1. Polynomial Long and Synthetic Division Pre-Calculus

2. Plan for the section 2.3 2.3  Polynomial Division  Synthetic Division  Remainder Theorem  Factor Theorem  Using Division to Factor / Find Zeros Homework

3. Use long division to divide polynomials by other polynomials. Use synthetic division to divide polynomials by binomials of the form (x – k). Use the Remainder Theorem and the Factor Theorem. What You Should Learn

4. When our factoring techniques do not easily work… Analyzing and Graphing a Function  Let’s say we want to analyze this function and graph it: f(x) = x 7 - 8x 5 - 2x 4 - 21x 3 + 10x 2 + 108x + 72  We know the left and right behavior  We know the y- intercept  To get a good approximation of the graph, we need to know the x-intercepts or the “zeros”. To find all the real zeros of the function we must factor it completely. Determining if one polynomial is a factor of another polynomial Factoring a polynomial Polynomial division will help with this.

5. Today We are going to learn about the process of division Learn about a couple of theorems to help in factoring and solving higher level polynomials

6. Division of Polynomials In this section, we will study two procedures for dividing polynomials. These procedures are especially valuable in factoring and finding the zeros of polynomial functions.

7. Polynomial Division Polynomial Division is very similar to long division. Example:

8. Polynomial Division Subtract!! Subtract!! Subtract!!

9. Example: Notice that there is no x term. However, we need to include it when we divide.

10. Polynomial Division

11. Try This Example: Answer:

12. What does it mean if a number divides evenly into another?? No remainder Remainder = 0 The Quotient is the FACTOR of the polynomial (Factor Theorem)

13. Now let’s look at another method to divide… Why??? Sometimes it is easier…

14. Synthetic Division Synthetic Division is a ‘shortcut’ for polynomial division that only works when dividing by a linear factor (x - k). It involves the coefficients of the dividend, and the zero of the divisor.

15. Synthetic Division The pattern for synthetic division of a cubic polynomial is summarized as follows. (The pattern for higher-degree polynomials is similar.)

16. Synthetic Division This algorithm for synthetic division works only for divisors of the form x – k. Remember that x + k = x – (–k).

17. Example  Divide:  Step 1:  Write the coefficients of the dividend in a upside-down division symbol. 156

18. Example  Step 2:  Take the zero of the divisor, and write it on the left.  The divisor is x – 1, so the zero is 1. 1561

19. Example  Step 3:  Carry down the first coefficient. 1561 1

20. Example  Step 4:  Multiply the zero by this number. Write the product under the next coefficient. 1561 1 1

21. Example  Step 5:  Add. 1561 1 1 6

22. Example  Step etc.:  Repeat as necessary 1561 1 1 6 6 12

23. Example The numbers at the bottom represent the coefficients of the answer. The new polynomial will be one degree less than the original. 1561 1 1 6 6 12

24. Using Synthetic Division Use synthetic division to divide x 4 – 10x 2 – 2x + 4 by x + 3. Solution: You should set up the array as follows. Note that a zero is included for the missing x 3 -term in the dividend.

25. Example – Solution Then, use the synthetic division pattern by adding terms in columns and multiplying the results by –3. So, you have. cont’d

26. Try These Examples:  (x 4 + x 3 – 11x 2 – 5x + 30)  (x – 2)  (x 4 – 1)  (x + 1) [Don’t forget to include the missing terms!] Answers:  x 3 + 3x 2 – 5x – 15  x 3 – x 2 + x – 1

27. Application of Long Division To begin, suppose you are given the graph of f (x) = 6x 3 – 19x 2 + 16x – 4.

28. Long Division of Polynomials Notice that a zero of f occurs at x = 2. Because x = 2 is a zero of f, you know that (x – 2) is a factor of f (x). This means that there exists a second-degree polynomial q (x) such that f (x) = (x – 2)  q(x). To find q(x), you can use long division.

29. Example - Long Division of Polynomials Divide 6x 3 – 19x 2 + 16x – 4 by x – 2, and use the result to factor the polynomial completely.

30. Example 1 – Solution Think Multiply: 6x 2 (x – 2). Subtract. Multiply: 2(x – 2). Subtract. Multiply: –7x(x – 2). Subtract.

31. Example – Solution From this division, you can conclude that 6x 3 – 19x 2 + 16x – 4 = (x – 2)(6x 2 – 7x + 2) and by factoring the quadratic 6x 2 – 7x + 2, you have 6x 3 – 19x 2 + 16x – 4 = (x – 2)(2x – 1)(3x – 2). cont’d

32. Example 1 – Solution Note that this factorization agrees with the graph shown in Figure 2.28 in that the three x-intercepts occur at x = 2, x =, and x =. cont’d Figure 2.28

33. Long Division of Polynomials In Example 1, x – 2 is a factor of the polynomial 6x 3 – 19x 2 + 16x – 4, and the long division process produces a remainder of zero. Often, long division will produce a nonzero remainder. For instance, if you divide x 2 + 3x + 5 by x + 1, you obtain the following.

34. Long Division of Polynomials In fractional form, you can write this result as follows. This implies that x 2 + 3x + 5 = (x + 1)(x + 2) + 3 which illustrates the following theorem, called the Division Algorithm. Multiply each side by (x + 1).

35. Long Division of Polynomials

36. The Remainder and Factor Theorems

37. The remainder obtained in the synthetic division process has an important interpretation, as described in the Remainder Theorem. The Remainder Theorem tells you that synthetic division can be used to evaluate a polynomial function. That is, to evaluate a polynomial function f (x) when x = k, divide f (x) by x – k. The remainder will be f (k).

38. Example Using the Remainder Theorem Use the Remainder Theorem to evaluate the following function at x = –2. f (x) = 3x 3 + 8x 2 + 5x – 7 Solution: Using synthetic division, you obtain the following.

39. Example – Solution Because the remainder is r = –9, you can conclude that f (–2) = –9. This means that (–2, –9) is a point on the graph of f. You can check this by substituting x = –2 in the original function. Check: f (–2) = 3(–2) 3 + 8(–2) 2 + 5(–2) – 7 = 3(–8) + 8(4) – 10 – 7 = –9 r = f (k) cont’d

40. The Remainder and Factor Theorems Another important theorem is the Factor Theorem, stated below. This theorem states that you can test to see whether a polynomial has (x – k) as a factor by evaluating the polynomial at x = k. If the result is 0, (x – k) is a factor.

41. Example – Factoring a Polynomial: Repeated Division Show that (x – 2) and (x + 3) are factors of f (x) = 2x 4 + 7x 3 – 4x 2 – 27x – 18. Then find the remaining factors of f (x). Solution: Using synthetic division with the factor (x – 2), you obtain the following. 0 remainder, so f (2) = 0 and (x – 2) is a factor.

42. Example – Solution Take the result of this division and perform synthetic division again using the factor (x + 3). Because the resulting quadratic expression factors as 2x 2 + 5x + 3 = (2x + 3)(x + 1) the complete factorization of f (x) is f (x) = (x – 2)(x + 3)(2x + 3)(x + 1). 0 remainder, so f (–3) = 0 and (x + 3) is a factor. cont’d

43. The Remainder and Factor Theorems For instance, if you find that x – k divides evenly into f (x) (with no remainder), try sketching the graph of f. You should find that (k, 0) is an x-intercept of the graph.

44. Use long division to divide polynomials by other polynomials. Use synthetic division to divide polynomials by binomials of the form (x – k). Use the Remainder Theorem and the Factor Theorem. Can you…