Algebra II
Unit 4
Roots and Zeros
CCSS: A. APR.3
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Standards for Mathematical
Practice

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





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.
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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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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Students shall be able to

Determine and find
the number and
type of roots for a
polynomial
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
5
Find the zeros of a
polynomial
equation.
Agenda
Warm up
 Home Work
 Lesson
 Practice

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Homework Review
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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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Concept Summary:
f x   an x  an1 x    a1 x  ao
n
n 1
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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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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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.
COROLLARY:
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.
Real/Imaginary Roots
If a polynomial has ‘n’ complex roots will its
graph have ‘n’ x-intercepts?
3
y  x  4x
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 3
x-intercepts.
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.
y  x 3  2x 2  x  4
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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Solve
of roots.
State the number and type
Original equation
Factor.
or
Zero Product Property
Solve each equation.
Answer: This equation has two real roots, –8 and 6.
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Solve
of roots.
State the number and type
Original equation
Factor out the GCF.
Use the Zero Product Property.
or
Subtract 6 from
each side.
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Square Root
Property
Answer: This equation has one real root at 0,
and two imaginary roots at
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Solve
State the number and type of roots.
Original equation
Factor differences
of squares.
Factor differences
of squares.
or
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or
Zero Product
Property
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Solve each
equation.
Answer: This equation has two real roots, –2 and 2,
and two imaginary roots, 2i and –2i.
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Solve each equation. State the number and type
of roots.
a.
Answer: This equation has exactly one root at –3.
b.
Answer: This equation has exactly two roots, –3 and 4.
c.
Answer: This equation has one real root at 0 and two
imaginary roots at
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d.
Answer: This equation has two real roots, –3 and 3,
and two imaginary roots, 3i and –3i.
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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!
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 +
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yes
+ to –
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no
– to –
no
– to –
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
no
– to –
no
– to –
yes
– to +
1
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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Answer:
Number of
Positive Real
Zeros
Number of
Negative Real
Zeros
Number of
Imaginary
Zeros
Total
2
0
2
0
2
2
0
0
2
4
4
6
6
6
6
6
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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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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
no
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yes
no
no
yes
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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
1
–4
14
–38
–2
1
–3
8
–12
–1
1
–2
4
0
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Each row in the table
shows the coefficients
of the depressed
polynomial and the
remainder.
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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Simplify.
Simplify.
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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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Find all of the zeros of
Answer:
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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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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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Square x – 4
and replace i2
with –1.
Simplify.
Multiply using
the Distributive
Property.
Combine like
terms.
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Answer:
is a polynomial
function of least degree with integral coefficients whose
zeros are 4, 4 – i, and 4 + i.
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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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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!
3
f (x)  x  4x
 x(x 2  4)
 x(x  2)(x  2)
The roots are: 0, -2, 2
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.
f (x)  x  5x  2x  10
one root is x  5
3
2
x2  2
x  5 x 3  5x 2  2x  10
(x - 5) is a factor
We can solve the resulting
polynomial to get the other 2 roots:
x  2,  2

x 3  5x 2
 2x  10
2x  10
0
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.
f (x)  x 3  5x 2  7x  51
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
Example (con’t)
3
2
Ex: Find all the roots of f (x)  x  5x  7x  51
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:
 x   4  i    x   4  i    x 2  x  4  i   x  4  i    4  i  4  i 
 x 2  4 x  xi  4 x  xi  16  i 2
 x 2  8 x  16  (1)
 x 2  8 x  17
Example (con’t)
3
2
Ex: Find all the roots of f (x)  x  5x  7x  51
If one root is 4 - i.
2
If the product of the two non-real factors is x  8x 17
then the third factor (that gives us the neg. real root) is
2
the quotient of P(x) divided by x  8x 17 :
x 3
2
3
2
x  8x  17 x  5x  7x  51

x 3  5x 2  7x  51
0
The third root
is x = -3
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.
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) = 0.1729t + 0.1522t² - 0.0374t³,
where t is the time in seconds. Find the
volume of air held by the lungs at3
seconds.
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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) = - 0.006 x^4 - 0.140x³ - 0.053x² + 1.79x,
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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Questions?
See Agenda for HW.
Work with the practice provided
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