Download 5-7 Roots and Zeros 12-4

Five-Minute Check (over Lesson 5–6)
CCSS
Then/Now
Concept Summary: Zeros, Factors, Roots, and Intercepts
Key Concept: Fundamental Theorem of Algebra
Example 1: Determine Number and Type of Roots
Key Concept: Corollary to the Fundamental Theorem of Algebra
Key Concept: Descartes’ Rule of Signs
Example 2: Find Numbers of Positive and Negative Zeros
Example 3: Use Synthetic Substitution to Find Zeros
Key Concept: Complex Conjugates Theorem
Example 4: Use Zeros to Write a Polynomial Function
Over Lesson 5–6
Use synthetic substitution to find f(2) for
f(r) = 3r4 + 7r2 – 12r + 23.
Use synthetic substitution to find f(6) for
f(c) = 2c3 + 19c2 + 2.
Given a polynomial and one of its factors, find the
remaining factors of the polynomial.
k4 + 7k3 + 9k2 – 7k – 10; k + 2
Given a polynomial and one of its factors, find the
remaining factors of the polynomial.
6p3 + 11p2 – 14p – 24; p + 2
What value of k would give a remainder of 6 when
x2 + kx + 18 is divided by x + 4?
Over Lesson 5–6
Use synthetic substitution to find f(2) for
f(r) = 3r4 + 7r2 – 12r + 23.
A. 21
B. 75
C. 855
D. 4091
Over Lesson 5–6
Use synthetic substitution to find f(6) for
f(c) = 2c3 + 19c2 + 2.
A. 94
B. 727
C. 1118
D. 1619
Over Lesson 5–6
Given a polynomial and one of its factors, find the
remaining factors of the polynomial.
k4 + 7k3 + 9k2 – 7k – 10; k + 2
A. (k + 5), (k + 1)
B. (k + 1), (k – 1)
C. (k + 5), (k + 1), (k – 1)
D. (k + 5), (k – 5), (k + 1), (k – 1)
Over Lesson 5–6
Given a polynomial and one of its factors, find the
remaining factors of the polynomial.
6p3 + 11p2 – 14p – 24; p + 2
A. (2p – 3)(2p + 3)
B. (2p – 3)(3p + 4)
C. (3p + 4)(3p – 4)
D. (2p + 3)(3p – 4)
Over Lesson 5–6
The function f(x) = x3 – 6x2 – x + 30 can be used to
describe the relative stability of a small boat
carrying x passengers, where f(x) = 0 indicates that
the boat is extremely unstable. With three
passengers, the boat tends to capsize. What other
passenger loads could cause the boat to capsize?
A. 6 passengers
B. 5 passengers
C. 4 passengers
D. 2 passengers
Over Lesson 5–6
What value of k would give a remainder of 6 when
x2 + kx + 18 is divided by x + 4?
A. 7
B. –1
C. 1
D. 3
Content Standards
N.CN.9 Know the Fundamental Theorem of
Algebra; show that it is true for quadratic
polynomials.
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.
Mathematical Practices
6 Attend to precision.
You used complex numbers to describe
solutions of quadratic equations.
• Determine the number and type of roots for
a polynomial equation.
• Find the zeros of a polynomial function.
Determine Number and Type of Roots
A. Solve x2 + 2x – 48 = 0. State the number and type
of roots.
Answer: This equation has two real roots, –8 and 6.
Determine Number and Type of Roots
B. Solve y4 – 256 = 0. State the number and types
of roots.
y4 – 256 = 0
(y2 + 16) (y2 – 16) = 0
(y2 +16) (y + 4)(y – 4) = 0
y2 + 16 = 0 or y + 4 = 0 or y – 4 = 0
y2 = –16
y = –4
y=4
Answer: This equation has two real roots, –4 and 4,
and two imaginary roots, 4i and –4i.
A. Solve x2 – x – 12 = 0. State the number and type
of roots.
A. 2 real: –3 and 4
B. 2 real: 3 and –4
C. 2 real: –2 and 6
D. 2 real: 3 and 4;
2 imaginary: 3i and 4i
B. Solve a4 – 81 = 0. State the number and type
of roots.
A. 2 real: –3 and 3
B. 2 real: –3 and 3
2 imaginary: 3i and –3i
C. 2 real: –9 and 9
2 imaginary: 3i and –3i
D. 2 real: –9 and 9
2 imaginary: 9i and –9i
Find Numbers of Positive and Negative Zeros
State the possible number of positive real zeros,
negative real zeros, and imaginary zeros of
p(x) = –x6 + 4x3 – 2x2 – x – 1.
p(x) = –x6 + 4x3
–
2x2 –
x
–
1
yes
– to +
yes
+ to –
no
– to –
no
– to –
Two sign changes → 2 or 0 positive real zeros.
p(–x) = –(–x)6 + 4(–x)3 – 2(–x)2 – (–x) –
–x6 –
no
– to –
4x3
–
no
– to –
2x2
+
yes
– to +
x
Two sign changes → 2 or 0 negative real zeros.
–
yes
+ to –
1
1
Find Numbers of Positive and Negative Zeros
State the possible number of positive real zeros,
negative real zeros, and imaginary zeros of
p(x) = –x6 + 4x3 – 2x2 – x – 1.
Descartes: 2,2,2 or 0,2,4 or 2,0,4 or 0,0,6
State the possible number of positive real zeros,
negative real zeros, and imaginary zeros of
p(x) = x4 – x3 + x2 + x + 3.
A. 2,0,2 or 0,0,4
B. 0,4,0 or 0,2,2 or 0,0,4
C. 4,0,0 or 2,0,2 or 0,0,4
D. 2,2,0 or 2,0,2 or 0,2,2 or 0,0,4
Identify Possible Zeros
A. List all of the possible rational zeros of
f(x) = 3x4 – x3 + 4.
p
q
The factors of p are ±1, ±2, ±4
The factors of q are ±1, ±3
Possibilities:
Answer:
1 −1 1 −1 2 −2 2 −2 4 −4 4 −4
, , , , , , , … , , ,
1 1 −1 −1 1 1 −1 −1 3 3 −3 −3
Identify Possible Zeros
B. List all of the possible rational zeros of
f(x) = x4 + 7x3 – 15.
Since the coefficient of x4 is 1, the possible zeros must
be a factor of the constant term –15.
Answer: So, the possible rational zeros are ±1, ±3, ±5,
and ±15.
A. List all of the possible rational zeros of
f(x) = 2x3 + x + 6.
A.
B.
C.
D.
B. List all of the possible rational zeros of
f(x) = x3 + 3x + 24.
A.
B.
C.
D.
Use Synthetic Substitution to Find Zeros
Find all of the zeros of f(x) = x3 – x2 + 2x + 4.
Since f(x) has degree of 3, the function has three zeros.
Use Descartes:
f(x) = x3
–
x2
+ 2x
+ 4
yes
f(–x) = –x3
–
no
yes
x2
–
no
Descartes: 2,1,0 or 0,1,2 zeros
no
2x
+
yes
4
Use Synthetic Substitution to Find Zeros
Find all of the zeros of f(x) = x3 – x2 + 2x + 4.
From the Rational Root theorem: ±1, ±2, ±4
1
1
1
-1
1
1
-1
1
0
2
0
2
4
2
6
-1
-1
-2
2
2
4
4
-4
0
Descartes: 2,1,0 or 0,1,2 zeros
Use Synthetic Substitution to Find Zeros
Find all of the zeros of f(x) = x3 – x2 + 2x + 4.
From the Rational Root theorem: ±1, ±2, ±4
So 𝑥 3 − 𝑥 2 + 2𝑥 + 4 = (𝑥 + 1)(𝑥 2 − 2𝑥 + 4)
-1
1
1
-1
-1
-2
2
2
4
4
-4
0
Descartes: 2,1,0 or 0,1,2 zeros
Use Synthetic Substitution to Find Zeros
Find all of the zeros of f(x) = x3 – x2 + 2x + 4.
From the Rational Root theorem: ±1, ±2, ±4
So 𝑥 3 − 𝑥 2 + 2𝑥 + 4 = (𝑥 + 1)(𝑥 2 − 2𝑥 + 4)
Use quadratic formula for (𝑥 2 − 2𝑥 + 4)
−𝑏 ± 𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
−(−2) ± (−2)2 −4(1)(4) 2 ± −12 2 ± 2𝑖 3
𝑥=
=
=
2(1)
2
2
𝑥 = −1
𝑥 =1±𝑖 3
Descartes: 2,1,0 or 0,1,2 zeros
Use Synthetic Substitution to Find Zeros
Answer: Thus, this function has one real zero at –1
and two imaginary zeros at
The graph of the function verifies that there
is only one real zero.
.
What are all the zeros of f(x) = x3 – 3x2 – 2x + 4?
A.
B.
C.
D.
Use Zeros to Write a Polynomial Function
Write a polynomial function of least degree with
integral coefficients, the zeros of which include
4 and 4 – i.
Understand
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.
Plan
Write the polynomial function as a
product of its factors.
f(x) = (x – 4)[x – (4 – i)][x – (4 + i)]
Use Zeros to Write a Polynomial Function
Write a polynomial function of least degree with
integral coefficients, the zeros of which include
4 and 4 – i.
f(x) = (x – 4)[x – (4 – i)][x – (4 + i)]
= (x – 4)[(x – 4) + i)][(x – 4) – i)]
= (x – 4)[(x – 4)2 – i2]
Plan
Write the polynomial function as a
product of its factors.
f(x) = (x – 4)[x – (4 – i)][x – (4 + i)]
Use Zeros to Write a Polynomial Function
Write a polynomial function of least degree with
integral coefficients, the zeros of which include
4 and 4 – i.
f(x) = (x – 4)[x – (4 – i)][x – (4 + i)]
= (x – 4)[(x – 4) + i)][(x – 4) – i)]
= (x – 4)[(x – 4)2 – i2]
What is a polynomial function of least degree with
integral coefficients the zeros of which include
2 and 1 + i?
A. x2 – 3x + 2 – xi + 2i
B. x2 – 2x + 2
C. x3 – 4x2 + 6x – 4
D. x3 + 6x – 4