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# 2.8: The Fundamental Theorem of Algebra

Difficulty Level: At Grade Created by: CK-12

## Learning objectives

• Understand the statement of the theorem and how to apply it to various functions
• Understand the conjugate root theorem

## Complex Roots of Polynomial Functions

A polynomial function with real coefficients does not necessarily have real zeros. You may recall that the quadratic equation \begin{align*}f(x)=ax^{2}+bx+c\end{align*}, where \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} are real numbers has real zeros if and only if the discriminant \begin{align*}b^{2}-4ac\geq 0\end{align*}. Otherwise, the quadratic equation has complex roots. For example, the zeros of the quadratic equation

\begin{align*}f(x)=x^{2}+3x+4\end{align*}

can be found by using the quadratic formula ((insert cross reference?)) as follows

\begin{align*}x & = \frac{-3 \pm \sqrt{9-16}}{2}\\ & = \frac{-3 \pm \sqrt{-7}}{2}\\ & = \frac{-3 \pm \sqrt{7}i}{2}\end{align*}

Therefore the zeros are \begin{align*}x=-\frac{3}{2}+\frac{\sqrt{7}}{2}i\end{align*} and \begin{align*}x=-\frac{3}{2}-\frac{\sqrt{7}}{2}i\end{align*}.

Another example, the polynomial function

\begin{align*}f(x)=2x^{3}-5x^{2}+4x-3\end{align*}

can be factored using synthetic division into

\begin{align*}f(x)=(x-3)(2x^{2}+x+1)\end{align*}

Using the quadratic formula on the second term, we find all the zeros to be,

\begin{align*}3,-\frac{1}{4}+\frac{\sqrt{7}}{4}i,-\frac{1}{4}-\frac{\sqrt{7}}{4}i\end{align*}

and thus \begin{align*}f(x)\end{align*} can be written as

\begin{align*}f(x) & = 2x^3 - 5x^2 + 4x -3\\ & = (x-3) \left [ x-\left ( -\frac{1}{4}+\frac{\sqrt{7}}{4}i \right ) \right ]\left [ x-\left ( -\frac{1}{4}-\frac{\sqrt{7}}{4}i \right ) \right ]\end{align*}

Written in this form, \begin{align*}f(x)\end{align*} is a complex polynomial function written in factored form.

Example 1

Write \begin{align*}g(x)=x^{2}+x+1\end{align*} as a complex polynomial in factored form.

Solution

Notice that \begin{align*}g(x)\end{align*} has no real roots. You can see this in the graph of \begin{align*}g(x)\end{align*}, or by looking at the discriminant, \begin{align*}b^{2}-4ac=1-4=-3\end{align*}.

<insert graph>

Using the quadratic formula, the roots of \begin{align*}g(x)\end{align*} are

\begin{align*}x & = \frac{-b \pm \sqrt{b^2-4ac}}{2z}\\ & = \frac{-1 \pm \sqrt{-3}}{2}\\ & = - \frac{1}{2} + i \frac{\sqrt{3}}{2} \ or \ - \frac{1}{2} - i \frac{\sqrt{3}}{2}\end{align*}

Finally, writing \begin{align*}g(x)\end{align*} in factored form,

\begin{align*}g(x)=\left[x-\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\right]\left[x-\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\right]\end{align*}

## Fundamental Theorem of Algebra

From the results above, we conclude that all polynomials are “factorable” as products of linear factors. These factors are based on the zeros of the polynomial functions. We present the following important four theorems in the theory of complex zeros of polynomial functions:

The Fundamental Theorem of Algebra

If \begin{align*}f(x)\end{align*} is a polynomial of degree \begin{align*}n\ge 1\end{align*}, then \begin{align*}f(x)\end{align*} has at least one zero in the complex number domain. In other words, there is at least one complex number \begin{align*}c\end{align*} such that \begin{align*}f(c)=0\end{align*}.

There is no rigorous proof for the fundamental theorem of algebra. Some mathematicians even believe that such proof may not exist. However, the theorem is considered to be one of the most important theorems in mathematics. A corollary of this important theorem is the factorization theorem,

Theorem 2: The Factorization Theorem

If

\begin{align*}f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}\end{align*}

where \begin{align*}a_{n} \ne 0\end{align*}, and \begin{align*}n\end{align*} is a positive integer, then

\begin{align*}f(x)=a_{n}(x-c_{1})(x-c_{2})\cdots(x-c_{0})\end{align*}

where the numbers \begin{align*}c_{i}\end{align*} are complex numbers.

Theorem 3: The \begin{align*}n-\end{align*}Roots Theorem

If \begin{align*}f(x)\end{align*} is a polynomial of degree \begin{align*}n\end{align*}, where \begin{align*}n\ne 0\end{align*}, then \begin{align*}f(x)\end{align*} has, at most, \begin{align*}n\end{align*} zeros.

Notice that this theorem does not restrict that the zeros must be distinct. In other words, multiplicity of the zeros is allowed. For example, the quadratic equation \begin{align*}f(x)=x^{2}+6x+9\end{align*} has one zero, -3, and we say that the function has -3 as a double zero or one zero with multiplicity \begin{align*}k=2\end{align*}. In general, if

\begin{align*}f(x)=(x-c)^{k}q(x)\quad\text{and} \quad q(c)\ne0\end{align*}

then \begin{align*}c\end{align*} is a zero of the polynomial \begin{align*}f\end{align*} and of multiplicity \begin{align*}k\end{align*}. For example,

\begin{align*}f(x)=(x-2)^{3}(x+5)\end{align*}

has 2 as one zero with \begin{align*}k=3\end{align*} and -5 as a zero with \begin{align*}k=1\end{align*}.

## Conjugate Pairs Theorem

Theorem: The Conjugate Root Theorem

If \begin{align*}f(z)\end{align*} is a polynomial of degree \begin{align*}n\end{align*}, with \begin{align*}n\ne0\end{align*} and with real coefficients, and if \begin{align*}f(z_{0})=0\end{align*}, where \begin{align*}z_{0}=a+bi\end{align*}, then \begin{align*}f(z_{0}^{*})=0\end{align*}. Where \begin{align*}z_{0}^{*}\end{align*} is the complex conjugate of \begin{align*}z_{0}\end{align*}.

This is a fascinating theorem! It says basically that if a complex number is a zero of a polynomial, then its complex conjugate must also be a zero of the same polynomial. In other words, complex roots (or zeros) exist in conjugate pairs for the same polynomial. For example, the polynomial function

\begin{align*}f(x)=x^{2}-2x+2\end{align*}

has two zeros: one is the complex number \begin{align*}1+i\end{align*}. By the conjugate root theorem, \begin{align*}1-i\end{align*} is also a zero of \begin{align*}f(x)=x^{2}-2x+2\end{align*}. We can easily prove that by multiplication:

\begin{align*}\left[x-(1+i)\right]\left[x-(1-i)\right] & = (x-1-i)(x-1+i)\\ & = x^2 -x +ix - x +1 - i - ix + i +1\\ & = x^2 -2x + 2\end{align*}

Example 2

What is the form of the polynomial \begin{align*}f(x)\end{align*} if it has the following numbers as zeros: \begin{align*}\frac{-1}{3}, 1-i\end{align*} and \begin{align*}2i\end{align*}?

Solution

Since the numbers \begin{align*}2i\end{align*} and \begin{align*}1+i\end{align*} are zeros, then they are roots of \begin{align*}f(x)=0\end{align*}. It follows that they must satisfy the conjugate root theorem. Thus \begin{align*}-2i\end{align*} and \begin{align*}1-i\end{align*} must also be roots to \begin{align*}f(x)\end{align*}. Therefore,

\begin{align*}f(x)= \left ( x+\frac{1}{3} \right )[x-(1-i)][x-(1+i)][x-(2i)][x-(-2i)]\end{align*}

Simplifying,

\begin{align*}f(x) = \left ( x+\frac{1}{3} \right )(x-1+i)(x-1-i)(x-2i)(x+2i)\end{align*}

After multiplying we get,

\begin{align*}f(x)=\frac{1}{3}(3x^{5}-5x^{4}+13x^{3}-19x^{2}+4x+4)\end{align*}

which is a fifth degree polynomial. Notice that the total number of zeros is also 5.

Example 3

What is the multiplicity of the zeros to the polynomial

\begin{align*}g(x)=x^{4}-6x^{3}+18x^{2}-54x+81\end{align*}

Solution

With the help of the rational zero theorem and the synthetic division, we find that \begin{align*}x=3\end{align*} is a zero of \begin{align*}g(x)\end{align*},

\begin{align*}& \ 3 \ \big ) \overline{1 \ -6 \ \ 18 \ -54 \ \ \ \ 81\;}\\ & \quad \ \ \underline{\downarrow \ \ \ 3 \ -9 \ \ \ \ 27 \ -81}\\ & \quad \ \ 1 \ -3 \ \ \ 9 \ -27 \ \ \ \ \ 0\end{align*}

\begin{align*}g(x)=x^{4}-6x^{3}+18x^{2}-54x+81=(x-3)(x^{3}-3x^{2}+9x-27)\end{align*}

Using synthetic division on the quotient, we find that 3 is again a zero:

\begin{align*}& \ 3 \ \big ) \overline{1 \ -3 \ \ 9 \ -27}\\ & \quad \ \ \underline{\downarrow \ \ \ 3 \ \ \ 9 \ -27}\\ & \quad \ \ 1 \ \ \ \ 0 \ \ \ 9 \ \ \ \ \ 0\end{align*}

or from the \begin{align*}n-\end{align*}Root Theorem (Theorem 3), we write our solution as

\begin{align*}g(x) & = (x-3)(x-3)(x^2+9)\\ & = (x-3)^2(x-3i)(x+3i)\end{align*}

So 3 is a double zero \begin{align*}(k=2)\end{align*} and \begin{align*}3i\end{align*} and \begin{align*}-3i\end{align*} are each of \begin{align*}k=1\end{align*}.

## Exercises

In problems 1-5, find a polynomial function with real coefficients that has the given numbers as its zeros.

1. \begin{align*}1, 2, i\end{align*}
2. \begin{align*}2, 2, 1-i\end{align*}
3. \begin{align*}i, i, 0, 2i\end{align*}
4. \begin{align*}1, 1, \left(1-i\sqrt{3}\right)\end{align*}
5. \begin{align*}0, 0, 2i\end{align*}
6. If \begin{align*}i-1\end{align*} is a root of the polynomial \begin{align*}f(x)=x^{4}+2x^{3}-4x-4\end{align*}, find all other roots of \begin{align*}f\end{align*}.
7. If \begin{align*}-2i\end{align*} is a zero of the polynomial \begin{align*}f(x)=x^{4}+x^{3}-2x^{2}+4x-24\end{align*}, find all other zeros of \begin{align*}f\end{align*}.

In Problems 8-10, determine whether the given number is a zero of the given polynomial. If so, determine its multiplicity.

1. \begin{align*}f(x)=9x^{4}-12x^{3}+13x^{2}-12x+4, x=\frac{2}{3}\end{align*}
2. \begin{align*}f(x)=x^{4}-4x^{3}+5x^{2}-4x+4, x=2\end{align*}
3. \begin{align*}f(x)=3x^{5}-4x^{4}+2x^{3}-\frac{3}{4}x^{2}+2x+12, x=-\frac{2}{3}\end{align*}

1. \begin{align*}x^{4}-3x^{3}+3x^{2}-3x+2\end{align*}
2. \begin{align*}x^{4}-6x^{3}+14x^{2}-16x+8\end{align*}
3. \begin{align*}x^{7}+6x^{5}+9x^{3}+4x\end{align*}
4. \begin{align*}x^{4}-4x^{3}+9x^{2}-10x+4\end{align*}
5. \begin{align*}x^{4}+4x^{2}\end{align*}
6. \begin{align*}-i-1,\sqrt{2},-\sqrt{2}\end{align*}
7. \begin{align*}2i,-3,2\end{align*}
8. Yes; \begin{align*}k=3\end{align*}
9. Yes; \begin{align*}k=2\end{align*}
10. No

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