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Day # 2 Review I. Polynomial Functions

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Day # 2 Review I. Polynomial Functions Day # 2 Review I. Polynomial Functions: 0 is a Polynomial Equation A root or solution is a value of for which the value of 0 Example #1 (Q) Determine whether 3 is a root of 2 560 (A) 3 3 23 53 6 27 18 15 6 0 00 Roots may be IMAGINARY! Complex Conjugates: : if is a root of with real coefficients, then is also a root. Fundamental Thm. of Algebra: A polynomial function of degree 0 has complex zeros. Some of these zeros may be repeated. Degree of polynomial indicates how many roots are possible. Linear Factorization Thm.: If is a polynomial function of degree 0, then has precisely linear factors and ## … Example #2 (Q) State the # of complex roots of 2 80. Find the roots. (A) 3 complex roots: 2 80. ON CALCULATOR: nd 28 0 Use 2 Calc. 42 0 Root Choose Upper & Lower Bounds 0,4 0,2 0 0,4,2 Example #3 (Q) Write with roots 3, 2, 2 (A) 0 3 2 2 3 4 → 1 3 4 3 412 Completing the Square: leading coefficient MUST be . Helps when the quadratic is not easily factorable. Example #4 (Q) Solve 0 10 50 1500 by completing the square. (A) 1500 10 50 5 25 6.25 1500 10 5 2 2 4 1500 106.25 10 5 6.25 1562.5 102.5 156.25 2.5 √156.25 2.5 12.5 2.5 12.5 2.5 10 12.5 2.5 15 Discriminant Nature of Roots 40 2 distinct Real roots 40 Exactly 1 Real root 40 No Real roots (2 imaginary) II. Synthetic Division and Remainder/Factor Theorems: Division Algorithm: Let and be polynomials with the degree of degree of , and 0. Then there are unique polynomials and , called the quotient and remainder, such that ⋅ . The summary statement above can also be written in fraction form: Remainder Thm: If a polynomial is divided by , then the remainder is . Example #5 (Q) Let 3 28 Find the remainder when is divided by 2. (A) 2 2 32 22 80 Factor Thm: A polynomial function has a factor 0. Example #6 (Q) Let 4 710. Determine if 5 is a factor of (A) 5 0 therefore 5 is a factor. Synthetic Division: A shortcut for the division of a polynomial by a linear divisor . Example #7 (Q) Divide 2 3 512 by 3 (A) Steps for Synthetic Division: 1. Use coefficients only; if a term is missing use 0. Write these numbers inside bracket. 2. Use c value outside bracket. 3. Bring down 1st coefficient. 4. Multiply by c and add to 2nd number inside bracket. 5. Multiply this number by c and add to 3rd number inside bracket. 6. Repeat the process until all coefficients have been used. 7. The last number is the remainder. 8. If necessary, rewrite the original function as . III. Rational Root Thm.: If a rational number , where and have NO common factors, is a root of the equation, then is a factor of (constant) and is a factor of (leading coefficient). If all roots are irrational you must use your calculator to find them. If the roots are rational and irrational / complex you synthetically divide by the rational roots to obtain a quadratic quotient function. Then solve this algebraically using the quadratic formula. Example #8 (Q) Find the roots if 6 10 3 , (A) 1,3 Pre-Calculus Name __________________ Day #2 Review Practice Date _______ Directions: Complete each problem. 1. Solve 9 6 algebraically. 2. Find the remainder after synthetically dividing 10 8 by 2. Is the divisor a factor of the dividend? Why or why not? 3. State the number of complex zeros for the polynomial functions listed below. List ALL possible rational zeros and find all that work using your calculator. If there are remaining irrational or complex zeros, find them after synthetically dividing by the rational zeros. If there are NO real zeros, state that only complex zeros exist. A. 34 56 B. 8 6 23 6 C. 27 8 14 8 4. Write a polynomial function with real coefficients with zeros 1,2 and degree 3. 5. Write the function 29 23 31 15 as a product of linear and irreducible quadratic factors all with real coefficients. You may use the graph of to aid in the location of possible real zeros. 6. Rewrite 5 25 12 in vertex form by completing the square. 7. Find the and for 23 5. 8. Divide 2 73 by 2, and write a summary statement in polynomial or quotient form. .
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