Polynomial Functions

1. Polynomial Functions

2. EVALUATING POLYNOMIAL FUNCTIONS an n n n– 1 a0 an 0 leading coefficient an constant term degree a0 n descending order of exponents from left to right. A polynomial function is a function of the form f(x) = an xn+ an– 1xn– 1+· · ·+ a1x + a0 Where an 0 and the exponents are all whole numbers. For this polynomial function, an is the leading coefficient, a0 is the constant term, and nis the degree. A polynomial function is in standard form if its terms are written in descending order of exponents from left to right.

3. Using the fundamental theorem of Algebra!!! Every polynomial equation with degree > 0 has at least one root in the set of complex numbers

4. How do we know if it is a root? A root is a solution to our equation. In other words, when you plug in a root for x, you answer should be zero.

5. Identifying Roots Decide whether 5 and 6 are roots of x3 – 5x2 – 3x – 18 = 0 SOLUTION f(5) = (5)3 – 5(5)2 – 3(5) – 18 = 125 – 125 – 15 – 18 = -33 NO it is not a root f(6) = (6)3 – 5(6)2 – 3(6) – 18 = 216 – 180 – 18 – 18 = 0 YES it is a root

6. Example If x = 3, x = 4 and x = -4 are roots, then the factors are (x – 3), (x – 4) and (x + 4). Write a polynomial equation of least degree with roots 3, 4, and -4. (x – 3)(x – 4)(x + 4) = 0 (x – 3)(x2 – 16) = 0 x3 - 3x2 – 16x + 48 = 0

7. Example If x = 2, x = 3i and x = -3i are roots, then the factors are (x – 2), (x – 3i) and (x + 3i). Write a polynomial equation of least degree with roots 2, 3i, and -3i. (x – 2)(x – 3i)(x + 3i) = 0 (x – 2)(x2 – 9i2) = 0 (x – 2)(x2 + 9) = 0 x3 – 2x2 + 9x – 18 = 0

8. Example There are 3 complex roots State the number of complex roots of the equation 32x3 – 32x2 + 4x – 4 = 0. Then find the roots. Factor to find the roots (use grouping) 32x3 – 32x2 / + 4x – 4 = 0 split equation in half 32x2 (x – 1) / 4 (x – 1) = 0 factor GCF (32x2 + 4)(x – 1) = 0 take GCF’s and common factors as the 2 factors 32x2 + 4 = 0 x – 1 = 0 SOLVE

9. Warm-Up(factoring) 1.) 6x2 + 15x = 2.) x2 + 10x + 25 = 3.) 4x2 – 9 = 4.) 2x2 – 5x – 12 =

10. The Remainder and Factor Theorems

11. When you divide a Polynomial f(x) by a divisor d(x), you get a quotient polynomial q(x) with a remainder r(x) written:f(x) = q(x) + r(x)d(x) d(x)

12. The degree of the remainder must be less than the degree of the divisor!

13. Polynomial Long Division: • You write the division problem in the same format you would use for numbers. If a term is missing in standard form …fill it in with a 0 coefficient. • Example: • 2x4 + 3x3 + 5x – 1 = • x2 – 2x + 2

14. 2x2 2x4 = 2x2 x2

15. 2x2 +7x +10 -( ) 2x4 -4x3 +4x2 7x3 - 4x2 +5x -( ) 7x3 - 14x2 +14x 10x2 - 9x -1 7x3 = 7x x2 -( ) 10x2 - 20x +20 11x - 21 remainder

16. The answer is written: • 2x2 + 7x + 10 + 11x – 21 x2 – 2x + 2 • Quotient + Remainder over divisor

17. Ex. 4 CHANGE SIGNS

18. (125y3 – 8)  (5y – 2) Ex. 5

19. (x4 – 9x2 – 5)  (x – 2) Ex. 6 You try…

20. (x3 + 9x2 – 5)  (x2 – 1) Ex. 7 You try…

21. Now you try one! • y4 + 2y2 – y + 5 = y2 – y + 1 • Answer: y2 + y + 2 + 3 y2 – y + 1

22. Remainder Theorem: • If a polynomial f(x) is divisible by (x – k), then the remainder is r = f(k). • Now you will use synthetic division (like synthetic substitution) • You can only use synthetic division when the divisor is to the 1st power.

23. f(x)= 3x3 – 2x2 + 2x – 5 Divide by x - 2 • Long division results in ?...... • 3x2 + 4x + 10 + 15 x – 2 • Synthetic Division: • f(2) = 3 -2 2 -5 2 6 8 20 3 4 10 15 Which gives you: + 15 x-2 3x2 + 10 + 4x

24. Synthetic Division Practice 1 • Divide x3 + 2x2 – 6x -9 by (a) x-2 (b) x+3 • (a) x-2 • 1 2 -6 -9 2 8 4 2 4 2 -5 1 Which is x2 + 4x + 2 + -5 x-2

25. Synthetic Division Practice cont. • (b) x+3 • 1 2 -6 -9 -3 3 9 -3 1 -1 -3 0 x2 – x - 3

26. Factor Theorem: • A polynomial p(x) has factor x – r iff p(r)=0 • note that r is a ZERO of the function because p(r)=0

27. Factoring a polynomial • Factor f(x) = 2x3 + 11x2 + 18x + 9 • Given f(-3)=0 • Since f(-3)=0 • x-(-3) or x+3 is a factor • So use synthetic division to find the others!!

28. Factoring a polynomial cont. • 2 11 18 9 • -3 -9 -15 -6 3 0 2 5 So…. 2x3 + 11x2 + 18x + 9 factors to: (x + 3)(2x2 + 5x + 3) Now keep factoring; it gives you: (x+3)(2x+3)(x+1)

29. Your turn! • Factor f(x)= 3x3 + 13x2 + 2x -8 • given f(-4)=0 • (x + 1)(3x – 2)(x + 4)

30. Determine the binomial factors • Use synthetic division to determine the roots of f(x) = 2x3 + 11x2 + 18x + 9 • Hint: Use graphing calculator to view TABLE and find where the y-values are zeros. • There are zeros when x = -3 and x = -1 • Use synthetic division to factor out -3 and -1 and you will be left with the remaining root.

31. Factoring a polynomial cont. • 2 11 18 9 • -3 -15 -9 -6 2 5 3 0

32. Factoring a polynomial cont. • 2 5 3 • -1 -3 -2 2 3 0 Now we have (2x + 3) = 0 so our last root is x = -3/2 Our factors are (x+1), (x+3)and (2x + 3)

33. Your turn! • Factor f(x)= 3x3 + 13x2 + 2x -8 • (x + 1)(3x – 2)(x + 4)

34. Finding the zeros of a polynomial function • f(x) = x3 – 2x2 – 9x +18. • One zero of f(x) is x=2 • Find the others! • Use synthetic div. to reduce the degree of the polynomial function and factor completely. • (x – 2)(x2 – 9) = (x-2)(x+3)(x-3) • Therefore, the zeros are x = 2,3,-3!!!

35. Your turn! • f(x) = x3 + 6x2 + 3x -10 • x = -5 is one zero, find the others! • The zeros are x = 2,-1,-5 • Because the factors are (x-2)(x+1)(x+5)

36. Assignment:page 356#16-20 even(LONG DIVISION)#28-32 even(SYNTHETIC DIVISION)#40-52 x4