Chapter 7.5 Roots and Zeros Standard & Honors Algebra II Mr. Gilbert Chapter 7.5 Roots and Zeros Standard & Honors 11/14/2018

Students shall be able to Determine and find the number and type of roots for a polynomial Find the zeros of a polynomial equation. 11/14/2018

Agenda Warm up Home Work Lesson Practice Homework 11/14/2018

Homework Review 11/14/2018

Communicate Effectively Theorem: A proposition that has been or is to be proved on the basis of explicit assumptions. The Fundamental Theorem of Algebra: Every polynomial equation with a degree greater than 0 has at least one root in the set of complex numbers. Complex Conjugates Theorem: For every polynomial, if there exists a complex root a+bi then a-bi also exists. 11/14/2018

Example 1 Determine Number and Type of Roots Example 2 Find Numbers of Positive and Negative Zeros Example 3 Use Synthetic Substitution to Find Zeros Example 4 Use Zeros to Write a Polynomial Function 11/14/2018 Lesson 5 Contents

Solve State the number and type of roots. Original equation Add 10 to each side. Answer: This equation has exactly one real root, 10. 11/14/2018 Example 5-1a

Solve State the number and type of roots. Original equation Factor. Zero Product Property or Solve each equation. Answer: This equation has two real roots, –8 and 6. 11/14/2018 Example 5-1b

Solve State the number and type of roots. Original equation Factor out the GCF. Use the Zero Product Property. or Subtract 6 from each side. 11/14/2018 Example 5-1c

Square Root Property Answer: This equation has one real root at 0, and two imaginary roots at 11/14/2018 Example 5-1d

Solve State the number and type of roots. Original equation Factor differences of squares. Factor differences of squares. or Zero Product Property 11/14/2018 Example 5-1e

Solve each equation. Answer: This equation has two real roots, –2 and 2, and two imaginary roots, 2i and –2i. 11/14/2018 Example 5-1f

Solve each equation. State the number and type of roots. Answer: This equation has exactly one root at –3. Answer: This equation has exactly two roots, –3 and 4. Answer: This equation has one real root at 0 and two imaginary roots at 11/14/2018 Example 5-1g

d. Answer: This equation has two real roots, –3 and 3, and two imaginary roots, 3i and –3i. 11/14/2018 Example 5-1h

State the possible number of positive real zeros, negative real zeros, and imaginary zeros of Since p(x) has degree 6, it has 6 zeros. However, some of them may be imaginary. Use Descartes Rule of Signs to determine the number and type of real zeros. Count the number of changes in sign for the coefficients of p(x). yes – to + yes + to – no – to – no – to – 11/14/2018 Example 5-2a

Since there are two sign changes, there are 2 or 0 positive real zeros Since there are two sign changes, there are 2 or 0 positive real zeros. Find p(–x) and count the number of sign changes for its coefficients. x 1 no – to – no – to – yes – to + yes + to – Since there are two sign changes, there are 2 or 0 negative real zeros. Make a chart of possible combinations. 11/14/2018 Example 5-2b

Answer: 2 6 4 Number of Positive Real Zeros Number of Negative Real Zeros Number of Imaginary Zeros Total 2 6 4 11/14/2018 Example 5-2c

State the possible number of positive real zeros, negative real zeros, and imaginary zeros of Answer: The function has either 2 or 0 positive real zeros, 2 or 0 negative real zeros, and 4, 2, or 0 imaginary zeros. 11/14/2018 Example 5-2d

Find all of the zeros of Since f (x) has degree of 3, the function has three zeros. To determine the possible number and type of real zeros, examine the number of sign changes in f (x) and f (–x). yes yes no no no yes 11/14/2018 Example 5-3a

The function has 2 or 0 positive real zeros and exactly 1 negative real zero. Thus, this function has either 2 positive real zeros and 1 negative real zero or 2 imaginary zeros and 1 negative real zero. To find the zeros, list some possibilities and eliminate those that are not zeros. Use a shortened form of synthetic substitution to find f (a) for several values of a. x 1 –1 2 4 –3 –4 14 –38 –2 8 –12 Each row in the table shows the coefficients of the depressed polynomial and the remainder. 11/14/2018 Example 5-3b

Replace a with 1, b with –2, and c with 4. From the table, we can see that one zero occurs at x = –1. Since the depressed polynomial, , is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation Quadratic Formula Replace a with 1, b with –2, and c with 4. 11/14/2018 Example 5-3c

Simplify. Simplify. 11/14/2018 Example 5-3d

Answer: Thus, this function has one real zero at –1 and two imaginary zeros at and The graph of the function verifies that there is only one real zero. 11/14/2018 Example 5-3e

Find all of the zeros of Answer: 11/14/2018 Example 5-3f

Short-Response Test Item Write a polynomial function of least degree with integer coefficients whose zeros include 4 and 4 – i. Read the Test Item • If 4 – i is a zero, then 4 + i is also a zero, according to the Complex Conjugate Theorem. So, x – 4, x – (4 – i), and x – (4 + i) are factors of the polynomial function. 11/14/2018 Example 5-4a

• Multiply the factors to find the polynomial function. Solve the Test Item • Write the polynomial function as a product of its factors. • Multiply the factors to find the polynomial function. Write an equation. Regroup terms. Rewrite as the difference of two squares. 11/14/2018 Example 5-4b

Square x – 4 and replace i2 with –1. Simplify. Multiply using the Distributive Property. Combine like terms. 11/14/2018 Example 5-4c

Answer: is a polynomial function of least degree with integral coefficients whose zeros are 4, 4 – i, and 4 + i. 11/14/2018 Example 5-4d

Short-Response Test Item Write a polynomial function of least degree with integer coefficients whose zeros include 2 and 1 + i. Answer: 11/14/2018 Example 5-4e

Homework See Syllabus 7.5 11/14/2018