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

A. 21

B. 75

C. 855

D. 4091

Use synthetic substitution to find f(2) for f(r) = 3r4 + 7r2 – 12r + 23.

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.

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

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

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?

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?

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.

Original equation

Factor.

Solve each equation.

Zero Product Property

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 equationy4 – 256 = 0

Factor.(y2 +16) (y + 4)(y – 4) = 0

Determine Number and Type of Roots

y2 = –16 y = –4 y = 4Solve each equation.

Answer: This equation has two real roots, –4 and 4, and two imaginary roots, 4i and –4i.

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.

A. 2 real: –3 and 3

B. 2 real: –3 and 32 imaginary: 3i and –3i

C. 2 real: –9 and 92 imaginary: 3i and –3i

D. 2 real: –9 and 92 imaginary: 9i and –9i

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

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 –

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

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.

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 ofp(x) = x4 – x3 + x2 + x + 3.

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

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.

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.

Use Synthetic Substitution to Find Zeros

Simplify.

Simplify.

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 ConjugateTheorem. So, x – 4, x – (4 – i), and x – (4 + i) are factors of the

polynomialfunction.Plan Write the polynomial function as aproduct of its factors.f(x) = (x – 4)[x – (4 – i)][x – (4 + i)]

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.

Use Zeros to Write a Polynomial Function

Square x – 4 and replace i2 with –1.

Simplify.

Multiply using the Distributive Property.

Combine like terms.

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.

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?