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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 Lesson Menu

Use synthetic substitution to find f(2) for f(r) = 3r4 + 7r2 – 12r + 23. A. 21 B. 75 C. 855 D. 4091 5-Minute Check 1

Use synthetic substitution to find f(6) for f(c) = 2c3 + 19c2 + 2. A. 94 B. 727 C. 1118 D. 1619 5-Minute Check 2

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) 5-Minute Check 3

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) 5-Minute Check 4

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 5-Minute Check 5

What value of k would give a remainder of 6 when x2 + kx + 18 is divided by x + 4? C. 1 D. 3 5-Minute Check 6

Mathematical Practices 6 Attend to precision. 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. CCSS

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. Then/Now

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A. Solve x2 + 2x – 48 = 0. State the number and type of roots. Determine Number and Type of Roots A. Solve x2 + 2x – 48 = 0. State the number and type of roots. Original equation Factor. Zero Product Property Solve each equation. Answer: This equation has two real roots, –8 and 6. Example 1

B. Solve y4 – 256 = 0. State the number and types of roots. Determine Number and Type of Roots B. Solve y4 – 256 = 0. State the number and types of roots. Original equation y4 – 256 = 0 Factor. (y2 + 16) (y2 – 16) = 0 Factor. (y2 +16) (y + 4)(y – 4) = 0 y2 + 16 = 0 or y + 4 = 0 or y – 4 = 0 Zero Product Property Example 1

y2 = –16 y = –4 y = 4 Solve each equation. 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. Example 1

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 Example 1

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 Example 1

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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 – Example 2

p(–x) = –(–x)6 + 4(–x)3 – 2(–x)2 – (–x) – 1 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. p(–x) = –(–x)6 + 4(–x)3 – 2(–x)2 – (–x) – 1 –x6 – 4x3 – 2x2 + 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. Example 2

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. Example 2

A. positive: 2 or 0; negative: 3 or 1; imaginary: 1, 3, or 5 State the possible number of positive real zeros, negative real zeros, and imaginary zeros of p(x) = x4 – x3 + x2 + x + 3. 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 Example 2

Find all of the zeros of f(x) = x3 – x2 + 2x + 4. 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). f(x) = x3 – x2 + 2x + 4 yes yes no f(–x) = –x3 – x2 – 2x + 4 no no yes Example 3

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. Example 3

Replace a with 1, b with –2, and c with 4. 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. Example 3

Use Synthetic Substitution to Find Zeros Simplify. Simplify. Example 3

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. Example 3

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

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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)] Example 4

Solve Multiply the factors to find the polynomial function. 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. Example 4

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

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. Example 4

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 Example 4

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