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Learn to solve equations, determine number and type of roots, and understand concepts such as Zero Product Property. Step-by-step examples included.

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  1. Splash Screen

  2. Concept

  3. Concept

  4. Original equation Factor. Zero Product Property Solve each equation. 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. Example 1

  5. y4 – 256 = 0 Original equation Factor. (y2 + 16) (y2 – 16) = 0 (y2 +16) (y + 4)(y – 4) = 0 Factor. 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 Example 1

  6. 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, 4iand –4i. Example 1

  7. 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

  8. B. Solve a4 – 81 = 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 Example 1

  9. Concept

  10. Concept

  11. yes – to + yes + to – no – to – no – to – 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 Example 2

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

  13. 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

  14. State the possible number of positive real zeros, negative real zeros, and imaginary zeros ofp(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

  15. yes yes no no no yes 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 f(–x) = –x3 – x2 – 2x + 4 Example 3

  16. 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

  17. Quadratic Formula 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. Example 3

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

  19. 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. Use Synthetic Substitution to Find Zeros Example 3

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

  21. Concept

  22. 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

  23. 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

  24. Square x – 4 and replacei2 with –1. Simplify. Multiply using the Distributive Property. Combine like terms. Use Zeros to Write a Polynomial Function Example 4

  25. 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

  26. 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

  27. Homework P. 363 # 3 – 42 (x3)

  28. End of the Lesson

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