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A

Anonymous

Jan 22, 2025

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

Approximate the real number solution(s) to the polynomial function ( f(x) = x^3 - 4x^2 + x + 6 ).

The candidate solutions to consider are as follows: - ( x = -3 ) - ( x = -1 ) - ( x = 2 ) - ( x = 3 ) - ( x = 1 ) - ( x = -2 )

Next, determine the quotient of the expression ( \frac{72y^3 + 12y^2 - 30y}{6y} ). The options are: 1. ( 12y^2 - 2y + 5 ) 2. ( 12y^2 + 2y - 5 ) 3. ( 12y^3 + 2y^2 - 5y ) 4. ( 12y^2 - 2y - 5 )

Question 7 (Multiple Choice Worth 1 point): What is the quotient of ( \frac{2x^2 + 12x + 18}{x + 3} )? 1. ( 2x + 6 ) 2. ( 2x - 6 ) 3. ( 2x + 4 ) 4. ( 2x - 4, r = 1 )

Question 8 (Multiple Choice Worth 1 point): What is the quotient of ( \frac{6x^4 - 15x^3 + 10x^2 - 10x + 4}{3x^2 + 2} )? 1. ( 2x^2 - 5x + 2 ) 2. ( 2x^2 + 5x - 2 ) 3. ( 2x^2 - 5x - 2 ) 4. ( 2x^2 + 5x + 2 )

Question 9 (Multiple Choice Worth 1 point): Using synthetic division, what is the quotient of ( \frac{2x^3 + 3x - 22}{x - 2} )? 1. ( 2x^2 + 4x - 11 ) 2. ( 2x^2 - 4x + 11 + \frac{3}{x - 2} ) 3. ( 2x^2 - 4x - 11 ) 4. ( 2x^2 + 4x + 11 )

Question 10 (Multiple Choice Worth 1 point): What is the remainder when ( 3x^4 + 2x^3 - x^2 + 2x - 9 ) is divided by ( x + 2 )? 1. 0 2. 5 3. 10 4. 15

Question 11 (Multiple Choice Worth 1 point): What are the zeros of the polynomial function ( f(x) = (x - 4)(2x + 1)(x + 7) )? 1. 4, (\frac{1}{2}), -7 2. -4, -(\frac{1}{2}), 7 3. 4, -(\frac{1}{2}), -7 4. -4, (\frac{1}{2}), 7

Question 12 (Multiple Choice Worth 1 point): According to the Fundamental Theorem of Algebra, how many zeros does the function ( f(x) = 17x^{15} + 41x^{12} + 13x^3 - 10 ) have? 1. 15 2. 12 3. 30 4. 3

Question 13 (Multiple Choice Worth 1 point): What are the zeros of the polynomial function ( f(x) = x^3 - 10x^2 + 24x )? 1. -6, 0, 4 2. 6, 0, -4 3. -6, 0, -4 4. 6, 0, 4

Question 14 (Multiple Choice Worth 1 point): What are the possible rational zeros of ( f(x) = x^4 + 2x^3 - 3x^2 - 4x + 18 )? 1. ( \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 7, \pm 8, \pm 9, \pm 10, \pm 11, \pm 12, \pm 13, \pm 14, \pm 15, \pm 16, \pm 17, \pm 18 ) 2. ( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 ) 3. 1, 2, 3, 6, 9, 18 4. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18

Question 15 (Multiple Choice Worth 1 point): What are the possible number of positive, negative, and complex zeros of ( f(x) = 3x^4 - 5x^3 - x^2 - 8x + 4 )? 1. Positive: 2; Negative: 2; Complex: 4, 2, or 0 2. Positive: 1; Negative: 3 or 1; Complex: 2 or 0 3. Positive: 3 or 1; Negative: 1; Complex: 2 or 0 4. Positive: 2 or 0; Negative: 2 or 0; Complex: 4, 2, or 0

Question 16 (Multiple Choice Worth 1 point): What are the possible number of positive, negative, and complex zeros of ( f(x) = -x^6 + x^5 - x^4 + 4x^3 - 12x^2 + 12 )? 1. Positive: 4, 2, or 0; Negative: 2 or 0; Complex: 6, 4, 2, or 0 2. Positive: 3 or 1; Negative: 3 or 1; Complex: 4, 2, or 0 3. Positive: 5, 3, or 1; Negative: 1; Complex: 4, 2, or 0 4. Positive: 2 or 0; Negative: 2 or 0; Complex: 6, 4, or 2

Question 17 (Multiple Choice Worth 1 point): Which of the following represent the zeros of ( f(x) = 6x^3 + 25x^2 - 24x + 5 )? 1. -5, (\frac{1}{3}), (\frac{1}{2}) 2. 5, -(\frac{1}{3}), (\frac{1}{2}) 3. 5, (\frac{1}{3}), -(\frac{1}{2}) 4. 5, (\frac{1}{3}), (\frac{1}{2})

Question 18 (Multiple Choice Worth 1 point): Which of the following is a polynomial with roots (-\sqrt{3}), (\sqrt{3}), and -2? 1. ( x^3 - 2x^2 - 3x + 6 ) 2. ( x^3 + 2x^2 - 3x - 6 ) 3. ( x^3 - 3x^2 - 5x + 15 ) 4. ( x^3 + 3x^2 - 5x - 15 )

Question 19 (Multiple Choice Worth 1 point): Which of the following is a polynomial with roots -2, -2i, and 2i? 1. ( x^3 - 4x^2 + 4x - 16 ) 2. ( x^3 + 4x^2 + 4x + 16 ) 3. ( x^3 + 2x^2 + 4x + 8 ) 4. ( x^3 - 2x^2 + 4x - 8 )

Question 20 (Essay Worth 1 point): 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

A
Anonymous

Jan 19, 2025

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

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