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Algebra 2 · N.CN.C.9

The fundamental theorem of algebra

The real numbers ran out of answers at x^2 + 1 = 0, so i was invented. Will some cubic or quintic need yet another new kind of number?

You need
Adding, subtracting and multiplying complex numbersAlgebra 2The remainder and factor theoremsAlgebra 2
Where you are 3 in Remainder and factor theorems
Algebra9
Polynomials10
Remainder and factor theorems3
The remainder and factor theorems
The fundamental theorem of algebra
The rational root theorem

Why this exists

Every time a new kind of equation turned up with no answer, mathematicians built new numbers. x + 5 = 2 has no answer among the counting numbers, so negatives were admitted. 3x = 2 needed fractions. x^2 = 2 needed irrational numbers, and x^2 + 1 = 0 needed i and the complex numbers.

So the worry is natural: maybe x^5 + x + 3 = 0, or some degree 50 equation, needs a still larger number system. It does not. The complex numbers are enough, for every polynomial equation of every degree, and even when the coefficients themselves are complex. That is the fundamental theorem of algebra, and it turns “how many roots?” into a question with an exact answer: as many as the degree.

Definition

The fundamental theorem of algebra. Every polynomial of degree n \ge 1, with complex coefficients, has at least one complex root.

Real numbers are complex numbers with imaginary part 0, so this covers every polynomial with real or integer coefficients too, and the root it promises may happen to be real.

A root r has multiplicity k when (x - r)^k is a factor of the polynomial and (x - r)^{k+1} is not. Counting r that many times is called counting with multiplicity. On a graph, multiplicity decides whether the curve crosses or touches the axis.

Intuition

The theorem promises only one root. The The remainder and factor theorems turns one root into a factor, and the factor theorem works for a complex number c exactly as for a real one: its proof divides by x - c and never uses that c is real, only the arithmetic of complex multiplication. The same goes for synthetic division by x - c. So a root c_1 of a degree n polynomial p gives

p(x) = (x - c_1)\,q_1(x),

with q_1 of degree n - 1. If n - 1 \ge 1, the theorem applies again, to q_1: it has a root c_2, and q_1(x) = (x - c_2)\,q_2(x). Each step peels off one linear factor and lowers the degree by one. After n steps only a constant is left, the leading coefficient a, and the polynomial is a product of n linear factors. A product of factors is zero only when a factor is, so those n numbers are all the roots there are.

A graph shows only the real roots. Figure 1 is the graph of a degree 5 polynomial that meets the x-axis at one point. The other roots exist; they are just not on the real number line.

x
y
0
y = x^5 + 9x^3
Figure 1The graph of y = x^5 + 9x^3 meets the x-axis once, at 0. The theorem still counts five roots.

Results

Linear factorization. A polynomial p of degree n \ge 1 with leading coefficient a can be written

p(x) = a(x - c_1)(x - c_2)\cdots(x - c_n),

where c_1, \ldots, c_n are complex numbers, not necessarily different.

Exactly n roots. Counted with multiplicity, a polynomial of degree n \ge 1 has exactly n complex roots. Without multiplicity, it has at least 1 and at most n different roots.

Real and non-real. The real roots and the non-real roots together number n, counted with multiplicity.

Real coefficients. When every coefficient is real, the non-real roots come in conjugate pairs, so there is an even number of them. The number of real roots is then n, n - 2, n - 4, and so on down to 0 or 1. In particular, a polynomial of odd degree with real coefficients has at least one real root.

Why it is true

No proof here. Every known proof of the theorem itself needs tools from beyond this course: the calculus of functions of a complex variable, or careful facts about continuous functions. What this course can prove is the step from “at least one root” to “exactly n roots”, and the Intuition section above is that proof: the factor theorem applied n times.

A short history. Mathematicians of the 1600s and 1700s believed the count of n roots and tried to prove it; attempts by d’Alembert, Euler and Lagrange each left gaps. Carl Friedrich Gauss made it the subject of his doctoral thesis in 1799 and returned to it through his life, publishing four proofs in all. The name “fundamental” dates from a time when algebra meant, above all, solving polynomial equations.

Worked examples

Example 1

Counting the roots of a quintic

How many roots does x^5 + 9x^3 = 0 have, counted with multiplicity? List them.

Solution

The polynomial has degree 5, so the theorem says there are exactly 5 roots counted with multiplicity. To find them, factor out the common factor x^3:

x^5 + 9x^3 = x^3(x^2 + 9).

The factor x^3 is (x - 0)^3, so 0 is a root of multiplicity 3. The factor x^2 + 9 is zero when x^2 = -9, that is, x = 3i or x = -3i, since (3i)^2 = 9i^2 = -9.

The roots are 0 (multiplicity 3), 3i and -3i. The count is 3 + 1 + 1 = 5, which matches the degree. Figure 1 shows only the root 0: 3i and -3i are not real, so they leave no mark on the x-axis.

Example 2

Four fourth roots

Find every solution of z^4 = 16.

Solution

Rewrite as z^4 - 16 = 0. A degree 4 equation has 4 roots counted with multiplicity, so the search is not over at z = 2. Factor as a difference of squares, twice where possible:

\begin{aligned} z^4 - 16 &= (z^2 - 4)(z^2 + 4) \\ &= (z - 2)(z + 2)(z^2 + 4). \end{aligned}

The last factor is zero when z^2 = -4, so z = 2i or z = -2i, and z^2 + 4 = (z - 2i)(z + 2i) because -(2i)^2 = -4i^2 = 4. The four roots are 2, -2, 2i and -2i, each once.

Check 2i: (2i)^4 = 16\,i^4 = 16 \cdot 1 = 16. Plotted in the complex plane (Figure 2), the four roots sit at equal distances from 0 and equally spaced around it; Roots of a complex number and roots of unity shows that this always happens for z^n = c with c \ne 0.

\text{real}
\text{imaginary}
0
2
-2
2i
-2i
Figure 2The four roots of z^4 = 16. Only 2 and -2 lie on the real axis.
Example 3

Reading the count from factors

Let p(x) = (x - 1)^3(x^2 + 4)(x + 5). Find its degree, all of its roots with their multiplicities, and how many are real and how many are not.

Solution

Degrees add when polynomials are multiplied, so the degree is 3 + 2 + 1 = 6. There must be 6 roots counted with multiplicity.

The factor (x - 1)^3 gives the root 1 with multiplicity 3. The factor x^2 + 4 gives 2i and -2i, as in Example 2. The factor x + 5 gives -5. The count is 3 + 1 + 1 + 1 = 6.

Real roots: 1, three times, and -5, so 4 real roots counted with multiplicity, but only 2 different real numbers. Non-real roots: 2i and -2i, a conjugate pair. The graph of p has two x-intercepts, crossing at -5 and crossing with a flattened shape at 1, since 3 is odd. Four of the six roots show on it as intercepts; 2i and -2i do not.

Example 4

Knowing when you are finished

Find all the roots of p(x) = x^4 + 5x^3 + 5x^2 - 25x - 26 and write p as a product of linear factors.

Solution

The degree is 4, so we are looking for exactly 4 roots counted with multiplicity. The The rational root theorem limits the rational candidates to \pm 1, \pm 2, \pm 13, \pm 26. Testing:

\begin{aligned} p(2) &= 16 + 40 + 20 - 50 - 26 = 0, \\ p(-1) &= 1 - 5 + 5 + 25 - 26 = 0. \end{aligned}

Synthetic division by x - 2 leaves x^3 + 7x^2 + 19x + 13, and dividing that by x + 1 leaves x^2 + 6x + 13. So

p(x) = (x - 2)(x + 1)(x^2 + 6x + 13).

The quadratic has discriminant 6^2 - 4 \cdot 13 = -16, so the quadratic formula gives

x = \frac{-6 \pm 4i}{2} = -3 \pm 2i.

That makes four roots, 2, -1, -3 + 2i and -3 - 2i, and the theorem says there are no more: there is no need to test \pm 13 or \pm 26. As a product of linear factors,

\begin{aligned} p(x) = {} &(x - 2)(x + 1) \\ &\cdot (x + 3 - 2i)(x + 3 + 2i). \end{aligned}
Example 5

How many real roots can a quintic have?

A polynomial of degree 5 has real coefficients. How many real roots, counted with multiplicity, can it have?

Solution

It has 5 roots in all, counted with multiplicity. Because the coefficients are real, the non-real roots come in conjugate pairs, so their number is even: 0, 2 or 4. (It cannot be 6; there are only 5 roots.) The real roots make up the rest, so there are 5, 3 or 1 of them.

Each case happens:

  • x(x^2 + 1)(x^2 + 4) has roots 0, \pm i, \pm 2i: one real.
  • x(x^2 - 4)(x^2 + 1) has roots 0, \pm 2, \pm i: three real.
  • x(x^2 - 1)(x^2 - 9) has roots 0, \pm 1, \pm 3: five real.

The answer is 1, 3 or 5, never 0, 2 or 4. So every real quintic has at least one real root, and its graph crosses the x-axis at least once.

Example 6

Complex coefficients

Show that 2 is a root of p(x) = x^2 - (3 + i)x + 2 + 2i and find the other root.

Solution

Substitute:

p(2) = 4 - (6 + 2i) + 2 + 2i = 0.

The degree is 2, so there is exactly one more root, counted with multiplicity. Synthetic division by x - 2 on the coefficients 1, -(3 + i), 2 + 2i: bring down 1; then 2 \cdot 1 = 2 and -3 - i + 2 = -1 - i; then 2(-1 - i) = -2 - 2i and 2 + 2i - 2 - 2i = 0. The quotient is x - 1 - i, so

p(x) = (x - 2)(x - (1 + i)).

The roots are 2 and 1 + i. This quadratic has exactly one real root, which a quadratic with real coefficients never can, since its non-real roots would pair up. There is no contradiction: the coefficients -(3 + i) and 2 + 2i are not real, so the pairing rule does not apply, while the count of n roots still does.

Boundaries

  • Degree at least 1. A nonzero constant such as 7 has degree 0 and no roots. The zero polynomial has every number as a root, and is given no degree.
  • Complex numbers are needed. Among the real numbers the theorem is false: x^2 + 1 has no real root. Among the rational numbers it fails too: x^2 - 2 has no rational root.
  • Count with multiplicity. (x - 1)^3 has degree 3 but one different root. “Exactly n” is true only when repeated roots are counted as often as they repeat.
  • Polynomials only. The theorem says nothing about 2^x = 0 or \displaystyle \sqrt{x} = -1, which are not polynomial equations and have no solutions.
  • Existence, not a method. The theorem says the roots exist and how many there are; it does not say how to find them. General formulas built from the coefficients with +, -, \times, \div and radicals (square roots, cube roots and so on) exist for degrees 2, 3 and 4. For degree 5 no such general formula exists: Paolo Ruffini gave an argument in 1799 that had gaps, and Niels Henrik Abel gave a complete proof in 1824. The same holds for every higher degree. Some particular quintics can still be solved this way, but the roots of a quintic like x^5 + x + 3 are in practice found by approximation.

Where this goes next

Because every root gives a factor and there are exactly n of them, a polynomial is fixed by its roots and its leading coefficient; Building a polynomial from its zeros uses this to build a polynomial from given zeros and one point. Conjugate pairs then decide which roots must appear together when the coefficients are real. And z^n = c, a polynomial equation of degree n, has n solutions, which Roots of a complex number and roots of unity finds exactly, spaced evenly around a circle.

Exercises

Counting roots

  1. How many complex roots, counted with multiplicity, does 4x^7 - x + 2 have?
  2. List every root of (x + 3)^2(x^2 + 1)(x - 4) with its multiplicity, and check the total against the degree.
  3. List every root of x^6 + 49x^4 with its multiplicity.
  4. List every root of (x^2 - 9)^2 with its multiplicity.
  5. List every root of (x^2 + 1)^3 with its multiplicity.

Every root, every factor

  1. Solve x^4 - 625 = 0 and write x^4 - 625 as a product of four linear factors.
  2. Find all three roots of x^3 + 125.
  3. Factor x^3 - 3x^2 + 16x - 48 by grouping, then into linear factors.
  4. 1 is a root of x^3 - 3x^2 + 7x - 5. Find the other two.
  5. Find all four roots of x^4 - 2x^3 - x^2 - 4x - 6.
  6. Write 2x^3 - x^2 + 8x - 4 as its leading coefficient times three linear factors.

Real and non-real roots

  1. A polynomial of degree 4 has real coefficients. How many real roots, counted with multiplicity, can it have?
  2. A polynomial of degree 7 with real coefficients has exactly 2 non-real roots. How many real roots does it have, counted with multiplicity?
  3. How many real and how many non-real roots does x^4 + 10x^2 + 9 have?
  4. How many real and how many non-real roots does x^4 - 3x^2 - 4 have?
  5. A polynomial of degree 6 with real coefficients has 1 + i and 3i among its roots. How many real roots can it have?
  6. Show that i is a root of x^2 + (1 - 2i)x - 1 - i, find the other root, and say why the two not being conjugates breaks no rule.

Explaining

  1. Pankaj says that x^2 + 1 has no roots, so the fundamental theorem of algebra is false. What is wrong with his claim?
  2. Can a cubic with real coefficients have 1 and i as roots and no other roots? Explain.
  3. (x - 2)^4 has degree 4. Explain how this fits “exactly 4 roots”.
  4. Folasade is given p(x) = x^5 - x + 1. (a) How many roots does it have, counted with multiplicity? (b) Show that neither rational candidate is a root. (c) Explain why p still has a real root.
  5. Dividing x^2 + 1 by x - i gives the quotient x + i. Explain why the argument “peel off a root, then apply the theorem to the quotient” needs the theorem for complex coefficients, not only real ones.
  6. Modesta claims a polynomial p of degree 3 with p(1) = p(2) = p(3) = p(4) = 0. Explain why no such polynomial exists.
Answers
  1. 7. The degree is 7, and the theorem gives exactly as many roots as the degree.
  2. -3 (multiplicity 2), i, -i and 4 (each 1). The total 2 + 1 + 1 + 1 = 5 matches the degree 2 + 2 + 1 = 5.
  3. x^6 + 49x^4 = x^4(x^2 + 49): 0 (multiplicity 4), 7i and -7i; 4 + 1 + 1 = 6.
  4. (x^2 - 9)^2 = (x - 3)^2(x + 3)^2: 3 (multiplicity 2) and -3 (multiplicity 2).
  5. (x^2 + 1)^3 = (x - i)^3(x + i)^3: i (multiplicity 3) and -i (multiplicity 3).
  6. x = 5, -5, 5i, -5i. x^4 - 625 = (x^2 - 25)(x^2 + 25), which is (x - 5)(x + 5)(x - 5i)(x + 5i).
  7. -5 and \displaystyle \frac{5}{2} \pm \frac{5\sqrt{3}}{2}i. x^3 + 125 = (x + 5)(x^2 - 5x + 25), and the quadratic has discriminant 25 - 100 = -75, so \displaystyle x = \frac{5 \pm 5i\sqrt{3}}{2}.
  8. x^2(x - 3) + 16(x - 3) = (x - 3)(x^2 + 16), which is (x - 3)(x - 4i)(x + 4i).
  9. 1 + 2i and 1 - 2i. Dividing by x - 1 leaves x^2 - 2x + 5, with discriminant 4 - 20 = -16, so \displaystyle x = \frac{2 \pm 4i}{2}.
  10. 3, -1, \displaystyle i\sqrt{2} and \displaystyle -i\sqrt{2}. The rational root test finds 3 and -1; dividing by both leaves x^2 + 2.
  11. \displaystyle 2(x - \frac{1}{2})(x - 2i)(x + 2i). Grouping gives x^2(2x - 1) + 4(2x - 1) = (2x - 1)(x^2 + 4), and \displaystyle 2x - 1 = 2(x - \frac{1}{2}).
  12. 0, 2 or 4. The non-real roots pair up, so there are 0, 2 or 4 of them, and the real roots are the rest of the 4.
  13. 5. All 7 roots are real or non-real, and 7 - 2 = 5.
  14. 0 real and 4 non-real: x^4 + 10x^2 + 9 = (x^2 + 1)(x^2 + 9), with roots \pm i and \pm 3i.
  15. 2 real and 2 non-real: x^4 - 3x^2 - 4 = (x^2 - 4)(x^2 + 1), with roots \pm 2 and \pm i.
  16. 0 or 2. The conjugates 1 - i and -3i are forced, so at least 4 of the 6 roots are non-real. The last 2 are either both real or another conjugate pair.
  17. The other root is -1 + i. Substituting, i^2 + (1 - 2i)i - 1 - i = -1 + i + 2 - 1 - i = 0. Synthetic division by x - i on 1, 1 - 2i, -1 - i: bring down 1; i \cdot 1 = i and 1 - 2i + i = 1 - i; i(1 - i) = 1 + i and -1 - i + 1 + i = 0. The quotient x + 1 - i gives the root -1 + i. The coefficients are not real, so the conjugate pairing rule does not apply.
  18. The theorem promises complex roots, not real ones. x^2 + 1 has the two complex roots i and -i, exactly as many as its degree.
  19. No. With real coefficients, the root i forces its conjugate -i, giving at least three roots 1, i, -i. A cubic has exactly three, so -i must be among them and “no other roots” is false.
  20. The only root is 2, with multiplicity 4, since (x - 2)^4 is the factor x - 2 four times. Counted with multiplicity that is 4 roots.
  21. (a) 5. (b) The only candidates are \pm 1, and p(1) = 1 - 1 + 1 = 1 and p(-1) = -1 + 1 + 1 = 1, neither zero. (c) Its degree is odd and its coefficients are real, so the non-real roots pair up and cannot use all 5 roots; at least one is real (and irrational).
  22. The quotient after peeling off a non-real root can have non-real coefficients, as x + i does. To keep going, the theorem must apply to that quotient, so it has to hold for complex coefficients. A version for real coefficients alone would stop after the first step.
  23. A polynomial of degree 3 has exactly 3 roots counted with multiplicity, so it cannot be zero at four different numbers. Each of 1, 2, 3, 4 would give a factor, and (x - 1)(x - 2)(x - 3)(x - 4) already has degree 4.
kg1128 · updated 2026-10-14 · CC-BY-SA-4.0