You have learned that a quadratic has at most two real zeroes and a cubic has at most three real zeros. You may have noticed that the number of real zeros is always less than or equal to the degree of the polynomial.

Concept Extension: Consider the graph or a parabola and draw three parabolas on three separate x and y axes crossing the x axis 0, 1, or 2 times. Explain the x-intercepts found vs. possible roots of your graph. Which values are complex numbers with an imaginary part? Do these zeros cross the x-axis? Can you see them visually on a graph?

#### Watch This

http://www.youtube.com/watch?v=NeTRNpBI17I James Sousa: Complex Numbers

#### Guidance

A real number is any rational or irrational number. When a real number is squared, it will always produce a non-negative value. Complex numbers include real numbers and another type of number called imaginary numbers. Unlike real numbers, imaginary numbers produce a negative value when squared. The square root of negative one is defined to be the imaginary number \begin{align*}i\end{align*}

\begin{align*}i=\sqrt{-1}\end{align*}

Complex numbers are written with a real component and an imaginary component. All complex numbers can be written in the form \begin{align*}a+bi\end{align*}

The **Fundamental Theorem of Algebra** states that an \begin{align*}n^{th}\end{align*}**Fundamental Theorem of Algebra** to be applied, an nth degree polynomial with real or complex coefficients must be considered.

Multiplicity refers to when a root counts more than once. For example, in the following function the only root occurs at \begin{align*}x=3\end{align*}

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

The Fundamental Theorem of Algebra states that this \begin{align*}2^{nd}\end{align*}

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

This function has 6 roots. The first two roots \begin{align*}x=1\end{align*} and \begin{align*}x=-2\end{align*} have multiplicities of 1 because the power of each of their binomial factors is 1. The third root \begin{align*}x=3\end{align*} has a multiplicity of 4 because the power of its binomial factor is 4. Keep in mind that all polynomials can be written in factorized form like the above polynomial, due to a theorem called the **Linear Factorization Theorem**. It states if f is a polynomial with degree *n* where n is greater than zero, then f has precisely *n* linear factors. In other words any polynomial can be factored in to one or more (mx + b) pieces.

**Example A**

Identify the zeroes of the following complex polynomial.

\begin{align*}f(x)=x^2+9\end{align*}

**Solution:** Set \begin{align*}y=0\end{align*} and solve for \begin{align*}x\end{align*}. This will give you the two zeros.

\begin{align*}0 &= x^2+9 \\ -9 &= x^2 \\ \pm 3i &= x\end{align*}

Notice there are two solutions. Thus the linear factorization of the function is:

\begin{align*}f(x)=(x-3i)(x+3i)\end{align*}

Consider the graph \begin{align*}f(x)=x^2+9\end{align*} . How would you graph f(x)? Use a calculator to graph the function. State the x-intercepts, y-intercept, domain, and range. What are the graph's solutions?

**Example B**

Examine the following graph and make conclusions about the number and type of zeros of this \begin{align*}7^{th}\end{align*} degree polynomial.

**Solution:** A \begin{align*}7^{th}\end{align*} degree polynomial has 7 roots. Three real roots are visible. The root at \begin{align*}x=-2\end{align*} has multiplicity 2 and the root at \begin{align*}x=2\end{align*} has multiplicity 1. The other four roots appear to be imaginary and the clues are the relative maximums and minimums that do not cross the \begin{align*}x\end{align*} axis. Recognize that the up and down path of the graph below the x-axis is the reason we know there are imaginary roots.

**Example C**

Form a polynomial that has the following five roots. \begin{align*}x=0, 2, 3, \pm \sqrt{5}i\end{align*}. Where does the graph hit the x-axis? What are the roots?

**Solution:** Write the function in factorized form.

\begin{align*}f(x)=(x-0)(x-2)(x-3)(x- \sqrt{5}i)(x+ \sqrt{5}i)\end{align*}

When you multiply through, it will be helpful to do the complex conjugates first. The complex conjugates are \begin{align*}(x- \sqrt{5}i) (x+ \sqrt{5}i)\end{align*}.

\begin{align*}f(x)&=x(x^2-5x+6) (x^2-5 \cdot (-1)) \\ f(x)&=(x^3-5x^2+6x) (x^2+5) \\ f(x)&=x^5-5x^4+6x^3+5x^3-25x^2+30x \\ f(x)&=x^5-5x^4+11x^3-25x^2+30x\end{align*}

The graph hits the x-axis at 0, 2, and 3. It's roots are 0, 2, 3, \begin{align*}\sqrt{5}, and-\sqrt{5}\end{align*}

#### Vocabulary

A ** complex number** is a number written in the form \begin{align*}a+bi\end{align*} where both \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are real numbers. When \begin{align*}b=0\end{align*}, the result is a real number and when \begin{align*}a=0\end{align*} the result is an imaginary number.

An ** imaginary number** is the square root of a negative number. \begin{align*}\sqrt{-1}\end{align*} is defined to be the imaginary number \begin{align*}i\end{align*}.

** Complex conjugates** are pairs of complex numbers with real parts that are identical and imaginary parts that are of equal magnitude but opposite signs. \begin{align*}1+3i\end{align*} and \begin{align*}1-3i\end{align*} or \begin{align*}5i\end{align*} and \begin{align*}-5i\end{align*} are examples of complex conjugates.

#### Guided Practice

1. Write the polynomial that has the following roots: 4 (with multiplicity 3), 2 (with multiplicity 2) and 0.

2. Factor the following polynomial into its linear factorization and state all of its roots. Use factors of 6 (the constant) in the form (x - zero), with long division and see what factors give a remainder of zero.

\begin{align*}f(x)=x^4-5x^3+7x^2-5x+6\end{align*}

3. Can you create a polynomial with real coefficients that has one imaginary root? Why or why not?

**Answers:**

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

2. You can use polynomial long division to obtain the following factorization.

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

Also take into consideration the end behavior of the graph. The graph looks like:

3. No, if a polynomial has real coefficients then either it has no imaginary roots, or the imaginary roots come in pairs of complex conjugates (so that the imaginary portions cancel out when the factors are multiplied).

### Homework

For 1 - 4, find the polynomial with the given roots.

1. 2 (with multiplicity 2), 4 (with multiplicity 3), \begin{align*}1, \sqrt{2}i, -\sqrt{2}i\end{align*}.

2. 1, -3 (with multiplicity 3), \begin{align*}-1, \sqrt{3}i, -\sqrt{3}i\end{align*}

3. 5 (with multiplicity 2), -1 (with multiplicity 2), \begin{align*}2i, -2i\end{align*}

4. \begin{align*}i, -i, \sqrt{2}i, -\sqrt{2}i\end{align*}

For each polynomial, factor into its linear factorization and state all of its roots.

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

6. \begin{align*}g(x)=x^4-1\end{align*}

7. \begin{align*}h(x)=x^3+5x^2\end{align*}

8. \begin{align*}j(x)=x^4-7x^3+17x^2-17x+6\end{align*}

9. \begin{align*}k(x)=x^3-64\end{align*}

10. \begin{align*}m(x)=x^6-2x^4-9x^2+18\end{align*}

11. \begin{align*}n(x)=2x^3-8x^2-3x-12\end{align*}

12. \begin{align*}p(x)=-x^3+2x^2-25x+50\end{align*}

13. How can you tell the number of roots that a polynomial has from its equation?

14. Explain the meaning of the term “multiplicity”.

15. A polynomial with real coefficients has one root that is \begin{align*}\sqrt{3}i \end{align*}. What other root(s) must the polynomial have?