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Review 4.1-4.4

Review 4.1-4.4. Pre-Calculus. 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). Rearrange the terms to have zero on one side: . Factor: .

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Review 4.1-4.4

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  1. Review 4.1-4.4 Pre-Calculus

  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. 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

  7. 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.

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

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

  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.

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

  12. Solve by factoring. Original equation Add 9 to each side. Factor. Zero Product Property or Solve each equation. 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 – Find the roots/zeros for this polynomial Given -2 is a zero, use synthetic division to find the remaining roots. Don’t forget your remainder should be zero -2 10 9 -19 6 -20 22 -6 The new, smaller polynomial is: 10 -11 3 0 This quadratic can be factored into: (5x – 3)(2x – 1) Therefore, the zeros to the problem

  17. Find all the zeros of each polynomial function Example: If we were to graph this equation to check you could see the zero From looking at the graph you can see that there is a zero at -2 ZERO ZERO ZERO

  18. Find All the Rational Zeros EXAMPLE 3 We have already known that the possible rational zeros are: So 1, 2, -2, and -3 are rational roots

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

  20. Given 1 ± 3i and 2 are roots, find the remaining roots. 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

  21. 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

  22. 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

  23. 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

  24. (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:

  25. (x – 2)(x – 2)(x2+4x+ 6) = 0 This was obtainedfrom the second synthetic division. The solution set of the original equation is {2, -2– i, -2 + i}. 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. Ise Quadratic Formula

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

  27. More Finding of Zeros

  28. 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

  29. 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.

  30. 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!

  31. 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 Determine the possible number of positive and negative real zeros of f(x)= x3+ 2x2 + 5x + 4. EXAMPLE:Using Descartes’ Rule of Signs

  32. 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 0 positive roots, and 3 negative roots

  33. 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:

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

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

  36. 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!)

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

  38. 2 6 2 1 -13 Find the zeros of Hint: 2 is a zero -6 4 10 -3 5 2 0 X

  39. 4 -20 1 0 -11 Find the zeros of Hint: 4 is a zero 20 4 16 5 4 1 0 X

  40. 2 6 1 -2 -3 Find the zeros of Hint: 2 is a zero -6 2 0 -3 0 1 0 X

  41. 2 1 5 -24 -2 Find the zeros of Hint: 2 is a zero 2 14 24 12 0 7 1 X

  42. -2 6 1 -2 -5 No Calculator Given –2 is a zero of find ALL the zeros of the function. -6 -2 8 3 -4 1 0

  43. 5 20 1 -5 -4 No constant, so 0 is a zero: No Calculator Given 5 is a zero of find ALL the zeros of the function. -20 5 0 -4 0 1 0

  44. 3 -1 1 -9 -3 23 -36 No Calculator Given -1 and 3 are zeros of find ALL the zeros of the function. 36 10 -33 -1 0 1 -10 33 -36 36 3 -21 12 -7 1 0

  45. /2 -6 2 -9 13 3 No Calculator Given is a zero of find ALL the zeros of the function. 3 -9 6 0 4 -6 2 2 -3 1

  46. /3 6 3 -8 -5 2 No Calculator Given is a zero of find ALL the zeros of the function. 2 -4 -6 -9 0 -6 3 -2 -3 1

  47. 2 -10 1 -6 13 No Calculator Given 2 is a zero of find ALL the zeros of the function. 10 2 -8 -4 5 1 0

  48. -3 3 1 3 1 No Calculator Given –3 is a zero of find ALL the zeros of the function. -3 -3 0 1 0 1 0

  49. No Calculator Find a polynomial function with real coefficients which has zeros of 1, -2, and 3.

  50. No Calculator Find a polynomial function with real coefficients which has zeros of 0, 2, -2, and 5.

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