Roots & Zeros of Polynomials

1. Roots & Zeros of Polynomials How the roots, solutions, zeros, x-intercepts and factors of a polynomial function are related.

2. Factors, Roots, Zeros For our Polynomial Function: The Factors are: (x + 5) & (x - 3) The Roots/Solutions are: x = -5 and 3 The Zeros are at: (-5, 0) and (3, 0)

3. Rearrange the terms to have zero on one side: • Factor: • Set each factor equal to zero and solve: Solving a Polynomial Equation The only way that x2 +2x - 15 can = 0 is if x = -5 or x = 3

4. x-Intercepts of a Polynomial The points where y = 0 are called the x-intercepts of the graph. The x-intercepts for our graph are the points... and (3, 0) (-5, 0)

5. Real/Imaginary Roots If a polynomial has ‘n’ complex roots will its graph have ‘n’ x-intercepts? In this example, the degree n = 3, and if we factor the polynomial, the roots are x = -2, 0, 2. We can also see from the graph that there are 3x-intercepts.

6. Real/Imaginary Roots Just because a polynomial has ‘n’ complex roots doesn’t mean that they are all Real! In this example, however, the degree is still n = 3, but there is only oneRealx-intercept or root at x = -1, the other 2 roots must have imaginary components.

7. EXAMPLE 1 Find the zeros of 2, -3 (d.r), 1, -4

8. Solve by factoring. Original equation Add 4x to each side. Factor the binomial. or Zero Product Property Solve the second equation. Example 3-1a Answer:The solution set is {0, –4}.

9. Example 3-1a Check Substitute 0 and –4 in for x in the original equation.

10. Solve by factoring. Original equation Subtract 5x and 2 from each side. Factor the trinomial. or Zero Product Property Solve each equation. Answer:The solution set is Check each solution. Example 3-1a

11. Solve each equation by factoring. a. b. Answer: Example 3-1b Answer:{0, 3}

12. Solve by factoring. Original equation Add 9 to each side. Factor. Zero Product Property or Solve each equation. Example 3-2a Answer:The solution set is {3}.

13. CheckThe graph ofthe related function, intersects the x-axis only once. Since the zero of the function is 3, the solution of the related equation is 3. Example 3-2a

14. Solve by factoring. Example 3-2b Answer:{–5}

15. Factor Theorem 1. If f(c)=0, that is c is a zero of f, then x - c is a factor of f(x). 2. Conversely if x - c is a factor of f(x), then f(c)=0.

16. Example 2: Find all the zeros of each polynomial function First, graph the equation to find the first zero ZERO From looking at the graph you can see that there is a zero at -2

17. Example 2 Continued Second, use the zero you found from the graph and do synthetic division to find a smaller polynomial -2 10 9 -19 6 Don’t forget your remainder should be zero -20 22 -6 10 -11 3 0 The new, smaller polynomial is:

18. Example 2 Continued: Third, factor or use the quadratic formula to find the remaining zeros. This quadratic can be factored into: (5x – 3)(2x – 1) Therefore, the zeros to the problem are:

19. EXAMPLE 3 Find All the Rational Zeros Find the rational zeros of We have already known that the possible rational zeros are:

20. EXAMPLE 3 Thus, -3 is a zero of f and x + 3 is a factor of f. Thus, -2 is a zero of f and x + 2 is a factor of f.

21. EXAMPLE 3 Thus f(x) factors as:

22. Example 4: The zeros of a third-degree polynomial are 2, 2, and -5. Write a polynomial. First, write the zeros in factored form (x – 2)(x – 2)(x – (-5)) = (x – 2)(x – 2)(x+5) Second, multiply the factors out to find your polynomial

23. Example 4 Continued (x – 2)(x – 2)(x+5) First FOIL or box two of the factors Second, box your answer from above with your remaining factors to get your polynomial: X 5 ANSWER

24. So if asked to find a polynomial that has zeros, 2 and 1 – 3i, you would know another root would be 1 + 3i. Let’s find such a polynomial by putting the roots in factor form and multiplying them together. EXAMPLE 5 If x = the root then x - the root is the factor form. Multiply the last two factors together. All i terms should disappear when simplified. -1 Now multiply the x – 2 through Here is a 3rd degree polynomial with roots 2, 1 - 3i and 1 + 3i

25. Conjugate Pairs • Complex Zeros Occur in Conjugate Pairs = If a + bi is a zero of the function, the conjugate a – bi is also a zero of the function (the polynomial function must have real coefficients) • EXAMPLES: Find a polynomial with the given zeros • -1, -1, 3i, -3i • 2, 4 + i, 4 – i

26. Now write a polynomial function of least degree that has real coefficients, a leading coeff. of 1 and 1, -2+i, -2-i as zeros. • f(x)= (x-1)(x-(-2+i))(x-(-2-i)) • f(x)= (x-1)(x+2 - i)(x+2+ i) • f(x)= (x-1)[(x+2) - i] [(x+2)+i] • f(x)= (x-1)[(x+2)2 - i2] Foil • f(x)=(x-1)(x2 + 4x + 4 – (-1)) Take care of i2 • f(x)= (x-1)(x2 + 4x + 4 + 1) • f(x)= (x-1)(x2 + 4x + 5) Multiply • f(x)= x3 + 4x2 + 5x – x2 – 4x – 5 • f(x)= x3 + 3x2 + x - 5

27. Now write a polynomial function of least degree that has real coefficients, a leading coeff. of 1 and 4, 4, 2+i as zeros. • Note: 2+i means 2 – i is also a zero • F(x)= (x-4)(x-4)(x-(2+i))(x-(2-i)) • F(x)= (x-4)(x-4)(x-2-i)(x-2+i) • F(x)= (x2 – 8x +16)[(x-2) – i][(x-2)+i] • F(x)= (x2 – 8x +16)[(x-2)2 – i2] • F(x)= (x2 – 8x +16)(x2 – 4x + 4 – (– 1)) • F(x)= (x2 – 8x +16)(x2 – 4x + 5) • F(x)= x4– 4x3+5x2 – 8x3+32x2 – 40x+16x2 – 64x+80 • F(x)= x4-12x3+53x2-104x+80

28. Ex: Find all the roots of If one root is 4 - i. Find Roots/Zeros of a Polynomial If the known root is imaginary, we can use the Complex Conjugates Thm. Because of the Complex Conjugate Thm., we know that another root must be 4 + i.

29. Ex: Find all the roots of If one root is 4 - i. Example (con’t) If one root is 4 - i, then one factor is [x - (4 - i)], and Another root is 4 + i, & another factor is [x - (4 + i)]. Multiply these factors:

30. If the product of the two non-real factors is Ex: Find all the roots of then the third factor (that gives us the real root) is the quotient of P(x) divided by If one root is 4 - i. Example (con’t) The third root is x = -3 So, all of the zeros are: 4 – i, 4 + i, and -3

31. Find Roots/Zeros of a Polynomial We can find the Roots or Zeros of a polynomial by setting the polynomial equal to 0 and factoring. Some are easier to factor than others! The roots are: 0, -2, 2

32. (x - 5) is a factor Find Roots/Zeros of a Polynomial If we cannot factor the polynomial, but know one of the roots, we can divide that factor into the polynomial. The resulting polynomial has a lower degree and might be easier to factor or solve with the quadratic formula. We can solve the resulting polynomial to get the other 2 roots:

33. (x – 2)(x – 2)(x2+4x+ 6) = 0 This was obtainedfrom the second synthetic division. EXAMPLE: Solving a PolynomialEquation Solve: x4-6x2- 8x + 24 = 0. SolutionNow we can solve the original equation as follows. x4-6x2+ 8x + 24 = 0This is the given equation. x – 2 = 0 orx – 2 = 0 or x2+4x+ 6 = 0Set each factor equal to zero. x= 2 x= 2 x2+4x+ 6 = 0Solve.

34. EXAMPLE: Solving a Polynomial Equation We use the quadratic formula because x2+4x+ 6 = 0 cannot be factored. Let a= 1, b= 4, and c= 6. Multiply and subtract under the radical. Simplify. The solution set of the original equation is {2, -2– i, -2 + i}. Solve: x4-6x2- 8x + 24 = 0. SolutionWe can use the quadratic formula to solve x2+4x+ 6 = 0.

35. FIND ALL THE ZEROS (Given that 1 + 3i is a zero of f) (Given that 5 + 2i is a zero of f)

36. More Finding of Zeros

37. SolutionThe constant term is –2 and the leading coefficient is 15. Divide 1 and 2 by 1. Divide 1 and 2 by 3. Divide 1 and 2 by 5. Divide 1 and 2 by 15. There are 16 possible rational zeros. The actual solution set to f(x)=15x3+ 14x2 - 3x – 2 = 0 is {-1, -1/3, 2/5}, which contains 3 of the 16 possible solutions. FIND ALL RATIONAL ROOTS: List all possible rational zeros of f(x)=15x3+ 14x2 - 3x – 2. EXAMPLE: Using the Rational Zero Theorem

38. EXAMPLE: Solving a Polynomial Equation FIND ALL POSSIBLE RATIONAL ROOTS Solve: x4-6x2- 8x + 24 = 0. SolutionRecall that we refer to the zeros of a polynomial function and the roots of a polynomial equation. Because we are given an equation, we will use the word "roots," rather than "zeros," in the solution process. We begin by listing all possible rational roots.

39. Descartes’ Rule of Signs Arrange the terms of the polynomial P(x) in descending degree: • The number of times the coefficients of the terms of P(x) change sign = the number of Positive Real Roots (or less by any even number) • The number of times the coefficients of the terms of P(-x) change sign = the number of Negative Real Roots (or less by any even number) In the examples that follow, use Descartes’ Rule of Signs to predict the number of + and - Real Roots!

40. Solution 1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f(x).Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f(-x). We obtain this equation by replacing x with -x in the given function. f(x)= x3+ 2x2 + 5x + 4This is the given polynomial function. Replace xwith -x. f(-x)= (-x)3+2(-x)2+ 5(-x) + 4 =-x3+ 2x2 - 5x + 4 3.4: Zeros of Polynomial Functions Determine the possible number of positive and negative real zeros of f(x)= x3+ 2x2 + 5x + 4. EXAMPLE:Using Descartes’ Rule of Signs

41. 1 2 3 3.4: Zeros of Polynomial Functions Determine the possible number of positive and negative real zeros of f(x)= x3+ 2x2 + 5x + 4. EXAMPLE:Using Descartes’ Rule of Signs Solution Now count the sign changes. f(-x)=-x3+ 2x2 - 5x + 4 There are three variations in sign. The number of negative real zeros of fis either equal to the number of sign changes, 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or 3 - 2 = 1 negative real zero.

42. Find the Roots of a Polynomial For higher degree polynomials, finding the complex roots (real and imaginary) is easier if we know one of the roots. Descartes’ Rule of Signs can help get you started. Complete the table below:

43. List the Possible Rational Roots For the polynomial: All possible values of: All possible Rational Roots of the form p/q:

44. Narrow the List of Possible Roots For the polynomial: Descartes’ Rule: All possible Rational Roots of the form p/q:

45. Find a Root That Works For the polynomial: Substitute each of our possible rational roots into f(x). If a value, a, is a root, then f(a) = 0. (Roots are solutions to an equation set equal to zero!)

46. Descartes’s Rule of Signs • EXAMPLES: describe the possible real zeros

47. 2 6 2 1 -13 Find the zeros of Hint: 2 is a zero -6 4 10 -3 5 2 0 X

48. 4 -20 1 0 -11 Find the zeros of Hint: 4 is a zero 20 4 16 5 4 1 0 X

49. 2 6 1 -2 -3 Find the zeros of Hint: 2 is a zero -6 2 0 -3 0 1 0 X

50. 2 1 5 -24 -2 Find the zeros of Hint: 2 is a zero 2 14 24 12 0 7 1 X