1. Finding Zeros of Polynomial Functions Taylor Johnson (Taylor.Johnson@kctcs.edu) Elizabethtown Community & Technical College

2. Tools for Searching for Zeros (1) Remainder Theorem (2) Factor Theorem (3) Intermediate Value Theorem (4) Synthetic Division (5) Rational Zero Theorem (6) Signs of Roots (Descartes’s Rule) (7) Product of Roots (8) Sum of Roots (9) Boundaries: upper and lower

3. (1) Remainder Theorem A polynomial function f with real coefficients divided by ( x – c ) yields the remainder = f ( c ). The division algorithm states: f ( x ) = ( x – c ) * q ( x ) + r (E1) Substitute x = c into (E1) to yield: f ( c ) = ( c – c ) * q ( c ) + r = r.

4. (2) Factor Theorem Let f ( x ) be a polynomial function with real coefficients. (a) If f ( c ) = 0, then ( x – c ) factors f ( x ). (b) If ( x – c ) factors f ( x ), then f ( c ) = 0. Recall from the Remainder Theorem that f ( c ) = r and is zero, thus ( x – c ) is a factor.

5. (3) Intermediate Value Theorem (IVT) Given polynomial function f with real coefficients and distinct real numbers a and b, if the sign of f ( a ) is the opposite of the sign of f ( b ), then at least one number c exists between a and b for which f ( c ) = 0, a root of the function f.

6. Example (IVT) Let f ( x ) = x 2 + 7 x – 60. f (0) = –60 f (8) = 64 + 56 – 60 = 60 The Remainder Theorem locates two points on opposite sides of the x-axis. Divide and conquer: f (4) = 16 + 28 – 60 = –16 f (6) = 36 + 42 – 60 = 18 f (5) = 25 + 35 – 60 = 0

7. Example (IVT) - (cont) f (0) = = –60 f (5) = 25 + 35 – 60 = 0 root at x = 5 f (8) = 64 + 56 – 60 = 60 This example illustrates how to combine these three theorems to find roots of a function.

8. (4) Synthetic Division This technique abbreviates long division of polynomials for divisors of the form (x – c). The “dividend” is the polynomial coefficients, including zeros, in row one. The “divisor” is the root c. Let f(x) = x 2 – 7x + 12 as an example, let c = 5 and start from left coefficient with 0 in row 2 and the sum of row one and row 2 in row 3.

9. 5 | 1 -7 12 0 1 Working left to right by coefficient record the product of row 3 value and the divisor on row 2 under the next coefficient and record the sum of row 1 and row 2 into row 3, until a sum appears in row 3 under the rightmost coefficient.

10. 5 | 1 -7 12 0 5 -10 1 -2 2 Row three is the result of the division. The rightmost number in row 3 is the remainder and the other numbers are the coefficients of the quotient which is q ( x ) = ( x – 2). The result follows: (x 2 – 7x + 12)/(x – 5) = (x – 2) + 2/(x – 5)

11. The result follows: (x 2 – 7x + 12)/(x – 5) = (x – 2) + 2/(x – 5) Notice that f (5) = 5*5 – 7 *5 + 12 = 25 – 35 + 12 = 2 as expected from the Remainder Theorem.

12. Performing the division (x 2 – 7x + 12)/(x – 4) = (x – 3) using synthetic division appears in this form: 4 | 1 -7 12 0 4 -12 1 -3 0 The zero remainder indicates the divisor is a root of f.

13. (5) Rational Zero Theorem (RZT) If f(x) = a n x n + a n-1 x n-1 + … + a 1 x + a 0 with integer coefficients and a rational zero of p/q, where (p, q) = 1, then p divides a 0 and q divides a n. Substitute x = p/q into a n x n + a n-1 x n-1 + … + a 1 x + a 0 = 0 and subtract a 0 on both sides to yield a n (p/q) n + a n-1 (p/q) n-1 + … + a 1 (p/q) = –a 0

14. Since p is a factor of each term on the left side, then it is a factor of a 0. Similarly solving for a n and dividing by x n shows that q is a factor of a n. For example, the function f(x) = 6x 2 + 7x – 5 has a n = 6 and a 0 = – 5.

15. In this example f(x) = 6x 2 + 7x – 5: p factors a 0 = – 5 into ± 1, ± 5. q factors a n = 6 into ± 1, ± 2, ± 3, ± 6. Possible rational zeros are: A leading coefficient of 1 eliminates the options from the denominator.

16. In this example g(x) = x 3 – 12x 2 + 41x – 42: p factors a 0 = – 42 into ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21, ± 42. q factors a n = 1 into ± 1 and has no impact. The p factors are the possible rational zeros. Checking x = 1 or x = –1 is fairly simple: g( 1) = 1 – 12 + 41 – 42 = –12 g(-1) = – 1 – 12 – 41 – 42 = – 94 Neither one or minus one is a root by the Factor Theorem.

17. Using synthetic division and the list of potential rational zeros the search for possible factors may begin. Let x = 2 and perform synthetic division: 2 | 1 -12 41 -42 0 2 -20 42 1 -10 21 0 With a remainder of zero start a list of roots with 2. Row three is the quotient: g(x) = x 2 – 10x + 21

18. After translating row three into the quotient, notice that it factors and reveals two more roots: g(x) = x 2 – 10x + 21 g(x) = ( x – 3 ) ( x – 7 ) The additional roots are 3 and 7. The search uncovered the three roots expected in a function of degree three, namely { 2, 3, 7 } The remaining topics present some number sense to help narrow the list of possible roots.

19. (6) Descartes’s Rules of Signs Let f ( x ) = a n x n + a n -1 x n -1 + … + a 1 x + a 0 be a polynomial with real coefficients. Count the number of sign changes in the function by adding one when contiguous terms have opposite signs.

20. Counting positive real zeros The number of positive real zeros is either: (a) the same as the number of sign changes in f ( x ) or (b) less than the number of sign changes by an even number.

21. Counting positive real zeros Examples f ( x ) = + x – 3 One sign change \ / One + root { 3 } g(x) = + x 2 – 10x + 21 Two sign changes \ / \ /Two + roots { 3, 7 }

22. Counting positive real zeros More Examples g(x) = +x 2 –10x+29 Two sign changes \/\ /Two pos. real roots or no real roots. g(x) = + x 2 – 10x + 25 = -4 (x - 5) 2 = -2Sq Root Prop x = 5 +/- 2i Roots { 5 - 2i, 5 + 2i }

23. Counting positive real zeros More Examples (2) g ( x ) = + x 3 – 12 x 2 + 41 x – 42Three sign chg \ / \ / \ /+ roots { 2, 3, 7 } h ( x ) = + x 3 + 8 x 2 + 1 x – 42One sign change \ /One positive root Now is the time to investigate negative roots.

24. Counting positive real zeros More Examples (2) g(x) = + x 3 – 12x 2 + 41x – 42Three sign chg \ / \ / \ /+ roots { 2, 3, 7 } h(x) = + x 3 + 8x 2 + 1x – 42One sign change \ /One positive root Now is the time to investigate negative roots.

25. Counting negative real zeros The number of negative real zeros is either: (a) the same as the number of sign changes in f (– x ) or (b) less than the number of sign changes by an even number.

26. Counting negative real zeros Examples Recall from last positive real zero example: h(x) = + x 3 + 8x 2 + 1x – 42One sign change \ /One positive root h(–x) = – x 3 + 8x 2 – 1x – 42Two sign changes \ /\ /Two neg. roots Or perhaps no negative roots. Hmm.

27. Counting negative real zeros Examples (cont) h(x) = + x 3 + 8x 2 + 1x – 42 Notice that this function will have the following possible rational roots: ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21, ± 42 Time to start the synthetic division tests! Or to stop and think about this.

28. (7) Product of Roots Let f ( x ) = ( x – r 1 )( x – r 2 ) … ( x – r n ) be the factorization of the function of degree n. Let f ( x ) = a n x n + a n -1 x n -1 + … + a 1 x + a 0 be a polynomial with real coefficients. Notice that (-1) n · a 0 is equivalent to the product of the roots from expanding the factorization. Both are the constant terms of two representations of the function.

29. (7) Product of Roots (cont.) Recall that the function f(x ) = + x 3 + 8 x 2 + 1 x – 42 has one positive root and possibly two negative roots. The product of the roots is (-1) 3 · -42 = 42. The prime factors of 42 are 2, 3, and 7. Or perhaps 1, 6, and 7.

30. f(x ) = + x 3 + 8 x 2 + 1 x – 42 Since 7 is a factor in both cases, check for 7 as a root using synthetic division: Let x = -7 (2 out 3 chances) and perform synthetic division: -7 | 1 8 1 -42 0 -7 -7 42 1 1 -6 0 And the quotient factors to yield 2 and -3. The solution set is { 2, -3, -7 }.

31. (8) Sum of Roots Let f ( x ) = ( x – r 1 )( x – r 2 ) … ( x – r n ) be the factorization of the function of degree n. Let f ( x ) = a n x n + a n -1 x n -1 + … + a 1 x + a 0 be a polynomial with real coefficients. The expansion of the factorization produces n terms of degree n -1 with a root as coefficient and their sum is the opposite of a n -1.

32. (8) Sum of Roots (cont) Again, recall that the function f(x ) = + x 3 + 8 x 2 + 1 x – 42 has one positive root and possibly two negative roots. The product of the roots is (-1)3· -42 = 42. The prime factors of 42 are 2, 3, and 7. The sum of the roots is – 8 = 2 – 3 – 7. No synthetic division required.

33. Another example from a previous exercise: g( x) = + x 3 + 12x 2 + 41x + 420 sgn chg 0 pos roots g(– x) = + x 3 – 12x 2 + 41x – 423 sgn chg \ / \ / \ / 3 neg roots – 12 = – 7 – 3 – 2 and (-7)(-3)(-2) = -42(-1) 3. No synthetic division required.

34. (9) Root Upper Boundary Let f ( x ) be a polynomial with real coefficients and a positive leading coefficient a. If p ( a ) > 0 and synthetic division yields row three with all positive values, then a is an upper bound for the roots of the polynomial.

35. (9) Boundaries (cont.) The leading coefficient of the quotient function q ( x ) is the same as p ( x ). Since a > 0 and all numbers on row three are positive, then q ( a ) > 0. The proof begins by selecting a root b and stating that p ( x ) = ( x – b )· q ( x ).

36. Let x equal a, then p ( a ) = ( a – b )· q ( a ). Dividing by q ( a ) yields ( a – b ) = p ( a )/ q ( a ) > 0. Since a > b and p ( a ) and q ( a ) are both positive conclude that a > b. An example follows.

37. Example (from earlier) p(x ) = + x 3 + 8 x 2 + 1 x – 42 has these roots: { –7, –3, 2 }. Given a = 3 > 0, p (3) = 60 > 0. and all row three numbers are positive, 3| 1 8 1 -42 0 3 33 102 1 11 34 60 the assumptions are satisfied. Thus 3 is an upper bound of the roots.

38. Example (from earlier) g(x) = + x 3 – 12x 2 + 41x – 42 has these roots: { 2, 3, 7 }. Given a = 8 > 0, p (8) = 30 > 0. but not all row three numbers are positive, 8| 1 -12 41 -42 0 8 32 72 1 -4 9 30 so not all assumptions are satisfied. The process does not guarantee least UB.

39. Increase the tested boundary to 13: Given a = 13 > 0, p (13) = 660 > 0. and all row three numbers are positive, 13| 1 -12 41 -42 0 13 13 702 1 1 54 660 so all assumptions are satisfied. The process does identify 13 as an upper bound of the roots. The sum coefficient greater than largest root distorts result.

40. (9) Root Lower Boundary Let f ( x ) be a polynomial with real coefficients and a negative leading coefficient a. If p ( a ) not equal to 0 and synthetic division yields row three with values of alternating signs, then a is a lower bound for the roots of the polynomial. The proof is similar to the upper bound.

41. Example of Lower Bound (earlier) g(x) = + x 3 + 12x 2 + 41x + 42 has these roots: { –2, – 3, – 7 }. Given a = -13 < 0, p (-13) = -660 ≠ 0, and all row three numbers have alternate signs, -13| 1 12 41 42 13 -13 13 -702 1 -1 54 -660 so all assumptions are satisfied. The process does not guarantee least UB.

42. One Last Example f(x) = x 3 – x – 60 This is what I call the Cardano polynomial. It led him to speculate about the imaginary unit. To find the roots set f ( x ) = 0 and begin: x 3 – x – 60 = 0

43. Add 60 on both sides and factor left side: x 3 – x = 60 x(x 2 – 1) = 60 (x – 1)x(x + 1) = 60 Three consecutive numbers with a product of 60 are 3, 4, and 5. Thus x = 4 is a root. f(x) = x 3 – x – 60 f(-x) = – x 3 + x – 60 The rules of signs indicate one positive root and two negative roots (or none).

44. f(x) = x 3 + 0x 2 – x – 60 The sum of the roots is zero and the product is (-60)(-1) 3 = 60. The positive solution is 4. Regardless of whether signed or not, the other two solutions are a conjugate pair, say (r + s) and (r – s ).

45. The sum of the roots is: 4 + (r + s) + (r – s ) = 0 4 + 2r = 0 r = -2 The product of the roots is: 4(r + s)(r – s ) = 60Divide by 4 and * r 2 – s 2 = 15Let r = -2 4 – s 2 = 15Solve for s s 2 = -11 s = [√(11)]i The roots are { 4, (-2- [√(11)]i), (-2+ [√(11)]i)}

46. Thank You Thank you for attending. If you have questions about this presentation, please contact me. Taylor Johnson Elizabethtown Community & Technical College taylor.johnson@kctcs.edu (270) 706 - 8564