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**Five-Minute Check (over Lesson 2-3)**Then/Now New Vocabulary Key Concept: Rational Zero Theorem Example 1: Leading Coefficient Equal to 1 Example 2: Leading Coefficient not Equal to 1 Example 3: Real-World Example: Solve a Polynomial Equation Key Concept: Upper and Lower Bound Tests Example 4: Use the Upper and Lower Bound Tests Key Concept: Descartes’ Rule of Signs Example 5: Use Descartes’ Rule of Signs Key Concept: Fundamental Theorem of Algebra Key Concept: Linear Factorization Theorem Example 6: Find a Polynomial Function Given Its Zeros Key Concept: Factoring Polynomial Functions Over the Reals Example 7: Factor and Find the Zeros of a Polynomial Function Example 8: Find the Zeros of a Polynomial When One is Known Lesson Menu**A.**B. C. D. Divide (2x4 – 3x3 – 13x2 + 36x – 45) ÷ (2x – 5) using long division. 5–Minute Check 1**A.**B. C. D. Divide (2x4 – 3x3 – 13x2 + 36x – 45) ÷ (2x – 5) using long division. 5–Minute Check 1**A.**B.x2 + 4x + 3 C. D.x2 – 4x + 1 Find (x3 + 2x2 – 5x – 6) ÷ (x – 2) using synthetic division. 5–Minute Check 2**A.**B.x2 + 4x + 3 C. D.x2 – 4x + 1 Find (x3 + 2x2 – 5x – 6) ÷ (x – 2) using synthetic division. 5–Minute Check 2**A.**B. C. D. Find (x4 – x3 + 2x + 5) ÷ (x + 2) using synthetic division. 5–Minute Check 3**A.**B. C. D. Find (x4 – x3 + 2x + 5) ÷ (x + 2) using synthetic division. 5–Minute Check 3**Use the Factor Theorem to determine if (x – 2) and (x + 1)**are factors of f(x) = x4 – 2x3 – 9x2 + 32x – 28. Use the binomials that are factors to write a factored form of f(x). A.yes, no; f(x) = (x – 2)(x3 – 9x + 14) B.no, yes; f(x) = (x + 1)(x3 – 3x2 – 6x + 28) C.yes, no; f(x) = (x – 2)2(x – 7) D.yes, yes; f(x) = (x – 2)(x + 1)(x2 + x – 8) 5–Minute Check 4**Use the Factor Theorem to determine if (x – 2) and (x + 1)**are factors of f(x) = x4 – 2x3 – 9x2 + 32x – 28. Use the binomials that are factors to write a factored form of f(x). A.yes, no; f(x) = (x – 2)(x3 – 9x + 14) B.no, yes; f(x) = (x + 1)(x3 – 3x2 – 6x + 28) C.yes, no; f(x) = (x – 2)2(x – 7) D.yes, yes; f(x) = (x – 2)(x + 1)(x2 + x – 8) 5–Minute Check 4**Find f(3) using synthetic substitution iff(x) = x6 – 5x5 +**6x4 + 4x3 + 12x2 + 8. A.–2422 B.8 C.80 D.224 5–Minute Check 5**Find f(3) using synthetic substitution iff(x) = x6 – 5x5 +**6x4 + 4x3 + 12x2 + 8. A.–2422 B.8 C.80 D.224 5–Minute Check 5**You learned that a polynomial function of degree n can have**at most n real zeros. (Lesson 2-1) • Find real zeros of polynomial functions. • Find complex zeros of polynomial functions. Then/Now**Rational Zero Theorem**• lower bound • upper bound • Descartes’ Rule of Signs • Fundamental Theorem of Algebra • Linear Factorization Theorem • Conjugate Root Theorem • complex conjugates • irreducible over the reals Vocabulary**Leading Coefficient Equal to 1**A. List all possible rational zeros of f(x) = x3–3x2 – 2x + 4. Then determine which, if any, are zeros. Step 1 Identify possible rational zeros. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros are ±1, ±2, and ±4. Example 1**Leading Coefficient Equal to 1**Step 2 Use direct substitution to test each possible zero. f(1) = (1)3 – 3(1)2 – 2(1) + 4 or 0 f(–1) = (–1)3 – 3(–1)2 – 2(–1) + 4 or 2 f(2) = (2)3 – 3(2)2 – 2(2) + 4 or –4 f(–2) = (–2)3 – 3(–2)2 – 2(–2) + 4 or –12 f(4) = (4)3 – 3(4)2 – 2(4) + 4 or 12 f(–4) = (–4)3 – 3(–4)2 – 2(–4) + 4 or –100 The only rational zero is 1. Answer: Example 1**Leading Coefficient Equal to 1**Step 2 Use direct substitution to test each possible zero. f(1) = (1)3 – 3(1)2 – 2(1) + 4 or 0 f(–1) = (–1)3 – 3(–1)2 – 2(–1) + 4 or 2 f(2) = (2)3 – 3(2)2 – 2(2) + 4 or –4 f(–2) = (–2)3 – 3(–2)2 – 2(–2) + 4 or –12 f(4) = (4)3 – 3(4)2 – 2(4) + 4 or 12 f(–4) = (–4)3 – 3(–4)2 – 2(–4) + 4 or –100 The only rational zero is 1. Answer: ±1, ± 2, ± 4; 1 Example 1**Leading Coefficient Equal to 1**B. List all possible rational zeros of f(x) = x3 – 2x – 1. Then determine which, if any, are zeros. Step 1 Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term –1. Therefore, the possible rational zeros of f are 1 and –1. Example 1**1 1 0 –2 –1**1 1 –1 1 1 –1 –2 –1 1 0 –2 –1 –1 1 1 1 –1 –1 0 Leading Coefficient Equal to 1 Step 2 Begin by testing 1 and –1 using synthetic substitution. Example 1**Leading Coefficient Equal to 1**Because f(–1) = 0, you can conclude that x = –1 is a zero of f. Thus f(x) = (x + 1)(x2 – x – 1). Because the factor x2 – x – 1 yields no rational zeros, we can conclude that f has only one rational zero, x = –1. Answer: Example 1**Leading Coefficient Equal to 1**Because f(–1) = 0, you can conclude that x = –1 is a zero of f. Thus f(x) = (x + 1)(x2 – x – 1). Because the factor x2 – x – 1 yields no rational zeros, we can conclude that f has only one rational zero, x = –1. Answer:±1; 1 Example 1**A.**B. C. D. List all possible rational zeros of f(x) = x4 – 12x2 – 15x – 4. Then determine which, if any, are zeros. Example 1**A.**B. C. D. List all possible rational zeros of f(x) = x4 – 12x2 – 15x – 4. Then determine which, if any, are zeros. Example 1**Leading Coefficient not Equal to 1**List all possible rational zeros of f(x) = 2x3– 5x2 – 28x + 15. Then determine which, if any, are zeros. Step 1 The leading coefficient is 2 and the constant term is 15. Possible rational zeros: Example 2**Testing each subsequent possible zero on the depressed**polynomial, you can determine that x = 5 and are also rational zeros. By the division algorithm, f(x) = (x + 3)(x – 5)(2x – 1) so the rational zeros are x = –3, x = 5, and . Check this result by graphing. Leading Coefficient not Equal to 1 Step 2 By synthetic substitution, you can determine that x = –3 is a rational zero. Example 2**Leading Coefficient not Equal to 1**Answer: Example 2**Answer:**Leading Coefficient not Equal to 1 Example 2**A.**B. C. D. List all possible rational zeros of f(x) = 4x3 – 20x2 + x – 5. Then determine which, if any, are zeros. Example 2**A.**B. C. D. List all possible rational zeros of f(x) = 4x3 – 20x2 + x – 5. Then determine which, if any, are zeros. Example 2**Solve a Polynomial Equation**WATER LEVEL The water level in a bucket sitting on a patio can be modeled by f(x) = x3+ 4x2 –2x + 7, where f(x) is the height of the water in millimeters and x is the time in days. On what day(s) will the water reach a height of 10 millimeters? Because f(x) represents the day when the water level, you need to solve f(x) = 10 to determine what day the water will reach a height of 10 millimeters. Example 3**Solve a Polynomial Equation**f(x) = 10 Write the equation. x3 + 4x2 – 2x + 7 = 10 Substitute x3 + 4x2 – 2x + 7 for f(x). x3 + 4x2 – 2x – 3 = 0 Subtract 10 from each side. Apply the Rational Zeros Theorem to this new polynomial function, f(x) = x3 + 4x2 – 2x – 3. Step 1 Possible rational zeros: factors of –3 = ±1, ±3. Example 3**1 1 4 –2 –3**1 5 3 1 5 3 0 Solve a Polynomial Equation Step 2 Because the number of the day cannot be negative, check each of the positive rational zeros using synthetic substitution. Doing so, you can determine that x = 1 is the only positive rational zero of f. Because x = 1 is a zero of f, x = 1 is a solution of f(x) = 0. So, it was day 1 when the water reached a height of 10 millimeters. Answer: Example 3**1 1 4 –2 –3**1 5 3 1 5 3 0 Solve a Polynomial Equation Step 2 Because the number of the day cannot be negative, check each of the positive rational zeros using synthetic substitution. Doing so, you can determine that x = 1 is the only positive rational zero of f. Because x = 1 is a zero of f, x = 1 is a solution of f(x) = 0. So, it was day 1 when the water reached a height of 10 millimeters. Answer:day 1 Example 3**A. 4 seconds, 10 seconds**B. 4 seconds C. 5 seconds, seconds D. 5 seconds PHYSICS The path of a ball is given by the function f(x) = –4.9x2 + 21.5x + 40, where x is the time in seconds and f(x) is the height above the ground in meters. After how many seconds will the ball reach a height of 25 meters? Example 3**A. 4 seconds, 10 seconds**B. 4 seconds C. 5 seconds, seconds D. 5 seconds PHYSICS The path of a ball is given by the function f(x) = –4.9x2 + 21.5x + 40, where x is the time in seconds and f(x) is the height above the ground in meters. After how many seconds will the ball reach a height of 25 meters? Example 3**Use the Upper and Lower Bound Tests**Determine an interval in which all real zeros of f(x) = x4– 4x3 – 11x2 – 4x – 12 must lie. Explain your reasoning using the upper and lower bound tests. Then find all the real zeros. Step 1 Graph f(x). From this graph, it appears that the real zeros of this function lie in the interval [–3, 7]. Example 4**–3 1 –4 –11 –4 –12**–3 21 –30 102 1 –7 10 –34 90 Values alternate signs in the last line, so –3 is a lower bound. 7 1 –4 –11 –4 –12 7 21 70 462 1 3 10 66 450 Values are all nonnegative in last line, so 7 is an upper bound. Use the Upper and Lower Bound Tests Step 2 Test a lower bound of c = –3 and an upper bound of c = 7. Example 4**Use the Upper and Lower Bound Tests**Step 3 Use the Rational Zero Theorem. Possible rational zeros: Factors of 12 = ±1, ±2, ±3, ±4 , ±6, ±12 . Because the real zeros are in the interval [–3, 7], you can narrow this list to just –1, –2, –3, 1, 2, 3, 4, and 6. From the graph it appears that only –2 and 6 are reasonable. Example 4**6 1 –4 –11 –4 –12**6 12 6 12 1 2 1 2 0 –2 1 2 1 2 –2 0 –2 1 0 1 0 Use the Upper and Lower Bound Tests Begin by testing 6. Now test –2 in the depressed polynomial. Example 4**Use the Upper and Lower Bound Tests**By the division algorithm, f(x) = (x – 6)(x + 2)(x2 + 1). Notice that the factor x2 + 1 has no real zeros associated with it because x2 + 1 = 0 has no real solutions. So f has two real solutions that are both rational, x = –2 and x = 6. The graph of f(x) = x4 – 4x3 – 11x2 – 4x – 12 supports this conclusion. Answer: Example 4**Use the Upper and Lower Bound Tests**By the division algorithm, f(x) = (x – 6)(x + 2)(x2 + 1). Notice that the factor x2 + 1 has no real zeros associated with it because x2 + 1 = 0 has no real solutions. So f has two real solutions that are both rational, x = –2 and x = 6. The graph of f(x) = x4 – 4x3 – 11x2 – 4x – 12 supports this conclusion. Answer:Upper and lower bounds may vary. Sample answer: [–3, 7]; With synthetic division, the values alternate signs when testing –3, and are all nonnegative when testing 7. So, –3 is a lower bound and 7 is an upper bound. The zeros are –2 and 6. Example 4**A. [0, 4]; 1,2**B. [–1, 2]; 1, C. [–3, 5]; 1, D. [–2, 1]; 1, Determine an interval in which all real zeros of f(x) = 2x4 – 5x3 – 13x2 + 26x – 10 must lie. Then find all the real zeros. Example 4**A. [0, 4]; 1,2**B. [–1, 2]; 1, C. [–3, 5]; 1, D. [–2, 1]; 1, Determine an interval in which all real zeros of f(x) = 2x4 – 5x3 – 13x2 + 26x – 10 must lie. Then find all the real zeros. Example 4**f(x) = x4 – 3x3 – 5x2 + 2x + 7**+ to – – to + = x4 + 3x3 – 5x2 – 2x + 7 + to – – to + Use Descartes’ Rule of Signs Describe the possible real zeros of f(x) = x4– 3x3 – 5x2 + 2x + 7. Examine the variations of sign for f(x) and for f(–x). f(–x) = (–x)4 – 3(–x)3 – 5(–x)2 + 2(–x) + 7 Example 5**Use Descartes’ Rule of Signs**The original function f(x) has two variations in sign, while f(–x) also has two variations in sign. By Descartes' Rule of Signs, you know that f(x) has either 2 or 0 positive real zeros and either 2 or 0 negative real zeros. Answer: Example 5**Use Descartes’ Rule of Signs**The original function f(x) has two variations in sign, while f(–x) also has two variations in sign. By Descartes' Rule of Signs, you know that f(x) has either 2 or 0 positive real zeros and either 2 or 0 negative real zeros. Answer:2 or 0 positive real zeros, 2 or 0 negative real zeros Example 5**Describe the possible real zeros of g(x) = –x3 + 8x2 –**7x + 9. A. 3 or 1 positive real zeros, 1 negative real zero B. 3 or 1 positive real zeros, 0 negative real zeros C. 2 or 0 positive real zeros, 0 negative real zeros D. 2 or 0 positive real zeros, 1 negative real zero Example 5