamental Theorem of Algebra, $p$ has at least one complex root; call it $r_n$ . By the Factor Theorem, $(x - r_n)$ divides $p$ , so there is a polynomial $q$ with complex coefficients satisfying $p(x) = (x - r_n)\,q(x)$ .

Compare degrees on both sides. Degrees add when polynomials multiply, so $n = 1 + \deg q$ , giving $\deg q = n - 1$ . Compare leading terms as well. The leading term of $(x - r_n)q(x)$ is $x$ times the leading term of $q$ , so $q$ must have the same leading coefficient $a$ that $p$ does.

Since $n \ge 2$ , the quotient has degree $n - 1 \ge 1$ , and its coefficients are complex, so the induction hypothesis applies to it exactly: $q(x) = a(x - r_1)(x - r_2)\cdots(x - r_{n-1})$ for some complex numbers $r_1, \ldots, r_{n-1}$ . Substituting this into $p(x) = (x - r_n)q(x)$ gives

p(x) = a(x - r_1)(x - r_2)\cdots(x - r_{n-1})(x - r_n),

which is the claim for degree $n$ . The induction is complete, and every polynomial of degree $n \ge 1$ factors into exactly $n$ linear factors over the complex numbers.

Building a polynomial backwards from its roots

The factorization runs backwards as well as forwards. Given the roots and their multiplicities, you can write down a polynomial that has exactly those roots, and no others, by writing one linear factor per copy.

Worked example 3 Build the polynomial of degree $3$ with leading coefficient $2$ whose roots are $3$ (multiplicity $2$ ) and $-i$

Run the factorization backwards. Each root of multiplicity $m$ contributes the factor $(x - r)$ repeated $m$ times, and the leading coefficient goes out front:

p(x) = 2(x - 3)^2\bigl(x - (-i)\bigr) = 2(x - 3)^2(x + i).

Check the count before expanding. Two factors of $(x - 3)$ plus one factor of $(x + i)$ is three linear factors, so the degree is $3$ , and $2 + 1 = 3$ multiplicities as required.

Expanding, $(x - 3)^2 = x^2 - 6x + 9$ , and multiplying by $(x + i)$ gives

(x^2 - 6x + 9)(x + i) = x^3 + (i - 6)x^2 + (9 - 6i)x + 9i.

Multiplying through by the leading coefficient $2$ ,

p(x) = 2x^3 + (2i - 12)x^2 + (18 - 12i)x + 18i.

Look at what came out. The coefficients are not real, and that is allowed: the Fundamental Theorem and the Root Counting Theorem never asked for real coefficients. Notice too that $-i$ is a root while $i$ is not. A non-real root does not automatically drag its conjugate along; that only happens under an extra hypothesis, which is exactly what the next lesson is about.

A bit of history (optional)

Everybody believed this theorem long before anybody could prove it, and the failures are the interesting part.

The eighteenth century is a graveyard of attempted proofs. Pierre-Simon Laplace, a French scientist better known for his work on the planets, published one in 1795. Several others tried in the same decades. Every one of them failed, and they tended to fail in the same place.

The arguments began by supposing that the roots existed somewhere, in some larger system nobody had bothered to describe. From there they showed that such roots had to be complex. Read that twice. Whether the roots exist at all was the entire question, and each proof had quietly helped itself to the answer.

Carl Friedrich Gauss, a German student of twenty-two, made exactly that objection in his doctoral thesis of 1799, and supplied a proof of his own. The irony is that his argument had a hole in it too. At one step he assumed that two curves drawn a certain way must cross, which is obvious to look at and genuinely hard to establish. More than a century passed before anybody closed the gap.

Notice what all this labor bought, and what it did not: a guarantee that the roots exist, with no way to find a single one of them. Existence is one question, and finding an actual root is quite another.