Polynomial and Synthetic Division

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 x3? x2 times. 2. Multiply (x-1) by x2. 3. Change sign, Add. 4. Bring down 4x. 5. x goes into 2x2? 2xtimes. 6. Multiply (x-1) by 2x. 7. Change sign, Add 8. Bring down -6. 9. x goes into 6x? 6times. 10. Multiply (x-1) by 6. 11. Change sign, Add .

4. Long Division. Check

5. Divide.

6. Long Division. Check

7. Example = Check

8. Review Long Division. Check

9. Synthetic Division - divide a polynomial by a polynomial 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.

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

11. 5 0 -4 1 6 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

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

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

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

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

16. 15 45 123 372 15 41 124 378 5 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.

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

18. Synthetic Division Divide x4 – 10x2 – 2x + 4 by x + 3 1 0 -10 -2 4 -3 -3 +9 -3 3 -1 1 1 1 -3

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

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

21. The Factor Theorem • A polynomial f(x) has a factor (x-k) if and only if f(k)=0. • Show that (x-2) is a factor of: 2 7 -4 -27 -18 +2 4 22 18 36 9 0 2 11 18

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. • r = f(k) • If r = 0 then (x – k) is a factor of f(x). • 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 . . . x3 – 5x2 – 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: x3 – 5x2 – 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 • 24x3 – 22x2 – 5x + 6 = 0 • This equation factors to: • (2x+1)(3x-2)(4x-3)= 0 • The roots therefore are:

27. Take a closer look at the original equation and our roots: • 24x3 – 22x2 – 5x + 6 = 0 • This equation factors to: (2x+1)(3x-2)(4x-3)= 0 The roots therefore are: What do you notice? The numerators 1, 2, and 3 all go into the last term, 6! The denominators 2, 3, and 4 all go into the first term, 24!

28. This leads us to the Rational Root Theorem For a polynomial, If is a root of the polynomial, then p is a factor of and q is a factor of

29. Example (RRT) 1. For polynomial Here p = -3 and q = 1 Factors of -3 Factors of 1  Or 3,-3, 1, -1 Possible roots are ___________________________________ 2. For polynomial Here p = 12 and q = 3 Factors of 12 Factors of 3  Possible roots are ______________________________________________ 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 Note that + 4 is listed twice; we only consider it as one answer That is where our 9 possible answers come from!

31. Let’s Try One Find the POSSIBLE roots of 5x3-24x2+41x-20=0

32. Let’s Try One 5x3-24x2+41x-20=0 The possible roots are:

33. That’s a lot of answers! • Obviously 5x3-24x2+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 -1 1 1 -3 -3 -3 3 1 -1 (-3)³ + (-3)² – 3(-3) – 3 = -12 (3)³ + (3)² – 3(3) – 3 = 24 3 -1 0 (1)³ + (1)² – 3(1) – 3 = -4 1 0 -3 0 (-1)³ + (-1)² – 3(-1) – 3 = 0 THIS IS YOUR ROOT BECAUSE WE ARE LOOKING FOR WHAT ROOTS WILL MAKE THE EQUATION =0 1x² + 0x -3

37. Step 7 – Rewrite • x³ + x² - 3x - 3 = (x + 1)(x² – 3) Step 8 – factor more and solve (x + 1)(x² – 3) (x + 1)(x –)(x + ) Roots are -1, ±

38. Using the Polynomial TheoremsFACTOR and SOLVE x³ – 5x² + 8x – 6 = 0 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. Using the Polynomial TheoremsFACTOR and SOLVE x³ – 5x² + 8x – 6 = 0 • 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. Using the Polynomial TheoremsFACTOR and SOLVE x³ – 5x² + 8x – 6 = 0 • Step 5 – Test each root • Step 6 – synthetic division X x³ – 5x² + 8x – 6 -450 78 0 -102 -2 -50 -2 -20 -6 6 3 -3 2 -2 1 -1 3 1 -5 8 -6 THIS IS YOUR ROOT 6 3 -6 1 -2 2 0 1x² + -2x + 2

41. Using the Polynomial TheoremsFACTOR and SOLVE x³ – 5x² + 8x – 6 = 0 • Step 7 – Rewrite • x³ – 5x² + 8x – 6 = (x - 3)(x² – 2x + 2) • Step 8– factor more and solve • (x - 3)(x² – 2x + 2) • Roots are 3, Quadratic Formula