1. Polynomial and Synthetic Division

2. What you should learn How to use long division to divide polynomials by other polynomials How to use synthetic division to divide polynomials by binomials of the form (x – k) How to use the Remainder Theorem and the Factor Theorem

3. 1. x goes into x 3 ?x 2 times. 2. Multiply (x-1) by x 2. 4. Bring down 4x. 5. x goes into 2x 2 ?2x times. 6. Multiply (x-1) by 2x. 8. Bring down -6. 9. x goes into 6x? 3. Change sign, Add. 7. Change sign, Add 6 times. 11. Change sign, Add. 10. Multiply (x-1) by 6.

4. Long Division. Check

5. Divide.

6. Long Division. Check

7. Example Check =

8. Review Long Division. Check

9. Synthetic Division - To use synthetic division: There must be a coefficient for every possible power of the variable. The divisor must have a leading coefficient of 1. divide a polynomial by a polynomial

10. Step #1: Write the terms of the polynomial so the degrees are in descending order. Since the numerator does not contain all the powers of x, you must include a 0 for the

11. Step #2: Write the constant r of the divisor x-r to the left and write down the coefficients. Since the divisor is x-3, r=3 50-416

12. 5 Step #3: Bring down the first coefficient, 5.

13. 5 Step #4: Multiply the first coefficient by r, so and place under the second coefficient then add. 15

14. 5 Step #5: Repeat process multiplying the sum, 15, by r; and place this number under the next coefficient, then add. 45 41

15. 5 15 45 41 Step #5 cont.: Repeat the same procedure. 123 124 372 378 Where did 123 and 372 come from?

16. Step #6: Write the quotient. The numbers along the bottom are coefficients of the power of x in descending order, starting with the power that is one less than that of the dividend. 5 15 45 41 123 124 372 378

17. The quotient is: Remember to place the remainder over the divisor.

18. Synthetic Division Divide x 4 – 10x 2 – 2x + 4 by x + 3 10-10-24 -3 1 +9 3 1 -3 1

19. 1-2-8 3 1 3 1 3 -5

20. The Remainder Theorem If a polynomial f(x) is divided by x – k, the remainder is r = f(k).

21. The Factor Theorem 27-4-27-18 +2 2 4 11 22 18 36 9 18 0

22. Uses of the Remainder in Synthetic Division The remainder r, obtained in synthetic division of f(x) by (x – k), provides the following information. 1.r = f(k) 2.If r = 0 then (x – k) is a factor of f(x). 3.If r = 0 then (k, 0) is an x intercept of the graph of f.

23. Theorems About Roots of Polynomial Equations Rational Roots

24. Consider the following... x 3 – 5x 2 – 2x + 24 = 0 This equation factors to: (x+2)(x-3)(x-4)= 0 The roots therefore are: -2, 3, 4

25. Take a closer look at the original equation and our roots: x 3 – 5x 2 – 2x + 24 = 0 The roots therefore are: -2, 3, 4 What do you notice? -2, 3, and 4 all go into the last term, 24!

26. Spooky! Let’s look at another

27. Take a closer look at the original equation and our roots:

28. This leads us to the Rational Root Theorem

29. 1. For polynomial Possible roots are ___________________________________ Here p = -3 and q = 1 Factors of -3 Factors of 1  2. For polynomial Possible roots are ______________________________________________ Here p = 12 and q = 3 Factors of 12 Factors of 3  Or 3,-3, 1, -1 Where did all of these come from?

30. Let’s look at our solutions Note that + 2 is listed twice; we only consider it as one answer Note that + 1 is listed twice; we only consider it as one answer That is where our 9 possible answers come from! Note that + 4 is listed twice; we only consider it as one answer

31. Let’s Try One Find the POSSIBLE roots of 5x 3 -24x 2 +41x-20=0

32. Let’s Try One 5x 3 -24x 2 +41x-20=0 The possible roots are:

33. That’s a lot of answers! Obviously 5x 3 -24x 2 +41x-20=0 does not have all of those roots as answers. Remember: these are only POSSIBLE roots. We take these roots and figure out what answers actually WORK.

34. Step 1 – find p and q p = -3 q = 1 Step 2 – by RRT, the only rational root is of the form… Factors of p Factors of q

35. Step 3 – factors Factors of -3 = ±3, ±1 Factors of 1 = ± 1 Step 4 – possible roots -3, 3, 1, and -1

36. Step 5 – Test each root Step 6 – synthetic division X X³ + X² – 3x – 3 -3 3 1 (-3)³ + (-3)² – 3(-3) – 3 = -12 (3)³ + (3)² – 3(3) – 3 = 24 (1)³ + (1)² – 3(1) – 3 = -4 (-1)³ + (-1)² – 3(-1) – 3 = 0 THIS IS YOUR ROOT BECAUSE WE ARE LOOKING FOR WHAT ROOTS WILL MAKE THE EQUATION =0 1 1 -3 -3 0 1-3 3 0 -1 0 1x² + 0x -3

37. Step 7 – Rewrite x³ + x² - 3x - 3 = (x + 1)(x² – 3)

38. Step 1 – find p and q p = -6 q = 1 Step 2 – by RRT, the only rational root is of the form… Factors of p Factors of q

39. Step 3 – factors Factors of -6 = ±1, ±2, ±3, ±6 Factors of 1 = ±1 Step 4 – possible roots -6, 6, -3, 3, -2, 2, 1, and -1

40. Step 5 – Test each root Step 6 – synthetic division X x³ – 5x² + 8x – 6 -6 6 3 -3 2 -2 1 THIS IS YOUR ROOT 3 1 -5 8 -6 -6 1 2 6 -2 3 0 1x² + -2x + 2 -450 78 0 -102 -2 -50 -2 -20

41. Step 7 – Rewrite x³ – 5x² + 8x – 6 = (x - 3)(x² – 2x + 2) Quadratic Formula