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Approximate the real number solution(s) to the polynomial function f(x) = x3 – 4×2 + x + 6?

x = –3, x = –1, x = 2 x = 3, x = 1, x = –2 x = 3, x = –1, x = 2 x = –3, x = 1, x = –2 What is the quotient: (72y3 + 12y2 – 30y) ÷ 6y ? 12y2 – 2y + 5 12y2 + 2y – 5 12y3 + 2y2 – 5y 12y2 – 2y – 5 Question 7 (Multiple Choice Worth 1 points)What is the quotient: (2x2 + 12x + 18) ÷ (x + 3) ? 2x + 6 2x – 6 2x + 4 2x – 4, r = 1 Question 8 (Multiple Choice Worth 1 points)What is the quotient: (6x4 – 15x3 + 10x2 – 10x + 4) ÷ (3x2 + 2) ? 2x2 – 5x + 2 2x2 + 5x – 2 2x2 – 5x – 2 2x2 + 5x + 2 Question 9 (Multiple Choice Worth 1 points)Using synthetic division, what is the quotient: (2x3 + 3x – 22) ÷ (x – 2) ? 2x2 + 4x – 11 2x2 – 4x + 11 + 3 over the quantity of x minus 2 2x2 – 4x – 11 2x2 + 4x + 11 Question 10 (Multiple Choice Worth 1 points)What is the remainder when (3x4 + 2x3 – x2 + 2x – 9) ÷ (x + 2) ? 0 5 10 15 Question 11 (Multiple Choice Worth 1 points)What are the zeros of the polynomial function: f(x) = (x – 4)(2x + 1)(x + 7)? 4, one–half, –7 –4, –one–half, 7 4, –one–half, –7 –4, one–half, 7 Question 12 (Multiple Choice Worth 1 points)According to The Fundamental Theorem of Algebra, how many zeros does the function f(x) = 17x15 + 41x12 + 13x3 – 10 have ? 15 12 30 3 Question 13 (Multiple Choice Worth 1 points)What are the zeros of the polynomial function: f(x) = x3 – 10x2 + 24x ? –6, 0, 4 6, 0, –4 –6, 0, –4 6, 0, 4 Question 14 (Multiple Choice Worth 1 points)What are the possible rational zeros of f(x) = x4 + 2x3 – 3x2 – 4x + 18 ? ± 1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 7, ± 8, ± 9, ± 10, ± 11, ± 12, ± 13, ± 14, ± 15, ± 16, ± 17, ±18 ± 1, ± 2, ± 3, ± 6, ± 9, ± 18 1, 2, 3, 6, 9, 18 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 Question 15 (Multiple Choice Worth 1 points)What are the possible number of positive, negative, and complex zeros of f(x) = 3x4 – 5x3 – x2 – 8x + 4 ? Positive: 2; Negative: 2; Complex: 4 or 2 or 0 Positive: 1; Negative: 3 or 1; Complex: 2 or 0 Positive: 3 or 1; Negative: 1; Complex: 2 or 0 Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0 Question 16 (Multiple Choice Worth 1 points)What are the possible number of positive, negative, and complex zeros of f(x) = –x6 + x5– x4 + 4x3 – 12x2 + 12 ? Positive: 4, 2, or 0; Negative: 2 or 0; Complex: 6, 4, 2, or 0 Positive: 3 or 1; Negative: 3 or 1; Complex: 4, 2, or 0 Positive: 5, 3, or 1; Negative: 1; Complex: 4, 2, or 0 Positive: 2 or 0; Negative: 2 or 0; Complex: 6, 4, or 2 Question 17 (Multiple Choice Worth 1 points)Which of the following represent the zeros of f(x) = 6x3 + 25x2 – 24x + 5 ? –5, one–third, one–half 5, –one–third, one–half 5, one–third, – one–half 5, one–third, one–half Question 18 (Multiple Choice Worth 1 points)Which of the following is a polynomial with roots – square root of 3, square root of 3, and –2 ? x3 – 2x2 – 3x + 6 x3 + 2x2 – 3x – 6 x3 – 3x2 – 5x + 15 x3 + 3x2 – 5x – 15 Question 19 (Multiple Choice Worth 1 points)Which of the following is a polynomial with roots –2, –2i, and 2i ? x3 – 4x2 + 4x – 16 x3 + 4x2 + 4x + 16 x3 + 2x2 + 4x + 8 x3 – 2x2 + 4x – 8 Question 20 (Essay Worth 1 points)Find a third–degree polynomial function such that f(0) = –12 and whose zeros are 1, 2, and 3. Using complete sentences, explain how you found it.

2 Answers

F(x) = x³ - 4x² + x + 6 → you wish to have to factorize it You need to in finding an convenient (- 2, - 1, 0, 1, 2) root to get zero. In case you alternative x via - 1: f(- 1) = (- 1)³ - 4(- 1)² + (- 1) + 6 f(- 1) = (- 1) - 4(1) + (- 1) + 6 f(- 1) = - 1 - 4 - 1 + 6 f(- 1) = 0 → so which you can factorize: [x - (- 1)] → (x + 1) = (x + 1)(ax² + bx + c) = ax³ + bx² + cx + ax² + bx + c = ax³ + ax² + bx² + bx + cx + c = ax³ + x²(a + b) + x(b + c) + c → you examine with: x³ - 4x² + x + 6 that you would be able to deduce that: c = 6 b + c = 1 → b = 1 - c → b = 1 - 6 → b = - 5 a + b = - four → a = - 4 - b → a = - 4 + 5 → a = 1 remember: (x + 1)(ax² + bx + c) x³ - 4x² + x + 6 = (x + 1)(x² - 5x + 6) Now, let's appear: (x² - 5x + 6) Polynomial like: ax² + bx + c, the place: a = 1 b = - 5 c = 6 Δ = b² - 4ac (discriminant) Δ = (- 5)² - 4(1 * 6) = 25 - 24 = 1 x1 = (- b - √Δ) / 2a = (5 - 1) / (2 * 1) = 2 x2 = (- b + √Δ) / 2a = (5 + 1) / (2 * 1) = 3 x² - 5x + 6 = (x - 2)(x - 3) f(x) = x³ - 4x² + x + 6 f(x) = (x + 1)(x - 2)(x - three) if you want to solve: f(x) = zero answer = - 1; 2; 3... Show More

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