Name: _________________________ Period: _________________________ Unit 6B Polynomials: Solving Higher Degree Algebra II 1/22/2013 1/21/13 1/22/13 1/23/13 1/24/13 1/25/13 1/28/13 1/29/13 1/30/13 1/31/13 2/1/13 Video Section 6.6 Section 6.7 Section 6.8 Section 6.9 Section 6.10 Classwork Section 6.6 Section 6.7 Section 6.8 Section 6.9 Section 6.10 Section 6.10 Worksheet Practice Test Unit 6B Exam Due Section 6.6 Section 6.7 Section 6.8 Section 6.9 Section 6.10 Practice Test Section 6.6 Evaluating Polynomial Functions Objective(s): Use synthetic substitution to evaluate polynomials. Essential Question: Homework: Assignment 6.6. #1 – 12 in the homework packet. Notes: Types of polynomials Leading Coefficient Degree Type Example 0 1 2 3 4 Constant Linear Quadratic Cubic Quartic f(x) = 4 f(x) = 2x + 1 f(x) = -x2 + 7x - 5 f(x) = 6x3 + x2 - 19 f(x) = -7x4 + 2x2 + 3 2 -1 6 -7 Depending on the polynomial, it can be easier to evaluate (plug in a number) using a process called synthetic substitution. Let’s suppose you had a polynomial f(x) = 2x4 + x2 – 12x + 6 and you wanted to evaluate it for x = 2. Traditionally, you would evaluate it in the following way. f ( x) 2 x 4 x 2 12 x 6 f (2) 2(2) 4 (2) 2 12(2) 6 2(16) 4 24 6 32 4 24 6 18 Sometimes it is quicker and easier to evaluate using synthetic substitution. 1. Rewrite the polynomial as f(x) = 2x4 + 0x3 + x2 – 12x + 6 (Hint: if the degree is 4, then you need FIVE terms – one for each power of x plus the constant.) 2. Take all of the coefficients and write them in a row. 3. Leaving space for a second row, draw a giant L. Reflection: 2 0 1 12 6 1 4. Write the 2 (evaluate for x = 2) on the outside of the giant L. 2 2 0 1 12 6 2 2 0 1 12 6 5. Drop the first number on the left of the top row. 2 6. Multiply this number by the number on the outside and put the result under the 0. 2 × 2 = 4 2 2 0 1 12 6 4 2 2 2 0 1 12 6 7. Now add the 0 and the 4. 0 + 4 = 4 and put this under the 0 and 4. 4 2 4 8. Multiply the 4 with the outside 2 and put the result under the 1. Then add. 2 2 0 1 12 6 4 8 2 4 9 2 2 0 1 12 6 9. Repeat for all of the numbers. 4 8 2 4 9 Notice how the last number in the bottom row is the same as f(2). This is NOT a coincidence. Use synthetic substitution to evaluate. Example 1: Reflection: If f(x) = -3x3+ 2x2 – 5, find f(4) (Are there any missing terms??? Remember to write a 0.) 2 Example 2: If g(x) = -x5 – 4x3 + 6x2 – x, find g(-2) Example 3: If f(x) = x4 – 6x3 + 8x2, find f(3). Example 4: If h(x) = x5 – 3x3 + 1, find h(-1). Example 5: For f (x) = x³ - 3x² + 4x -2 show that f(1) = 0 1 1 3 4 2 This means that 1 is a ______________of the polynomial x³ - 3x² + 4x -2 . x³ - 3x² + 4x -2 = 0 has ______zeros. What is one of them ? _________ If x = _______is a solution, then what is a factor of the polynomial? _____________________ Reflection: 3 Section 6.7 The Remainder Theorem Objective(s): Divide polynomials and relate the result to the remainder theorem and the factor theorem. Essential Question: If f(x) is a polynomial that has (x - 7) as a factor, what do you know about the value of f(7)? Homework: Assignment 6.7. #13 – 23 in the homework packet. Notes: If a polynomial f(x) is divided by x - k, then the remainder is r = f(k). Let f ( x) 3x3 2 x 2 2 x 5 If we use long division to divide x – 2 into our polynomial, we have x 2 3x3 2 x 2 2 x 5 3x 2 x 2 3x3 2 x 2 2 x 5 divide 3x3 by x and get 3x2, put on top and multiply 3x 2 x 2 3x3 2 x 2 2 x 5 3x 6 x 3 now subtract 2 What is your remainder? _________________ Now, what would happen if we were to use synthetic substitution? 2 3 2 2 5 Notice that the other numbers are the coefficients of the answer(quotient). Reflection: 4 The remainder theorem says that using synthetic substitution for x = 2 will give you the same remainder as dividing by x – 2. Find the remainder. Example 1: (x3 – x2 + 5 ) ÷ (x + 2) (Are there any missing terms??? Remember to write a 0.) Divide using synthetic substitution. Example 2: (10x4 + 5x3 + 4x2 – 9 ) ÷ (x + 1) Example 3: (2x4 – 6x3 + x2 – 3x – 3 ) ÷ (x – 3) Reflection: 5 Section 6.8 The Factor Theorem Objective(s): Divide polynomials and relate the result to the remainder theorem and the factor theorem. Essential Question: If f(x) is a polynomial that has (x - 7) as a factor, what do you know about the value of f(7)? Homework: Assignment 6.8. #24 – 39 in the homework packet. Notes: How do you know if 23 is a factor of 1766? The Factor Theorem states that a polynomial f(x) has a factor x - k if and only if f(k) = 0. If the remainder is zero, then x - k is a factor. In addition, k is a solution/root of the polynomial. Use synthetic substitution to determine whether the binomial is a factor of f(x). Example 1: f(x) = x3 – 8x2 + 21x – 18; x – 2 Yes No Example 2: f(x) = x3 + 9x2 + 14x – 24; x – 3 Yes No Using synthetic substitution, is the given number a root/zero of the polynomial? Example 3: Reflection: Is 5 a zero (or root) of f(x) = x3 – 10x2 + 11x + 70? Yes No 6 Example 4: Is 4 a zero (or root) of f(x) = x3 – 6x2 + 11x – 6? Yes No Factor the polynomial f(x) given one of the zeros. Example 5: f(x) = x3 – 4x2 – 17x + 60 given 3 is a zero/root Example 6: f(x) = x3 + 3x2 – 16x – 48 given 4 is a zero/root Given one zero/root, find the others. Example 7: f(x) = x3 – 12x2 + 44x – 48; x = 2 Example 8: f(x) = x3 + 11x2 + 36x + 36; x = -2 Reflection: 7 Sample CCSD Common Exam Practice Question(s): 1. What is x 3 3 x 2 9 x 2 divided by x 2 ? A. x 2 x 11 B. x 2 2 x 11 C. x2 5x 1 D. x 2 4 x 5 Reflection: 8 Section 6.9 Using the Fundamental Theorem of Algebra Objective(s): Use the Fundamental Theorem of Algebra to determine the number of zeros of a polynomial function. Solve polynomial equations. Essential Question: What is the conjugate of a complex number, and why is it important when finding all of the zeros of a polynomial function? Homework: Assignment 6.9. #40 – 46 in the homework packet. Notes: The Fundamental Theorem of Algebra states that the degree of the polynomial is the same as the TOTAL number of real and imaginary solutions/roots/zeros. For the function, find the number of zeros that the function has. Example 1: f(x) = x5 – 5x3 + 10x + 4 Number of zeros = ______________ Example 2: f(x) = x8 – 7x4 + 2x3 + 4x2 – 12x – 7 Number of zeros = ______________ The complex zeros of a polynomial function with real coefficients always occur in complex conjugate pairs. That is, if a + bi is a zero, then a – bi must also be a zero. Find the missing root. 5 are roots of f(x) = x4 – x3 – 11x2 + 5x + 30, what is the missing root? Example 3: If -2, 3, and Example 4: If -4 and 2 + 3i are roots of f(x) = x3 – 3x + 52, what is the missing root? Example 5: If 2i and 3i are roots of f(x) = x4 + 13x2 + 36, what are the missing roots? Example 6: If 7 is a zero of a polynomial, what is the factor? Example 7: If -2, 3, and 5 are zeros of a polynomial, what are the factors? Find a polynomial function with -2, 3, and 5 as zeros. (Hint: multiply it out) Reflection: 9 Example 8: Find a polynomial function with -1, 2i, and -2i as zeros. Example 9: Find a polynomial function with 3, 4i, and -4i as zeros. Sample CCSD Common Exam Practice Question(s): 1. According to the Fundamental Theorem of Algebra, how many complex zeros does the polynomial f x 5 x 4 2 x 3 x 1 have? A. B. C. D. 2 3 4 5 Reflection: 10 Section 6.10 Finding Rational Zeros Objective(s): Identify all rational zeros of a polynomial function by using the rational root theorem. Find rational zeros of a polynomial. Essential Question: If the leading coefficient of a polynomial with integer coefficients is 1, what type of numbers must any possible rational zeros be? Homework: Assignment 6.10. #47 – 58 in the homework packet. Notes: Rational Zero Theorem – If a polynomial has integer coefficients, then every rational zero of the polynomial has the following form all factors of constant all factors of lead coefficient For example, all the possible rational zeros of f(x) = x3 + 2x2 – 11x – 12 would be 1, 2, 3, 4, 6, 12 which would simplify to 1, 2, 3, 4, 6, 12 . 1 All the possible rational zeros of f(x) = 2x3 + 7x – 12 would be 1 3 1, 2, 3, 4, 6, 12 which would simplify to 1, , , 2, 3, 4, 6, 12 . 2 2 1, 2 List all possible rational zeros of f. Example 1: f(x) = 2x5 + 3x3 – 7 Example 2: f(x) = 2x4 + 8x3 – 7x2 – 9x + 8 Reflection: 11 Example 3: f(x) = 3x5 + 12x4 + 21x3 – 8x2 – x – 27 Example 4: Which is NOT a possible rational root of f(x) = 6x4 + 7x2 – 3x + 2 a) -2 b) 2/3 c) ½ d) -3 Procedure to find 3 roots 1. List all possible rational roots 2. Use synthetic substitution until you find a root (remainder = 0) 3. Use the new coefficients and find the remaining 2 roots by factoring or the quadratic formula. x b b2 4ac 2a Given the polynomial function f(x), find all real and complex zeros. Example 5: Reflection: f(x) = x3 – 7x + 6 12 Given the polynomial function f(x), find all real and complex zeros. Example 6: f(x) = x3 – 4x2 + 9x – 10 For the polynomial function, solve for all its zeros (both real and imaginary), and graph it - include a description of its degree, roots, and end behavior. Example 7: f(x) = – x3 – 3x2 + 13x + 15 Degree = _______ Possible roots = _________________________ Actual roots = __________________________ As x → + ∞ f(x) → ____________ As x → - ∞ f(x) → ____________ Reflection: 13 Example 8: f(x) = 3x3 + 7x2 – 7x – 3 Degree = _______ Possible roots = _________________________ Actual roots = __________________________ As x → + ∞ f(x) → ____________ As x → - ∞ f(x) → ____________ Reflection: 14 Example 9: f(x) = x4 + 2x3 – 9x2 – 2x + 8 Degree = _______ Possible roots = _________________________ Actual roots = __________________________ As x → + ∞ f(x) → ____________ As x → - ∞ f(x) → ____________ Reflection: 15