Chapter 11 Polynomial Functions How are polynomials classified and transformed and why are we interested in their x-intercepts?

Polynomials and Polynomial Functions 11-1 What is a root or zero of a function? The Remainder and Factor Theorems 11-2 How do you use the remainder and factor theorems?

Vocabulary Leading Coefficient—the numeric part of the term of highest degree Solution—the values that make an equation true Also called roots or zeros Why do we call it a zero?

Activation What procedures would you use to solve the following problem:

Can we translate this to algebraic equations Long division: Can we translate this to algebraic equations If a factor divides evenly with no remainder, then it is a factor of the polynomial and we could then find the root.

Divide 125x3 – 8 by 5x – 2 to determine if 5x – 2 is a factor of the polynomial.

Practice Polynomial Long Division Divide 9x3 – 48x2 + 13x + 3 by x - 5.

Example 2 If a factor divides evenly with no remainder, then it is a factor of the polynomial and we could then find the root.

Divide x3 – 1 by x – 1 to determine if x – 1 is a factor of the polynomial.

Polynomial Long Division Allows us to find factors. If the quotient divides evenly with no remainder, then it is a factor. Once we have factors roots are easy, just set each factor = 0 and solve it! Use polynomial long division for any factors including linear, quadratic, cubic, quartic, etc.

Assignment P. 483 13 – 18 all

Synthetic Division

When the divisor is linear (x + a), we can use a synthetic division When the divisor is linear (x + a), we can use a synthetic division. Divide the following using synthetic division. (x3 +3x2 – x – 3) ÷ (x – 1)

Example 2 (x3 + 27)÷(x – 1) (x3+27) ÷(x + 3) Both long division and synthetic give the other factor when the remainder = 0

Synthetic Division Allows us to find factors If we know the factor, we set the factor = 0 and can find the root Can only use it if the factor is in the form: x + a x – a

Is (x – 5 ) a factor to (6x3 - 25x2 - 31x + 30)?

The Remainder Theorem helps to determine roots as well but does not give the remaining factors/roots. Simply involves substitution. Ex. Given: f(x)= x3 + 4x2 + 4x, are -1, 0 or 4i roots? Sub in to see if equation = 0.

Summary How are polynomial long division, synthetic division and the remainder theorem similar?

Homework Page 483 Use the directions below and not the directions listed in the book! 2 - 4 Use remainder theorem 5 – 8 Use synthetic division 10 – Use Long Division (for extra practice)

Theorems About Roots 11-3 How can you find information about the roots of a polynomial?

Activation Where do the roots or solutions ALWAYS occur on the graph?

Activation – do not need to copy down! How many roots does/can each equation have? y = x5 – 3x - 1 y = x5-13x3 + 36x y = 2x4–9x3+4x2+3x-1 y = x4 -13x2 +36 y = x3 + 2x2 – x – 1 y = x3 y = x2 – 2x + 6 y = x2 - 3x - 4 y = 3x + 4

Fundamental Theorem of Algebra Every polynomial of degree n > 1 has at least one linear factor.

Irrational and Imaginary Root Theorems Equations have as many roots as their highest degree. Some of the roots are rational, some irrational and some imaginary. Irrational and Imaginary Root Theorems Irrational and imaginary roots occur in groups of two because they are generated from the quadratic formula.

If an equation has one complex root it has two complex roots which are conjugates of each other. Ex. If 1 – i is a root then 1 + i must also be a root. The same is true for irrational roots: Ex. If is a root then must also be a root!

Ex. Suppose a polynomial of degree 3 (or a cubic) has 3 – 4i and 9 as root. Find all the roots. Roots are: 3 – 4i, 3 + 4i and 9 which means x = 3 – 4i x = 3 + 4i x = 9 x – 3 + 4i = 0 x – 3 – 4i = 0 x – 9 = 0 (x – 3 + 4i)(x – 3 – 4i)(x – 9) = 0 (x2 – 3x – 4xi – 3x + 9 + 12i + 4xi – 12i – 16i2)(x - 9) = 0 (x2- 6x + 25)(x – 9) = 0 x3 - 15x2 + 79x – 225 = 0

Multiplicity If a factor occurs more than once, we say that the root is a root of multiplicity. Ex. Root Multiplicity -3 1 factor (x + 3) occurs 1 time 2 3 (x – 2) occurs 3 times (x-2)(x-2)(x-2) -1 2 (x + 1) occurs twice (x + 1)(x + 1) 0 1 monomial factor occurs 1 time

Suppose a polynomial of degree 5 with the following: 0 as a root of multiplicity 1 -1 as a root with multiplicity 3 4 as a root of multiplicity 1 Identify all roots and find a possible equation. Roots are: x = 0 x = -1 x = -1 x = -1 x = 4 x = 0 x + 1 = 0 x + 1 = 0 x + 1 = 0 x – 4 = 0 x(x + 1)3 (x - 4) = 0 Now we would need to multiply the left side together!

Ex. Find the roots and state the multiplicity of P(x) = x2 (x - 1) (x + 4)3 Roots are: x = 0 x = 0 x = 1 x = -4 x = -4 x = -4 Multiplicity is: 2 1 3

Ex. Given that P(x) = x 4 – 81 and 3i is root, find all the roots of the polynomial. Since 3i is a root, then -3i is a root. P(x) = (x – 3i)(x + 3i) * Q(x) for some Q(x) P(x) = (x2 + 9) * Q(x) Use long division to find Q(x): x2 – 81 P(x) = (x2 + 9)(x + 9)(x – 9) So the roots are 3i, -3i, 9 and -9

Ex. Find all the roots of the equation having the following roots: 0, 3 + i, 2 -

Ex. Find the 3rd degree polynomial with roots of 4 - 3i and 6.

Homework Page 495 2 – 16 evens (omit 10), 24 – 32 (omit 30), 40, 42 Copy the directions for each problem down in your homework!

Rational Roots 11-4 Descartes’ Rule of Signs 11-5 What is a rational root? Descartes’ Rule of Signs 11-5 What is Descartes Rule of Signs?

Activation Factor the following equation: 15x2 – 2x – 8 = 0

To find all the roots of any equation factor if possible or use the quadratic formula: x3 – 5x2 + 6x = 0 x(x2 – 5x + 6) = 0 x (x – 2)(x – 3) = 0 x = 0 x = 2 x = 3

Since factoring is not always possible we use the Rational Root Theorem It is a method of finding all rational roots of any polynomial equation including those that do not factor nicely. Just as in factoring a quadratic by grouping method, the potential factors come from the a and c, in ax2 + bx + c the factors for any equation come from the leading coefficient and the constants.

The constant is referred to as “p” and the leading coefficient as “q” The constant is referred to as “p” and the leading coefficient as “q”. To determine possible rational roots we following these steps: List all of the factors of p. List all of the factors of q. List all of the factors of p/q. One of these may be a possible rational root so we check it by synthetic division to see if it divides evenly. This Rational Root Theorem only finds the Rational Roots, there may be irrational or imaginary roots in addition to the rational roots. We will find them in a different way.`

Find the rational roots of x3 + x2 – 3x – 3 = 0

What if….. What if I asked you to find all rational roots to 4x3 – 3x + 24. How many p’s and q’s are there?

So…. there are a lot of p/q’s sometimes So….there are a lot of p/q’s sometimes. Isn’t there a way to minimize the number of possible roots that we need to check with synthetic division??? Sure there is! There are 2 tools we can use for this: 1. Upper and Lower Bound Theorem and 2. Descartes Rule of Signs.

The upper and lower bounds help to minimize the number of trials you have to do. When all the signs at the bottom of the synthetic division are positive, you have the upper bound meaning no values larger than the one you tested will work. When the signs at the bottom of the synthetic division alternate sign (positive to negative) you have a lower bound meaning no values smaller than the one you tested will work.

Descartes’ Rule of Signs – Positive Roots The rule uses the number of variations of +/- signs meaning the number of times coefficients have different signs. To determine the number of possible positive roots look at the signs of f(x). Then the number of variations of sign is the possible root or less by a positive even integer. Ex. P(x) = 2x5 – 5x2 + 3x + 6 Look at the sign changes in the coefficients! + to – to + to + The number of variations is 2 so the number of positive roots is 2 or 2 – 2 = 0.

Descartes’ Rule of Signs – Negative Roots To determine the number of possible negative roots look at the signs of f(-x) (means to replace x with –x). The number of sign changes determines the number of possible roots of that type or that number minus positive even integers. Ex. Determine number of negative real roots. P(x) = 5x4 + 3x3 + 7x2 + 12x + 4 P(-x) = 5(-x)4 + 3(-x)3 + 7(-x)2 + 12(-x) + 4 = 5x4 – 3x3 + 7x2 – 12x + 4 + to - to + to - to + 4 variations in sign so number of negative roots are 4, 2 or 0.

Use the rational root theorem to find all the roots of 2x3 – x2 + 2x – 1 = 0

Example: State the number of possible roots, find all the number of rational roots or show that none exist 5x4 – 4x3 + 19x2 - 16x – 4 = 0

Determine the possible number of positive and negative real zeros of f(x) = x3 + 2x2 + 5x + 4 For negative real zeros: Find f(-x) then count the sign changes. F(-x) = -x3 + 2x2 – 5x + 4 For positive real zeros: Count the sign changes. There are none, so there are no positive real zeros. 3 sign changes. # of negative real zeros is = 3 or less than 3 by an even integer: 3 – 2 = 1. Meaning there are either 3 negative real zeros of 1 negative real zero. What does this mean for us? Without using Descartes’ Rule of Signs the possible rational roots are: 1, -1, 2, -1, 4, -4 (found using Rational Root Theorem) However, Descartes’ Rule of Signs tells us that in the previous example there are no positive real zeros so this mean the equation has no positive real roots. Meaning we can eliminate the positive numbers from our list of possible rational roots and only include -1, -2, -4. Check using synthetic division to get the equation down to a quadratic which can them be solved by factoring or by quadratic equation.

State the number of possible positive and negative real roots, then find all the number of rational roots or show that none exist 6x7 + 2x2 + 5x + 4 = 0. No roots

State the number of possible positive/negative real roots, then find all the number of rational roots or show that none exist x5 - 5x4 + 5x3 + 15x2 – 36x + 20 = 0 -1 and 2

Homework Page 503 2 – 20 even Use DesCartes’ rule of signs to determine the number of possible positive and negative roots for each Page 499 2- 12 even Find all roots if possible or prove that none exist

Graphs of Polynomial Functions 11-6 How do you graph a higher degree polynomial function?

Activation How do we graph the following equation: x2 – 4x – 12 = 0 What do we know about the graph before we even begin? It’s a parabola, with 2 possible x-intercepts and the tail goes up on the right side of the graph

Cubic or Third Degree Polynomial Graphs Quartic or Fourth Degree Polynomial Graphs Can also look like a parabola

Graphing any polynomial: Look at the degree of the polynomial for the basic shape End Behavior - Look at the leading coefficient: if it is positive it goes up on the right side of the graph If it is negative it goes down on the right side of the graph Determine any symmetries: Replace f(x) with f(-x) – no change even (y-axis) - all change – odd (origin) Find the y intercept (x = 0) and any x intercepts (y = 0) (roots) Can use Descartes’ rule of signs to determine the number of possible positive and negative roots Use a t-chart for a few additional points Plot the points and connect them appropriately

Graph the following polynomial: f(x) = x3 – 3x2 + 1 1. It is a cubic so know basic shape. 2. End Behavior - The leading coef is +1 so graph ends up to the right. 3. Symmetries - none 4. Y-intercept means x = 0 so y = 1 or (0, 1). X-int means y = 0 – cannot factor so must Rational Root Theorem: p/q = ± 1 Do synthetic division to see if one is a root! 4a. Descartes Rule of Signs: only 2 roots so no need to use this! Create table. We will need to estimate the roots since they are either irrational or imaginary. X Y -2 -1 1 2

Estimating Roots: If you can’t find any rational roots the t-chart helps you determine between which x-values the graph crosses the x because of a change in sign of the y Use your calculator to find the values of y for increments of one tenth in x to find a close approximation to the irrational roots. f(x)= x3 – 3x2 + 1 Between -1 and 0 f (-.5) = Between 0 and 1 F(.5) = Between

Graph the following polynomial: f(x)= x4 – 6x2 + 8 1. It is 4th degree. 2. The leading coef is +1 so up to right 3. Symmetries - to y-axis 4. Y-int x = 0 y = 8 (0, 8) x-intercepts - rational root thm Rule of Signs: 2 or 0 positive 2 or 0 negative 5. Table of values between the roots. X Y 1.4 1.6 1.8 1.9 X Y 1.5 -.44 1.6 -.81 1.8 -.94 1.9 -.6 X Y 1 3 2 X Y 1 2 4 8

Homework Page 507 7, 9, 12, 15

Graph the following polynomial: f(x)= 1. Basic Shape 2. End Behavior 3. Symmetries 4. Y-int x = 0 X-int y=0 Descartes 5. Table X Y X Y

Graph the following polynomial: f(x)= 1. Basic Shape 2. End Behavior 3. Symmetries 4. Y-int x = 0 X-int y=0 Descartes 5. Table X Y X Y