Math

2. 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

3. Over Lesson 5–6
A. 21
B. 75
C. 855
D. 4091
Use synthetic substitution to find f(2) for
f(r) = 3r4
+ 7r2
– 12r + 23.

4. Over Lesson 5–6
A. 94
B. 727
C. 1118
D. 1619
Use synthetic substitution to find f(6) for
f(c) = 2c3
+ 19c2
+ 2.

5. Over Lesson 5–6
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)
Given a polynomial and one of its factors, find the
remaining factors of the polynomial.
k4
+ 7k3
+ 9k2
– 7k – 10; k + 2

6. Over Lesson 5–6
A. (2p – 3)(2p + 3)
B. (2p – 3)(3p + 4)
C. (3p + 4)(3p – 4)
D. (2p + 3)(3p – 4)
Given a polynomial and one of its factors, find the
remaining factors of the polynomial.
6p3
+ 11p2
– 14p – 24; p + 2

7. Over Lesson 5–6
A. 6 passengers
B. 5 passengers
C. 4 passengers
D. 2 passengers
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?

8. Over Lesson 5–6
A. 7
B. –1
C. 1
D. 3
What value of k would give a remainder of 6 when
x2
+ kx + 18 is divided by x + 4?

9. 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.

10. 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.

13. 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.
Original equation
Factor.
Solve each equation.
Zero Product Property

14. Determine Number and Type of Roots
B. Solve y4
– 256 = 0. State the number and types
of roots.
y2
+ 16 = 0 or y + 4 = 0 or y – 4 = 0 Zero Product
Property
Factor.
(y2
+ 16) (y2
– 16) = 0
Original equation
y4
– 256 = 0
Factor.
(y2
+16) (y + 4)(y – 4) = 0

15. Determine Number and Type of Roots
y2
= –16 y = –4 y = 4 Solve each
equation.
Answer: This equation has two real roots, –4 and 4,
and two imaginary roots, 4i and –4i.

16. 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
A. Solve x2
– x – 12 = 0. State the number and type
of roots.

17. 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
B. Solve a4
– 81 = 0. State the number and type
of roots.

20. 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.
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).
p(x) = –x6
+ 4x3
– 2x2
– x – 1
yes
– to +
yes
+ to –
no
– to –
no
– to –

21. Find Numbers of Positive and Negative 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.
Since there are two sign changes, there are 2 or 0
negative real zeros. Make a chart of possible
combinations.
p(–x) = –(–x)6
+ 4(–x)3
– 2(–x)2
– (–x) – 1
no
– to –
no
– to –
yes
– to +
yes
+ to –
–x6
– 4x3
– 2x2
+ x – 1

22. Find Numbers of Positive and Negative Zeros
Answer:
There are 2 or 0 positive real zeros, 2 or 0 negative
real zeros, and 6, 4, or 2 imaginary zeros.

23. A. positive: 2 or 0; negative: 3 or 1;
imaginary: 1, 3, or 5
B. positive: none; negative: none;
imaginary: 6
C. positive: 2 or 0; negative: 0;
imaginary: 6 or 4
D. positive: 2 or 0; negative: 2 or 0;
imaginary: 6, 4, or 2
State the possible number of positive real zeros,
negative real zeros, and imaginary zeros of
p(x) = x4
– x3
+ x2
+ x + 3.

24. 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.
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
f(x) = x3
– x2
+ 2x + 4
f(–x) = –x3
– x2
– 2x + 4

25. Use Synthetic Substitution to Find Zeros
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 synthetic substitution to
find f(a) for several values of a.
Each row in the table
shows the coefficients of
the depressed polynomial
and the remainder.

26. Use Synthetic Substitution to Find Zeros
From the table, we can see that one zero occurs at
x = –1. Since the depressed polynomial, x2
– 2x + 4, is
quadratic, use the Quadratic Formula to find the roots of
the related quadratic equation x2
– 2x + 4 = 0.
Quadratic Formula
Replace a with 1, b with
–2, and c with 4.

27. Use Synthetic Substitution to Find Zeros
Simplify.
Simplify.

28. 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.

29. What are all the zeros of f(x) = x3
– 3x2
– 2x + 4?
A.
B.
C.
D.

31. 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)]

32. Use Zeros to Write a Polynomial Function
Solve Multiply the factors to find the polynomial
function.
f(x) = (x – 4)[x – (4 – i)][x – (4 + i)] Write an equation.
= (x – 4)[(x – 4) + i)][(x – 4) – i)] Regroup terms.
= (x – 4)[(x – 4)2
– i2
] Rewrite as the
difference of two
squares.

33. Use Zeros to Write a Polynomial Function
Square x – 4 and
replace i2
with –1.
Simplify.
Multiply using the
Distributive
Property.
Combine like
terms.

34. Use Zeros to Write a Polynomial Function
Answer: f(x) = x3
– 12x2
+ 49x – 68 is a polynomial
function of least degree with integral
coefficients whose zeros are 4, 4 – i,
and 4 + i.

35. A. x2
– 3x + 2 – xi + 2i
B. x2
– 2x + 2
C. x3
– 4x2
+ 6x – 4
D. x3
+ 6x – 4
What is a polynomial function of least degree with
integral coefficients the zeros of which include
2 and 1 + i?