A.4 & 4.5 Polynomial Long Division, Synthetic Division, and Zeros of Polynomial Functions Vocabulary and Main Ideas: ο· In this section and the next we want to learn how to find the zeros of polynomial functions that we don’t already know how to factor (note: if we already know how to factor a polynomial then finding the zeros is easy!). Ex 1: Let π(π₯) = π₯ 3 − 7π₯ − 6. Even though we cannot factor π by any methods we’ve learned thus far, we still want to be able to find the zeros of π! To find these zeros, we need to learn how to divide polynomials. ο· Polynomials can be divided just like whole numbers can. This process is called polynomial long division. In fact, the process of polynomial long division mirrors the process of whole number long division Ex 2: Perform each division. a. (π₯ 3 − 5π₯ 2 − 23π₯ − 7) ÷ (π₯ + 4) o ο· b. (π₯ 5 − 3π₯ 2 − 9π₯ + 9) ÷ (π₯ 2 − 3) Notice in part b that we were able to find two zeros of the function π(π₯) = π₯ 5 − 3π₯ 2 − 9π₯ + 9. When dividing by a linear factor of the form π − π, the process of polynomial long division can be shortened significantly into what we call synthetic division. Ex 3: Use synthetic division to divide the polynomials. a. (π₯ 3 − 5π₯ 2 − 23π₯ − 7) ÷ (π₯ + 4) ο· b. (π₯ 4 − 3π₯ 3 − 7π₯ 2 − 10π₯ + 8) ÷ (π₯ − 2) The remainder theorem states that when doing a division problem of the formπ(π) ÷ (π − π), the remainder obtained from the division problem is the same as evaluating π(π). Ex 4: For each function, find the indicated function values and compare your answers to the remainders found in example 3. a. Let π(π₯) = π₯ 3 − 5π₯ 2 − 23π₯ − 7, find π(−4) b. Let π(π₯) = π₯ 4 − 3π₯ 3 − 7π₯ 2 − 10π₯ + 8, find π(2) Ex 5: Let π(π₯) = π₯ 3 − 7π₯ − 6. a. Use synthetic division to find π(−1). b. Use the result from part a to factor the polynomial function completely. c. Find all the zeros of π. ο· The rational zeros theorem gives us a method to find all the possible rational zeros of a polynomial function. o Let π be a polynomial function (so π(π₯) = ππ π₯ π + ππ−1 π₯ π−1 + β― + π1 π₯ + π0 ). Then all the rational π (“fractional”) zeros of π must be of the form π = π, where p is a factor of ππ and q is a factor of ππ . o The rational zeros theorem says nothing about irrational zeros (for example π₯ = √3 might be the zero of a polynomial function, but the rational zeros theorem would not reveal this). Ex 6: For each polynomial function, list all possible rational zeros. a. π(π₯) = 3π₯ 3 + 5π₯ 2 − 26π₯ + 8 b. π(π₯) = π₯ 7 − 3π₯ 4 + π₯ − 3 ο· To find all zeros of a polynomial function using division: 1. List all possible rational zeros. 2. Test possible rational zeros using synthetic division until you find an actual zero. 3. Use the actual zero and the quotient found from synthetic division to factor the polynomial. 4. Set each factor to zero and solve. Ex 7: Let π(π₯) = 3π₯ 3 + 5π₯ 2 − 26π₯ + 8. Find all the zeros of f. More Examples: 1. Perform the indicated division. Use synthetic division when possible. Write your answer as ππππππππ + a. (4π₯ 2 − 8π₯ + 6) ÷ (2π₯ − 1) π₯ 3 −27 b. π₯−3 2π₯ 5 −8π₯ 4 +2π₯ 3 +π₯ 2 c. 2π₯ 3 +1 πππππππ ππ . π ππππππ d. (π₯ 7 + π₯ 5 − 10π₯ 3 + 12) ÷ (π₯ + 2) 2. Let π(π₯) = 2π₯ 3 − 3π₯ 2 − 11π₯ + 6. Use synthetic division to find π(1), π(−3), and π(−2). 3. Use the results from #2 to find all the zeros of the function π(π₯) = 2π₯ 3 − 3π₯ 2 − 11π₯ + 6. 4. Use the rational zeros theorem and synthetic division to factor each polynomial completely. Then, find all the zeros of the function. a. π(π₯) = π₯ 3 − 2π₯ 2 − 11π₯ + 12 b. π(π₯) = π₯ 4 − 2π₯ 2 − 16π₯ − 15