The Fundamental Theorem of Algebra

Quick answer

The Fundamental Theorem of Algebra says that a polynomial of degree n ≥ 1 has exactly n zeros in the complex numbers, counted with multiplicity. Each zero r gives a factor x − r, so the polynomial splits into n linear factors. When the coefficients are real, nonreal zeros come in conjugate pairs a ± bi, so a polynomial of odd degree always has a real zero. These facts tell you how many zeros to look for, and they let you write a polynomial from its zeros.

What you'll learn

State the Fundamental Theorem of Algebra and count zeros with multiplicity
Find all real and complex zeros of a polynomial
Use conjugate pairs to find the remaining zeros
Write a polynomial with real coefficients from given zeros

How many zeros?

A quadratic always has two roots, once complex numbers are allowed and a repeated root counts twice. The lesson on quadratics with complex solutions showed why. The same holds for every degree:

Fundamental Theorem of Algebra. Every polynomial of degree $n \ge 1$ has exactly $n$ zeros in the complex numbers, counted with multiplicity.

Put another way, every such polynomial factors completely into linear factors:

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

Only the real zeros show up on a graph, as $x$ -intercepts. The cubic $x^3 - x^2 + 4x - 4$ crosses the $x$ -axis once, at $x = 1$ , yet it has three zeros.

y = x³ − x² + 4x − 4

y = x³ − x² + 4x − 4 crosses the x-axis once

Why the count is exactly n

The deep part of the theorem is that every nonconstant polynomial has at least one complex zero; its proof uses ideas beyond algebra. The exact count then follows from the Factor Theorem. A zero $r_1$ gives a factor $x - r_1$ , and dividing it out leaves a quotient of degree $n - 1$ . That quotient has a zero of its own, which gives another factor, and so on, until the quotient is a constant. There can be no extra zeros, because a product of factors is zero only when one factor is. Each zero peels off one linear factor and lowers the degree by one, so a polynomial of degree $n$ has room for exactly $n$ zeros.

Conjugate pairs

If a polynomial has real coefficients and $a + bi$ is a zero, then its conjugate $a - bi$ is a zero too. Conjugating the equation $p(a + bi) = 0$ changes every $i$ to $-i$ and leaves the real coefficients alone, so it becomes $p(a - bi) = 0$ . The pair multiplies to a real quadratic factor:

\bigl(x - (a + bi)\bigr)\bigl(x - (a - bi)\bigr) = (x - a)^2 + b^2

Because nonreal zeros pair up, a real polynomial of odd degree always has at least one real zero: an odd count cannot be split into pairs.

Worked examples

Common mistakes

Practice problems

How many complex zeros does $3x^5 - x^2 + 7$ have, counted with multiplicity?

Answer

$5$

Full solution

The degree is $5$ , and the Fundamental Theorem of Algebra gives exactly that many zeros.
Find all zeros of $x^3 + 9x$ .

Answer

$0$ , $3i$ and $-3i$

Full solution

$x^3 + 9x = x(x^2 + 9)$ . Then $x = 0$ or $x^2 = -9$ .
Find all zeros of $x^4 - 5x^2 - 36$ .

Answer

$3$ , $-3$ , $2i$ and $-2i$

Full solution

Factor as a quadratic in $x^2$ : $(x^2 - 9)(x^2 + 4)$ . Then $x^2 = 9$ or $x^2 = -4$ .
A polynomial with real coefficients has the zero $2 - i$ . Name another zero.

Answer

$2 + i$

Full solution

Nonreal zeros of a polynomial with real coefficients come in conjugate pairs.
Find all zeros of $x^3 - 2x^2 + x - 2$ .

Answer

$2$ , $i$ and $-i$

Full solution

Group: $x^2(x - 2) + (x - 2) = (x - 2)(x^2 + 1)$ .
Write the polynomial of least degree with real coefficients, leading coefficient $1$ , and zeros $1$ and $2i$ .

Answer

$x^3 - x^2 + 4x - 4$

Full solution

The zero $-2i$ comes with $2i$ : $(x - 1)(x^2 + 4) = x^3 - x^2 + 4x - 4$ .
Write the polynomial of least degree with real coefficients, leading coefficient $1$ , and zeros $3$ and $1 - i$ .

Answer

$x^3 - 5x^2 + 8x - 6$

Full solution

The conjugate $1 + i$ is also a zero, and the pair gives $(x - 1)^2 + 1 = x^2 - 2x + 2$ . Then $(x - 3)(x^2 - 2x + 2) = x^3 - 5x^2 + 8x - 6$ .
Given that $2$ is a zero of $x^3 - 6x^2 + 13x - 10$ , find the other zeros.

Answer

$2 + i$ and $2 - i$

Full solution

Dividing by $x - 2$ leaves $x^2 - 4x + 5$ . The quadratic formula gives $x = \tfrac{4 \pm \sqrt{16 - 20}}{2} = 2 \pm i$ .
A polynomial of degree $4$ with real coefficients has zeros $1$ , $-1$ and $2i$ . What is the fourth zero?

Answer

$-2i$

Full solution

The conjugate of $2i$ must be a zero, and there is room for exactly one more.
A student claims a cubic with real coefficients can have the zeros $1$ , $2$ and $i$ . What went wrong?

Hint

What must come with $i$ ?

Answer

The conjugate $-i$ would also be a zero, giving four zeros for a degree $3$ polynomial. No such cubic exists.

Full solution

With real coefficients, nonreal zeros come in pairs, so $i$ brings $-i$ . A cubic has exactly three zeros, so it cannot hold $1$ , $2$ , $i$ and $-i$ . A cubic whose coefficients are allowed to be nonreal could, though: $(x - 1)(x - 2)(x - i)$ .

Frequently asked questions

What does the Fundamental Theorem of Algebra say?

Every polynomial of degree n ≥ 1 has exactly n complex zeros, counted with multiplicity. Equivalently, it factors into n linear factors.

Does every polynomial have a real zero?

No. x² + 1 has only the zeros i and −i. But a polynomial with real coefficients and odd degree always has at least one real zero.

Why do complex zeros come in conjugate pairs?

For a polynomial with real coefficients, conjugating p(a + bi) = 0 leaves the coefficients unchanged and gives p(a − bi) = 0. So a − bi is a zero whenever a + bi is.

What does counted with multiplicity mean?

A zero that comes from a repeated factor is counted once per factor. In (x − 1)²(x + 2), the zero 1 counts twice, so the cubic has three zeros.

Can a cubic have zeros 1, 2 and i?

Not with real coefficients. The conjugate −i would also have to be a zero, which makes four zeros for a polynomial of degree 3.

What to learn next

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSN.CN.C.9The Complex Number System(+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials.
CCSS.MATH.CONTENT.HSN.CN.C.8The Complex Number System(+) Extend polynomial identities to the complex numbers.
CCSS.MATH.CONTENT.HSA.APR.B.3Arithmetic with Polynomials and Rational ExpressionsIdentify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.

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