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Essential Questions

Essential Questions How do we apply various theorems to find roots of polynomial functions and to solve polynomial equations? Standards MM3A3: Students will solve a variety of equations and inequalities.

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Essential Questions

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  1. Essential Questions • How do we apply various theorems to find roots of polynomial functions and to solve polynomial equations? • Standards • MM3A3: Students will solve a variety of equations and inequalities. • MM3A3a: Find real and complex roots of higher degree polynomial equations using the factor theorem, remainder theorem, rational root theorem, and fundamental theorem of algebra, incorporating complex and radical conjugates. • MM3A3b: Solve polynomial, exponential, and logarithmic equations analytically, graphically, and using appropriate technology. • MM3A3c: Solve polynomial, exponential, and logarithmic inequalities analytically, graphically, and using appropriate technology. Represent solution sets of inequalities using interval notation. • MM3A3d: Solve a variety of types of equations by appropriate means, choosing among mental calculation, pencil and paper, or appropriate technology.

  2. Activating Strategy: Factor: 1. f(x) = x2 – 5x – 24 2. g(x) = 2x2 – 9x – 5

  3. Activating Strategy: Factor and find the roots: f(x) = x2 – 5x – 24 (x 8) (x 3) (x – 8) (x + 3) g(x) = 2x2 – 9x – 5 (x 8) (x 3) (x – 8) (x + 3)

  4. Place the correct signs to give the middle term.  2x² + 7x − 15 = (2x − 3)(x + 5) b)  2x² − 7x − 15 = (2x + 3)(x − 5)  c)  2x² − x − 15 = (2x + 5)(x − 3)  d)  2x² − 13x + 15 = (2x − 3)(x − 5) 

  5. Place the correct signs to give the middle term.  2x² + 7x − 15 = (2x − 3)(x + 5) b)  2x² − 7x − 15 = (2x + 3)(x − 5)  c)  2x² − x − 15 = (2x + 5)(x − 3)  d)  2x² − 13x + 15 = (2x − 3)(x − 5) 

  6. Polynomial Roots Task Factoring Quadratics Solving polynomials that have a degree greater than those solved in Math I and Math II is going to require the use of skills that were developed when quadratic equations were solved in Math II. Let’s begin by taking a look at some second degree polynomials and the strategies used to solve them. These equations have the form ax2 + bx + c = 0, and when they are graphed the result is a parabola. Factoring is used to solve quadratics of the form when the roots are rational. 1. Find the roots of the following quadratic functions: a. f(x) = x2 – 5x – 14 b. f(x) = x2 – 64 c. f(x) = 6x2 + 7x - 3

  7. Polynomial Roots Task Factoring Quadratics Solving polynomials that have a degree greater than those solved in Math I and Math II is going to require the use of skills that were developed when quadratic equations were solved in Math II. Let’s begin by taking a look at some second degree polynomials and the strategies used to solve them. These equations have the form ax2 + bx + c = 0, and when they are graphed the result is a parabola. Factoring is used to solve quadratics of the form when the roots are rational. 1. Find the roots of the following quadratic functions: a. f(x) = x2 – 5x – 14 b. f(x) = x2 – 64 c. f(x) = 6x2 + 7x - 3 f(x) = (x + 2)(x – 7) f(x) = (x + 8)(x – 8) f(x) = (3x – 1)(2x + 3) x = -2, 7 x = -8, 8 x = 1/3, -3/2

  8. Using the Discriminant and Quadratic Formula Another option for solving a quadratic whether it is factorable, but particularly when it is not is to use the quadratic formula. Remember that this concept was developed during Math II. Remember that is the discriminant and gives us the ability to determine the nature of the roots. Calculate the discriminant to determine the number and nature of the roots, then determine the roots for each of the following. a. f(x) = 4x2 – 2x + 9 b. f(x) = 3x2 + 4x – 8 c. f(x) = x2 – 6x + 9

  9. Using the Discriminant and Quadratic Formula Another option for solving a quadratic whether it is factorable, but particularly when it is not is to use the quadratic formula. Remember that this concept was developed during Math II. Remember that is the discriminant and gives us the ability to determine the nature of the roots. Calculate the discriminant to determine the number and nature of the roots, then determine the roots for each of the following. a. f(x) = 4x2 – 2x + 9 b. f(x) = 3x2 + 4x – 8 c. f(x) = x2 – 6x + 9 = (-2)2 – 4(4)(9) = (4)2 – 4(3)(-8) = (-6)2 – 4(1)(9) = 4 – 144 = 16 + 96 = 36 - 36 = -140 = 112 = 0 2 Imaginary roots 2 real roots 1 real root x = 1 ± i√35 x = -1 + √7 x = 3 2 2

  10. Consider the situation of a polynomial that is one degree greater. Based on what we learned about polynomial graphs, think about what similarities a cubic function would have to a quadratic function when we want to find the roots. For example, suppose we want to find the roots of . By inspecting the graph of the function, we can see that one of the roots is distinctively 2. Since we know that x = 2 is a solution to , we also know that is a factor of the expression . This means that if we divide by there will be a remainder of zero. Polynomial Long Division Before we start trying to divide by (x – 2), let’s think about something we learned in elementary school, long division. a. Use long division to solve. b. Now, we are going to use the same idea to divide polynomials.

  11. Consider the situation of a polynomial that is one degree greater. Based on what we learned about polynomial graphs, think about what similarities a cubic function would have to a quadratic function when we want to find the roots. For example, suppose we want to find the roots of . By inspecting the graph of the function, we can see that one of the roots is distinctively 2. Since we know that x = 2 is a solution to , we also know that is a factor of the expression . This means that if we divide by there will be a remainder of zero. Polynomial Long Division Before we start trying to divide by (x – 2), let’s think about something we learned in elementary school, long division. a. Use long division to solve. b. Now, we are going to use the same idea 81 21/23 to divide polynomials. X2 + 4x + 3 -368 -(x3 – 2x2) 88 4x2 – 5x -46-(4x2 – 8x) 4242/46 = 21/23 3x – 6 46 –(3x – 6) 0

  12. Use polynomial long division to find each quotient and remainder. 1. (x4 + 3x2 – 4x + 2)  (x + 1) 5. (2x3 + 17x2 + 32x – 13)  (x + 5) 2. (x4 – 2x3 – 2x2 + 5x – 2)  (x2 – 3x + 2) 6. (5x3 + 13x2 – 24x – 18)  (5x + 3)

  13. 3. (x5 + 2x3 – 5x + 8)  (x – 3) 7. (3x4 + 2x – 5)  (x – 3) 4. (6x3 – 7x2 + 3x + 7)  (2x + 1) 8. (x4 – 2x3 + 4x2 – 2x + 3)  (x2 + 1)

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