1. Unit 4 Roots and Zeros CCSS: A. APR.3
Algebra II
Unit 4
Roots and Zeros
CCSS: A. APR.3
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2. Standards for Mathematical Practice
1. Make sense of problems and persevere in solving them.
2. Reason abstractly and quantitatively.
3. Construct viable arguments and critique the reasoning of others.  
4. Model with mathematics.
5. Use appropriate tools strategically.
6. Attend to precision.
7. Look for and make use of structure.
8. Look for and express regularity in repeated reasoning.

3. CCSS: A. APR.3
Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.
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4. ESSENTIAL QUESTION (s):
How do I find the roots of a polynomial function?
How does the Remainder Theorem help me to find the roots of a polynomial function?
How do I determine the equation and graph of a polynomial function given its roots?
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5. Students shall be able to
Determine and find the number and type of roots for a polynomial
Find the zeros of a polynomial equation.
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6. Agenda
Warm up
Home Work
Lesson
Practice
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7. Homework Review
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8. 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
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Lesson 5 Contents

9. Concept Summary: f(c ) = 0 c is a zero of the polynomial function f(x)
x – c is a factor of polynomial function f(x)
c is a root or solution of polynomial function f(x)
If c is a real number, then (c,0) is an intercept of the graph f(x)
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10. 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.
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11. Fundamental Theorem Of Algebra (your text)
Every Polynomial Equation with a degree higher than zero has at least one root in the set of Complex Numbers.
A Polynomial Equation of the form P(x) = 0 of degree ‘n’ with complex coefficients has exactly ‘n’ Roots in the set of Complex Numbers.
COROLLARY:

12. Real/Imaginary Roots
If a polynomial has ‘n’ complex roots will its graph have ‘n’ x-intercepts?
In this example, the degree n = 3, and if we factor the polynomial, the roots are x = -2, 0, 2. We can also see from the graph that there are x-intercepts.

13. Real/Imaginary Roots
Just because a polynomial has ‘n’ complex roots doesn’t mean that they are all Real!
In this example, however, the degree is still n = 3, but there is only one Real x-intercept or root at x = -1, the other 2 roots must have imaginary components.

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

15. 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.
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Example 5-1b

16. 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.
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Example 5-1c

17. Square Root Property
Answer: This equation has one real root at 0, and two imaginary roots at
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Example 5-1d

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

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

20. 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
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Example 5-1g

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

22. Descartes’ Rule of Signs
Arrange the terms of the polynomial P(x) in descending degree:
The number of times the coefficients of the terms of P(x) change sign = the number of Positive Real Roots (or less by any even number)
The number of times the coefficients of the terms of P(-x) change sign = the number of Negative Real Roots (or less by any even number)
In the examples that follow, use Descartes’ Rule of Signs to predict the number of + and - Real Roots!

23. 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 –
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Example 5-2a

24. 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.
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Example 5-2b

25. Answer: 2 6 4 Number of Positive Real Zeros
Number of Negative Real Zeros
Number of Imaginary Zeros
Total 2 6 4
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Example 5-2c

26. 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.
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Example 5-2d

27. 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
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Example 5-3a

28. 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.
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Example 5-3b

29. 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.
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Example 5-3c

30. Simplify.
Simplify.
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Example 5-3d

31. 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.
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Example 5-3e

32. Find all of the zeros of
Answer:
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Example 5-3f

33. 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.
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Example 5-4a

34. • 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.
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Example 5-4b

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

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

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

38. Find Roots/Zeros of a Polynomial
We can find the Roots or Zeros of a polynomial by setting the polynomial equal to 0 and factoring.
Some are easier to factor than others!
The roots are: 0, -2, 2

39. Find Roots/Zeros of a Polynomial
If we cannot factor the polynomial, but know one of the roots, we can divide that factor into the polynomial. The resulting polynomial has a lower degree and might be easier to factor or solve with the quadratic formula.
(x - 5) is a factor
We can solve the resulting polynomial to get the other 2 roots:

40. Find Roots/Zeros of a Polynomial
If the known root is imaginary, we can use the Complex Conjugates Theorem.
Ex: Find all the roots of
If one root is 4 - i.
Because of the Complex Conjugate Theorem, we know that another root must be 4 + i.
Can the third root also be imaginary?
Consider… Descartes: # of Pos. Real Roots = 2 or 0
Descartes: # of Neg. Real Roots = 1

41. Example (con’t) Ex: Find all the roots of If one root is 4 - i.
If one root is 4 - i, then one factor is [x - (4 - i)], and
Another root is 4 + i, & another factor is [x - (4 + i)].
Multiply these factors:

42. Example (con’t) Ex: Find all the roots of If one root is 4 - i.
If the product of the two non-real factors is
then the third factor (that gives us the neg. real root) is the quotient of P(x) divided by :
The third root is x = -3

43. Finding Roots/Zeros of Polynomials
We use the Fundamental Thm. Of Algebra, Descartes’ Rule of Signs and the Complex Conjugate Thm. to predict the nature of the roots of a polynomial.
We use skills such as factoring, polynomial division and the quadratic formula to find the zeros/roots of polynomials.

44. Real Life app: Physiology
During a five-second respiratory cycle, the volume of air in liters in the human lungs can be described by the function:
A(t) = t t² t³, where t is the time in seconds. Find the volume of air held by the lungs at3 seconds.
liter
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45. Real Life app: Medicine
Doctors can measure cardiac output in patients at high risk for a heart attack by monitoring the concentration of dye injected into a vein near the heart. A normal heart dye concentration is approximated by:
d(x) = x^ x³ x² x,
where x is the time in seconds.
a. Find all real zeros by graphing. Then verify them by using synthetic division.
b. Which root makes sense for an answer to this problem? Why?
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46. Work with the practice provided
Questions?
See Agenda for HW.
Work with the practice provided
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