The Fundamental Theorem of Algebra is really the foundation on which most study of Algebra is built. In simple terms it says that every polynomial has *zeros*. That means that *every* polynomial can be factored and set equal to zero (the Factorization Theorem).

That is an extremely broad statement! **Every** polynomial can be factored? What about functions like \begin{align*}x^{2} = -4\end{align*}? What about crazy big ones, like \begin{align*}x^{6} - 23x^{5} + \frac{1}{246}x^{4}-23x^{3}\end{align*}?

### Watch This

AutenA2Math: 5.7 The Fundamental Theorem of Algebra (Practice)

### Guidance

Here are four important theorems in the study 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.*

*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.

*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*

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(\overline{z_{0}})=0\end{align*}. Where \begin{align*}\overline{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 pairs theorem (also called 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 +xi - x +1 - i - xi + i +1\\ &= x^2 -2x + 2 \end{align}

#### Example A

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*}.

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} + \frac{\sqrt{3}}{2}i \ or \ - \frac{1}{2} - \frac{\sqrt{3}}{2}i \end{align}

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

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

#### Example B

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 of \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 C

What is the multiplicity of the zeros of 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 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*}Roots 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*}.

### Guided Practice

**Problems 1 - 3: Identify or estimate the values of the zeros from the graphs or equations and state their multiplicities.**

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

2) A 4th degree equation:

3) \begin{align*}g(x) = (x^2 + 6x + 9)(x^3 + 6x^2 + 12x + 8)\end{align*}

**Problems 4 - 5: Find a polynomial function with real coefficients that has the given values as its zeros.**

4) \begin{align*}2, 3, -3, 1\end{align*}

5) \begin{align*}1,-2, 1, -2i\end{align*}

*Answers*

1) To identify the roots and their multiplicities:

- First, set the function equal to 0: \begin{align*}(x + 2)(x + 2)(x - 1) = 0\end{align*}
- The roots then are \begin{align*}x = 1\end{align*} and \begin{align*}x = -2\end{align*}
- Since the \begin{align*}x = -2\end{align*} root appears twice, it has a multiplicity of 2, whereas the \begin{align*}x = 1\end{align*} root appears only once, so its multiplicity is 1.

**Note: The graph of this function (shown below) will pass through the axis at the root \begin{align*}x = 1\end{align*} and bounce off the axis at the root \begin{align*}x = -2\end{align*}.**

**If a root has an even multiplicity, it will "bounce" off of the axis, and if it has an odd multiplicity, it will pass through.**

2) Recall that the roots are locations where the graph contacts the \begin{align*}x-\end{align*}axis. The image indicates this happens at \begin{align*}x= -3, -2, \end{align*} and \begin{align*}1\end{align*}.

- Applying the rule from the solution to question 1 tells us that the root "-3" has an even multiplicity since it bounces off of the axis. The other 2 roots have odd multiplicities since they pass through.
- The question specifies that this is a 4th degree equation; therefore, the root "-3" has a multiplicity of 2 and the other two roots displayed each have a multiplicity of 1.

3) First, factor the polynomial:

- \begin{align*}g(x) = (x + 3)^2(x + 2)^3\end{align*}
- The roots are \begin{align*}x = -2\end{align*} and \begin{align*}x = -3\end{align*}
- The multiplicities stem from the multiples of the same binomial, so the root \begin{align*}x = -2\end{align*} has a multiplicity of 3 and \begin{align*}x = -3\end{align*} has a multiplicity of 2.
- A graph of this equation would show the line passing through \begin{align*}x = -3\end{align*} and bouncing off at \begin{align*}x = -2\end{align*}

4) To find a function with the specified zeros:

- Recall that the zeros of a function are the additive inverse of the constant term in each binomial of the factored polynomial, giving:
- \begin{align*}(x - 2)(x - 3)(x + 3)(x - 1)\end{align*}
- Distribute \begin{align*}(x^2 - 5x + 6)(x^2 +2x - 3)\end{align*}
- Multiply the polynomials \begin{align*}(x^4 - 3x^3 -7x^2 +27x - 18)\end{align*}

\begin{align*}\therefore (x^4 - 3x^3 -7x^2 +27x - 18)\end{align*} is the specified polynomial.

5) To identify a function with the given zeros:

- Write out binomials with additive inverses of the given zeros: \begin{align*}(x - 1)(x + 2)(x - 1)(x + 2i)\end{align*}
- Recall that
**all**complex zeros come in pairs, meaning that \begin{align*}(x + 2i)\end{align*} has a conjugate: \begin{align*}(x - 1)(x + 2)(x - 1)(x + 2i)(x - 2i)\end{align*}. The result is: - \begin{align*}(x - 1)(x + 2)(x - 1)(x + 2i)(x - 2i)\end{align*}
- Distribute: \begin{align*}(x^2 + x - 2)(x - 1)(x^2 +4)\end{align*}
- Distribute again: (In this case, the latter two binomials were distributed first.) \begin{align*}(x^2 + x - 2)(x^3 - x^2 + 4x -4)\end{align*}
- Multiply the polynomials: \begin{align*}x^5 + x^3 + 2x^2 - 12x + 8\end{align*}

\begin{align*}\therefore x^5 + x^3 + 2x^2 - 12x + 8\end{align*} is the polynomial

### Explore More

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

- \begin{align*}1, 2, i\end{align*}
- \begin{align*}2, 2, 1-i\end{align*}
- \begin{align*}i, i, 0, 2i\end{align*}
- \begin{align*}1, 1, \left(1-i\sqrt{3}\right)\end{align*}
- \begin{align*}0, 0, 2i\end{align*}
- 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*}.
- 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.**

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

**For problems 11 - 15, sketch the graph, properly indicating multiplicities.**

- \begin{align*}g(x) = x^2(x - 1)(x - 3)(x + 2)(x + 4)^2\end{align*}
- \begin{align*}f(x) = -x^2(x - 3)^3(x - 1)(x - 2)\end{align*}
- \begin{align*}f(x) = -x(x - 1)(x + 2)^3\end{align*}
- \begin{align*}g(x) = x^3(x + 3)^4(x - 2)\end{align*}
- \begin{align*}f(x) = (x + 3)^2(x - 1)(x + 1)\end{align*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 2.14.