Ani agustina (a1 c011007) polynomial

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Ani agustina (a1 c011007) polynomial

  1. 1. Polynomial by Ani Agustina Click here Welcome to Polynomial Start Exit
  2. 2. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES First Slide QUIZ
  3. 3. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES First Slide QUIZ
  4. 4. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES First Slide QUIZ
  5. 5. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES First Slide QUIZ
  6. 6. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES General Aspects of Polynomial Value of a polynomial Polynomial Dividing Remainder Theorem Factorization Theorem Roots of a Polynomial Equation First Slide QUIZ
  7. 7. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES EXERCISES 1 EXERCISES 2 EXERCISES 3 EXERCISES 4 EXERCISES 5 EXERCISES 6 First Slide QUIZ
  8. 8. Polynomial by Ani Agustina POLYNOMIAL MATTERS EXERCISES QUIZ QUIZ 1 QUIZ 2 First Slide
  9. 9. Polynomial by Ani Agustina General Aspects of Polynomial 1. Definition and Polynomial Component Definition and Polynomial Component Algebraic Operations of Polynomial Polynomial equation Algebraic forms, such as a linier equation and a quadratic equation are kinds of polynomial equation of degree 1 and 2. Consider the following algebraic forms: a. x3 + 4x2 – 7x + 10 b. 5x4 – 2x3 + 3x2 – x – 5 c. x5 + 6x4 – 3x3 + 2x2 + 10x – 3 the algebraic forms above are called polynomial in variabel x that accomadate in the polynomial. They consecutively have degree of 3, 4, and 5. 1 2
  10. 10. Polynomial by Ani Agustina General Aspects of Polynomial Definition and Polynomial Component Algebraic Operations of Polynomial Polynomial equation Definition The general form of polynimial in variabel x of degree n is: P(x) = anxn + an-1 xn-1 + an-2 xn-2 + . . . + a1x + a0 The polynomial are arranged based on descending order of the exponent of x, where: an , an-1 , . . . , a1 = real coefficients of the polynomial and an ≠ 0, a0 = real constant, n = degree of the polynomial, n is whole number. 1 2
  11. 11. Polynomial by Ani Agustina General Aspects of Polynomial Definition and Polynomial Component 2. Algebraic Operations of Polynomial Polynomial equation Algebraic Operations of Polynomial Algebraic operations of polynomial such as addition, subtraction, multiplication, and divison have similar characteristic and rulers as the algerbraic operations on real numbers. 1 2
  12. 12. Polynomial by Ani Agustina General Aspects of Polynomial Definition and Polynomial Component Algebraic Operations of Polynomial Polynomial equation Example : Given f(x) = x3 + 2x + 3 and g(x) = x5 + 3x4 – 7x2 – 3x + 1 Determine: a. f(x) + g(x) b. f(x) – g(x) Answer: a. f(x) + g(x) = (x3 + 2x + 3) + (x5 +3x4 – 7x2 – 3x + 1) = x5 + 3x4 + x3 – 7x2 – x + 4 b. f(x) – g(x) = (x3 + 2x + 3) - (x5 +3x4 – 7x2 – 3x + 1) = -x5 - 3x4 + x3 + 7x2 + 5x + 2 1 2
  13. 13. Polynomial by Ani Agustina General Aspects of Polynomial Definition and Polynomial Component Algebraic Operations of Polynomial Polynomial equation 3. Polynomial equation Definition Suppose given polynomial of f(x) and g(x), where f(x) = anxn + an-1 xn-1 + an-2 xn-2 + . . . + a1x + a0 g(x) = bnxn + bn-1 xn-1 + bn-2 xn-2 + . . . + b1x + b0 f(x) is equal to g(x) (written f(x) ≡ g(x)) if satisfies: an = bn , an-1 = bn-1 , an-2 = bn-2 , . . . , a2 = b2 , a1 = b1 , a0 = b0 1 2
  14. 14. Polynomial by Ani Agustina General Aspects of Polynomial Definition and Polynomial Component Algebraic Operations of Polynomial Polynomial equation Find the value of a and b of equation x3 + 4x2 – 7x + a = (x – 2) (x + 1) (x + b). Answer : X3 + 4x2 – 7x + a = (x – 2) (x + 1) (x + b) = (x2 – x – 2) (x + b) = x3 + (b – 1) x2 – (b + 2)x – 2b b – 1 = 4 → coefficient of x2 b + 2 = 7 → coefficient of x2 Based on the equationa above b = 5 a = -2b → a constant ↔ a = -2(5) = -10 Thus, the value of a = -10 and b = 5 1 2
  15. 15. Polynomial by Ani Agustina Value of a polynomial Determining the value of polynomial by substitution method Determining the value of a polynomial by Scheme (chart) 1. Determining the value of polynomial by substitution method Using this method, the value of the polynomial of P(x) for x = k, written P(k), is determined by substituting the value of k into the variable of x on the polynomial. Example Determine the value of polynomial P(x) = x5 – 2x4 + 3x3 + 4x2 – 10x + 3 for x = 1 Answer: P(1) = (1)5 – 2(1)4 + 3(1)3 + 4(1)2 – 10(1) + 3 = 1 – 2 + 3 + 4 – 10 + 3 = -1
  16. 16. Polynomial by Ani Agustina Value of a polynomial Determining the value of polynomial by substitution method Determining the value of a polynomial by Scheme (chart) 2. Determining the value of a polynomial by Scheme (chart) The first step of using scheme method is to write a polynomial from the right to the left starting from the highest exponent of the variable. To solve P(x) = x5 – 2x4 + 3x3 + 4x2 – 10x + 3 for x = 1, you must write each coefficient each term of the polynomial inthe chart. 1 2
  17. 17. Polynomial by Ani Agustina Value of a polynomial Consider the following chart. Determining the value of polynomial by substitution method Determining the value of a polynomial by Scheme (chart) Thus, the value of P(1) = -1 Description Symbol ↗ means multiply by the input number (in this case input number = 1 since we will find the value of the polynomial, P(x), with x = 1). 1 2
  18. 18. Polynomial by Ani Agustina Polynomial Dividing Definition Long Division Synthetic Division (Horner Method) Definition Consider the following expression. P(x) = Q(x) . H(x) + S(x) The expression above is a division of polynomials, where P(x) = the divided, Q(x) = the divisor, H(x) = the quotient, and S(x) = the remainder.
  19. 19. Polynomial by Ani Agustina Polynomial Dividing 1. Definition Long Division Synthetic Division (Horner Method) Long Division The procedure of dividing a polynomial using long division is similar to the procedures of dividing integers. Example Determine the quotient and the remainder of the division of P(x) = x3 + 5x2 – 4x – 20 by x + 3 Answers: P(x) has degree of 3, Q(x) has degree of 1, and the quotient of H(x) has degree of 3 – 1 = 2, the remainder of division S(x) has 1 2 degree of 1 – 1 = 0
  20. 20. Polynomial by Ani Agustina Polynomial Dividing Definition Long Division Synthetic Division (Horner Method) Using this method, we get the quotient of H(x) = x2 + 2x – 10, and the remainder, S(x) = 10. 1 2
  21. 21. Polynomial by Ani Agustina Polynomial Dividing Definition 2. Long Division Horner method is a short way to solve the division of polynomial. This method can solve several forms of division of polynomial. a. Dividing polynomial by (x – k) The principle of horner method is derived from the konsep of long division. P(x) = ax2 + bx + c Synthetic Division (Horner Method) Synthetic division (Horner Method) 1 2 3 4 5 6 7
  22. 22. Polynomial by Ani Agustina Polynomial Dividing Definition Long Division Synthetic Division (Horner Method) the qoutient is H(x) = ax + (b + ak) with S(x) = ak2 + bk + c 1 2 3 4 5 6 7
  23. 23. Polynomial by Ani Agustina Polynomial Dividing Definition Long Division Synthetic Division (Horner Method) b. Dividing a polynomial by (ax – b) The division of a polynomial by (ax – b) is the expansion of the division of polynomial by (x – k). The general form of dividing a polynomial by (ax – b) with the quotient H(x) and the remainder S(x) is: P(x) = (ax – b) . H(x) + S(x) 1 2 3 4 5 6 7
  24. 24. Polynomial by Ani Agustina Polynomial Dividing Definition Long Division Synthetic Division (Horner Method) 1 2 3 4 5 6 7
  25. 25. Polynomial by Ani Agustina Polynomial Dividing c. Definition Long Division Synthetic Division (Horner Method) Dividing polynomial by (ax2 + bx + c) with a ≠ 0 The division of the polynomial of P(x) by ax2 + bx + c using the horner method can be done if the divisor of (ax2 + bx + c) can be factorized. The general form of this division is: P(x) = (ax2 + bx + c) . H(x) + S(x) With H(x) and S(x) are consecutively the quotient and the remaider of the division. 1 2 3 4 5 6 7
  26. 26. Polynomial by Ani Agustina Polynomial Dividing Definition Long Division Synthetic Division (Horner Method) Suppose (ax2 + bx + c) can be factorized to be P1 . P2 therefore ax2 + bx + c = P1 . P2 , then : P(x) = (ax2 + bx + c) . H(x) + S(x) = P1 . P2 H(x) + S(x) 1 2 3 4 5 6 7
  27. 27. Polynomial by Ani Agustina Polynomial Dividing Definition Long Division Synthetic Division (Horner Method) The steps to determine the qoutient and the remainder of the divison of a polynomial are: 1. Divide the polynomial of P(x) by P1 to get H0(x) and S1 as the quotient and the remaider. 2. Divide H0(x) by P2 to get H(x) and S2 as the quotient and the remainder. 3. The quotient of dividing P(x) by (Q(x) = P1 . P2) is H(x), and the remainder is S(x) = P1 . S2 + S1. Consider if P1 or P2 is in the form of (ax – b), a ≠ 0, then you need to divide H0(x) or H(x) by a to get the quotient. Reconsider the division of a polynomial by (ax – b) that you have learned. 1 2 3 4 5 6 7
  28. 28. Polynomial by Ani Agustina Remainder Theorem 1. Division by (x – k) Division by (x – k) Division by (ax – b) Division by (x – a) (x – b) Theorem If polynomial of f(x) with degree of n divided by (x – k), then the remainder is S = f(k) Proof: f(x) = (x – k) H(x) + S for x – k = 0 x = k → f(k) = (k – k) H(k) + S f(k) = 0 + S f(x) = S then, the remainder of S = f(k), it is proved. Note: A polynomial of f(x) is divisible by (x – k) if f(k) = 0
  29. 29. Polynomial by Ani Agustina Remainder Theorem Division by (x – k) Division by (ax – b) Division by (x – a) (x – b) 2. Division by (ax – b) Theorem If polynomial of f(x) with degree of n divided by (ax – b), then the remainder is S = f(b/a).
  30. 30. Polynomial by Ani Agustina Remainder Theorem 3. Division by (x – k) Division by (ax – b) Division by (x – a) (x – b) Division by (x – a) (x – b) Dividing polynomial of f(x) by (x – a) (x – b) with the quotient of H(x) and the remainder of S(x), is written as: f(x) = (x – a) (x – b) H(x) + S(x) since the degree of the divisor is 2, the degree of S is at most 1. Let S = (px + q), then the division can written as follows. f(x) = (x – a) (x – b) H(x) + (px + q)
  31. 31. Polynomial by Ani Agustina Factorization Theorem f(x) = (x – k) H(x) + f(k), where H(x) = is the quotient. I. based on the equation above if f(k)n = 0 then f(x) = (x – k) H(x). In other words (x – k) is a factor of f(x). II. if (x – k) is a factor of f(x) then: f(x) = (x – k) H(x) for any H(x) such that f(x) = (k – k) H(x) = 0 from (I) and (II) we have f(k) = 0 if and only if (x – k) is a factor of f(x). Therefore, the division of a polynomial by one of its factor resulted a remainder which is equal to zero. 1 2 3 4
  32. 32. Polynomial by Ani Agustina Factorization Theorem There are several aspects that must be considered when factorizing a polynomial. 1. If the sum of all coefficients is aqual to zero, then (x – 1) is a factor of the polynomial 2. If the sum of all coeficients of variables with an even exponent is equal to the sum of all coefficients of variabel with an odd exponent, then (x + 1) is a factor of the polynomial. 3. If (1) and (2) are not satisfied, then solve as in example. 1 2 3 4
  33. 33. Polynomial by Ani Agustina Factorization Theorem Example Determine the linier factor of f(x) = x3 – 7x2 + 8x + 16 Answer : consider that the sum of all coefficients of variables with an even exponen is equal to the sum of all coefficients of variables with an odd exponent i.e. (-7 + 16 = 1 + 8). Thus (x + 1) is one of the factors. Next, we use Horner method to determine the other factors. 1 2 3 4
  34. 34. Polynomial by Ani Agustina Factorization Theorem f(x) = (x + 1) (x2 – 8x + 16) = (x + 1) (x – 4) (x – 4) Therefore, the linier factors of f(x) = x3 – 7x2 + 8x + 16 is (x + 1) and (x – 4) 1 2 3 4
  35. 35. Polynomial by Ani Agustina Roots of a Polinomial Equation Definition of Roots of a Polynomial Equation Rational Roots of a Polynomial 1. Definition of Roots of a Polynomial Equation Based on the theorem of factorization we have proved that (x – k) is a factor of f(x) if and only if f(k) = 0. Similarly, we can define the root of a polynomial equation. Theorem (x = k) is the root of a polynomial equation of f(x) = 0 if and only if f(k) = 0
  36. 36. Polynomial by Ani Agustina Roots of a Polinomial Equation Definition of Roots of a Polynomial Equation Rational Roots of a Polynomial 1 2
  37. 37. Polynomial by Ani Agustina Roots of a Polinomial Equation Definition of Roots of a Polynomial Equation Rational Roots of a Polynomial 1 2
  38. 38. Polynomial by Ani Agustina Exercises 1 1. The polynomial of (4x – 3) (2x2 + 5) (x-4) has degree of ... a. 2 b. 3 c. 4 d. 5 e. 6 2. The coefficients of x2 in the polynomial (4x -2) (x + 1) (3x + 1) is ... a. 4 b. 5 c. 6 d. 9 e. 10
  39. 39. Polynomial by Ani Agustina Solution (4x – 3) (2x2 + 5) (x-4) = (8x3 - 6x2 + 20x - 15) (x – 4) = 8x4 - 6x3 + 20x2 - 15x – 32x3 + 24x2 – 80x + 60 = 8x4 – 38x3 + 44x2 – 95x + 60 So, this answer is 4 (c) Back
  40. 40. Polynomial by Ani Agustina Solution (4x -2) (x + 1) (3x + 1) = (4x2 + 2x – 2) (3x + 1) = 12x3 + 4x2 + 6x2 + 2x – 6x – 2 = 12x3 + 10x2 -4x – 2 So, this answer is 10 (e) Back
  41. 41. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  42. 42. Polynomial by Ani Agustina Sorry, your answer is false !!! Back
  43. 43. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  44. 44. Polynomial by Ani Agustina Exercises 2 1. The value of polynomial f(x) = 4x3 + 2x2 – 3x + 5 for x = -2 is ... a. -13 b. -5 c. -10 d. 5 e. 6 2. The value of polynomial g(x) = 9x5 - 5x4 + 7x3 – 8x2 + 6x – 9 for x = -1 is ... a. -44 b. 43 c. 34 d. -34 e. -33
  45. 45. Polynomial by Ani Agustina Solution Way 1 Way 2 f(x) f(-2) So, the value of f(-2) = -13 (a) = 4x3 + 2x2 – 3x + 5 = 4(-2)3 + 2(-2)2 – 3(-2) + 5 = 4(-8) + 2(4) + 6 + 5 = -32 + 8 + 11 = -13 (a) Back
  46. 46. Polynomial by Ani Agustina Solution Way 1 So, the value of f(-1) = -44 (a) g(x) Way 2 = 9x5 - 5x4 + 7x3 – 8x2 + 6x – 9 = 9(-1)5 – 5(-1)4 + 7(-1)3 – 8(-1)2 + 6(-1) – 9 =-9–5–7–8–6–9 = - 44 (a) Back
  47. 47. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  48. 48. Polynomial by Ani Agustina Sorry, your answer is false !!! Back
  49. 49. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  50. 50. Polynomial by Ani Agustina Exercises 3 1. The qoutient and the remainder of division of f(x) = 2x3 + 5x2 – 3x + 6 by (x – 3) successively are ... a. 2x2 – 11x + 30 and 96 c. 2x2 + 11x + 30 and 86 b. 2x2 + 11x - 30 and 96 d. 2x2 + 11x + 30 and 96 e. 2x2 + 11x + 30 and 79 2. The qoutient and the remainder if f(x) = x3 – 5x2 + 2 is divisible by (x2 + 4x – 1) are ... a. x – 9 and 37x – 7 c. x – 9 and 37x + 7 b. x + 9 and 37x - 7 d. x + 9 and 37x + 7 e. x – 9 and 27x - 7
  51. 51. Polynomial by Ani Agustina Solution Using Horner method The quotient is H(x) = 2x2 + 11x + 30 with S(x) = 96, so the answer is d. Back
  52. 52. Polynomial by Ani Agustina Solution f(x) has degree of 3, g(x) has degree of 2, and the quotient of H(x) has degree of 3 – 2 = 1, the remainder of the division S(x) has degree of 2 – 1 = 1. Using this method, we get the quotient of H(x) = x – 9 and the remainder, S(x) = 37x – 7 (a) Back
  53. 53. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  54. 54. Polynomial by Ani Agustina Sorry, your answer is false !!! Back
  55. 55. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  56. 56. Polynomial by Ani Agustina Exercises 4 1. the polynomial of f(x) divided by (x -1) the remainder is 6, if divided by (x + 1) the remainder is -4. Determine the remainder if f(x) divided by (x2 – 1) is ... a. 5x - 1 c. 5x + 1 b. 5x + 2 d. 5x - 2 e. 5x – 3 2. If polynomial of f(x) divided by (x - 1) the remainder is 3, while if it is divided by (x + 2) the remainder is 6, determine the remainder if f(x) is divided by (x – 1) (x + 2) is ... a. - x + 4 c. x – 4 b. x + 5 d. x – 5 e. x + 6
  57. 57. Polynomial by Ani Agustina Solution f(x) : (x – 1) then remainder = 6 → f(1) = 6 f(x) : (x + 1) then remainder = -4 → f(-1) = -4 f(x) = (x – 1) (x + 1) H(x) + (px+q) so, the remainder of S(x) = px + q = 5x + 1 (c) Back
  58. 58. Polynomial by Ani Agustina Solution f(x) : (x – 1) then remainder = 3 → f(1) = 3 f(x) : (x + 2) then remainder = 6 → f(-1) = 6 f(x) = (x – 1) (x + 2) H(x) + (px+q) so, the remainder of S(x) = px + q = -x + 4 (a) Back
  59. 59. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  60. 60. Polynomial by Ani Agustina Sorry, your answer is false !!! Back
  61. 61. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  62. 62. Polynomial by Ani Agustina Exercises 5 1. If (x + 2) is one of the factors of f(x) = -5x5 + ax3 + x2 – 4, then the value of a is ... a. -4 c. 20 b. -2 d. 4 e. 16 2. The factorization of f(x) = 2x3 + 3x2 – 17x + 12 is ... a. (x + 1) (x + 4) (2x – 3) c. (x + 1) (x - 4) (2x – 3) b. (x - 1) (x + 4) (2x – 3) d. (x + 1) (x + 4) (2x + 3) e. (x - 1) (x - 4) (2x + 3)
  63. 63. Polynomial by Ani Agustina Solution using horner method -8a + 160 = 0 -8a = -160 a = 20 so, the value of a is 20 (c) Back
  64. 64. Polynomial by Ani Agustina Solution consider that the sum of all coefficients of variables with an even exponen is equal to the sum of all coefficients of variables with an odd exponent i.e. (2 + 3 – 17 + 12 = 0). Thus (x - 1) is one of the factors. Next, we use Horner method to determine the other factors. f(x) = (x – 1) (2x2 + 5x – 12) f(x) = (x – 1) (x + 4) (2x – 3) (b) Back
  65. 65. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  66. 66. Polynomial by Ani Agustina Sorry, your answer is false !!! Back
  67. 67. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  68. 68. Polynomial by Ani Agustina Exercises 6 1. the integer roots of x3 – 6x2 + 11x – 6 = 0 are ... a. 1, 2, and -3 c. -1, -2, and 3 b. -1, 2, and 3 d. 1, 2, and 3 e. 1, -2, and -3 2. If f(x) = x3 – 8x2 + ax + 10 = 0 has roots x = 2, the sum of the two other roots is ... a. -6 c. 7 b. 6 d. 9 e. 10
  69. 69. Polynomial by Ani Agustina Solution consider that the sum of all coefficients of variables with an even exponen is equal to the sum of all coefficients of variables with an odd exponent i.e. (1 6 + 11 - 6 = 0). Thus (x - 1) is one of the factors. Next, we use Horner method to determine the other factors. f(x) = (x – 1) (x2 - 5x + 6) f(x) = (x – 1) (x -2) (x – 3) thus, the roots of f(x) = 0 is x = 1, x = 2 and x = 3 (d) Back
  70. 70. Polynomial by Ani Agustina Solution Back
  71. 71. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  72. 72. Polynomial by Ani Agustina Sorry, your answer is false !!! Back
  73. 73. Polynomial by Ani Agustina Good, your answer is true !!! Solution
  74. 74. Polynomial by Ani Agustina Quiz 1
  75. 75. Polynomial by Ani Agustina Quiz 2 1. Using the remainder theorem, determine the remainder of the following division. a. f(x) = (3x2 – 7x + 1) : (x – 2) b. f(x) = (2x3 – 4x2 + 3x + 7) : (x + 1) 2. determine the value of a and b in the following statements. a. The polynomial of f(x) = (2x3 + 5x2 + ax + b) if it is divided by (x + 1) the remainder is 1, whereas if it is divided by (x – 2) the remainder is 43 b. The polynomial of f(x) = (ax3 – 7x + b) if it is divided by (x – 1) and (x + 2) successively is 1 and 4. 3. If x2 – 2x – 3 is a factor of f(x) = x4 – 2x3 – 16x2 – px + q, determine the value of p and q and other factors! 4. If 1 is the root of x3 + ax2 – 2x + 2 = 0, determine the value of a and the other roots! 5. Given equation of x4 + 4x3 – x2 – 16x – 12 = 0 has roots of a, b, c, and d. Determine the value of a2 + b2 + c2 + d2.
  76. 76. Polynomial by Ani Agustina Thank you very much !!!!

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