Zeros of p(x)

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Finding the zeros of Polynomial Function - Math 4 Topics

Text of Zeros of p(x)

  • 1. Finding the Zeros of a Polynomial Function Reynaldo B. Pantino, T2
  • 2. Objectives 1.) To determine the zeros of polynomial functions of degree greater than 2 by; a.) factor theorem b.) factoring c.) synthetic division d.) depressed equations 2.)To determine the zeros of polynomial functions of degree n greater than 2 expressed as a product of linear factors.
  • 3. Recapitulations What is remainder theorem? What is synthetic division? What is factoring? What is zero of a function?
  • 4. Discussions UNLOCKING OF DIFFICULTIES The zero of a polynomial function P(x) is the value of the variable x, which makes polynomial function equal to zero or P(x) = 0.
  • 5. Discussions UNLOCKING OF DIFFICULTIES The fundamental Theorem of Algebra states that Every rational polynomial function P(x) = 0 of degree n has exactly n zeros.
  • 6. Discussions UNLOCKING OF DIFFICULTIES When a polynomial is expressed as a product of linear factors, it is easy to find the zeros of the related function considering the principle of zero products.
  • 7. Discussions UNLOCKING OF DIFFICULTIES The principle of zero product state that, for all real numbers a and b, ab = 0 if and only if a = 0 or b = 0, or both.
  • 8. Discussions UNLOCKING OF DIFFICULTIES The degree of a polynomial function corresponds to the number of zeros of the polynomial.
  • 9. Discussions UNLOCKING OF DIFFICULTIES A depressed equation of P is an equation which has a degree less that of P.
  • 10. Discussions Illustrative Example 1 Find the zeros of P(x) = (x 3)(x + 2)(x 1)(x + 1). Solution: (Use the principle of zero products) P(x) = 0; that is x - 3 = 0 x + 2 = 0 x - 1 = 0 x + 1 = 0 x = 3 x = -2 x = 1 x = -1
  • 11. Discussions Illustrative Example 2 Find the zeros of P(x) = (x + 1)(x + 1)(x +1)(x 2) Solution: (By zero product principle) we have, P(x) = 0 the zeros are -1 and 2. The factor (x + 1) occurs 3 times. In this case, the zero -1 has a multiplicity of 3.
  • 12. Discussions Illustrative Example 3 Find the zeros of P(x) = (x + 2)3(x2 9). Solution: (By factoring) we have, P(x) = (x +2)(x+2)(x+2)(x 3)(x + 3). The zeros are; -2, 3, -3, where -2 has a multiplicity of 3.
  • 13. Discussions Illustrative Example 4 Function Zeros No. of Zeros P(x) = x 4 P(x) = x2 + 8x + 15 P(x) = x3 -2x2 4x + 8 P(x) = x4 2x2 + 1 4 1 -3, -5 2 2, -2, 2 3 1,1,-1,-1 4
  • 14. Discussions Illustrative Example 4 Solve for the zeros of P(x) = x3 + 8x2 + 19x + 12, given that one zero is -1. Solution: By factor theorem, x + 1 is a factor of x3 + 8x2 + 19x + 12. Then; P(x) = x3 + 8x2 + 19x + 12 = (x+1) Q(x).
  • 15. Discussions Illustrative Example 4 (Continuation of solution) To determine Q(x), divide x3 + 8x2 + 19x + 12 by (x + 1). By synthetic division; --11 11 88 1199 1122 11 --11 77 --77 1122 --1122 00
  • 16. Discussions Illustrative Example 4 (Continuation of solution) The equation x2 + 7x + 12 is a depressed equation of P(x). To find the remaining zeros use this depressed equation. By factoring we have; x2 + 7x + 12 = 0 (x +3)(x + 4) = 0 x = -3 and x = -4 Observe that a polynomial function of degree 3 has exactly three zeros. Therefore; the three zeros are -1, -3, and -4.
  • 17. Exercises 1. Solve for the other zeros of P(x) = x4 x3 11x2 + 9x + 18, given that one zero is -3. 2. Solve for the other zeros of P(x) = x3 2x2 3x + 10, given that 2 is a zero.
  • 18. Activity Numbers Which of the numbers -3, -2, -1, 0, 1, 2, 3 are zeros of the following polynomials? 1.) f(x) = x3 + x2 + x + 1 2.) g(x) = x3 4x2 + x + 6 3.) h(x) = x3 7x + 6 4.) f(x) = 3x3 + 8x2 2x + 3 5.) g(x) = x3 + 3x2 x 3
  • 19. Activity Factors Which of the binomials (x 1), (x + 1), (x 4), (x + 3) are factors of the given polynomials. 1.) x3 + x2 - 7x + 5 2.) 2x3 + 5x2 + 4x + 1 3.) 3x3 12x2 + 2x 8 4.) 4x4 - x3 + 2x2 + x 3 5.) 4x4 + 5x3 - 14x2 4x + 3
  • 20. Activity Zeros Find the remaining zeros of the polynomial function, real or imaginary, given one of its zeros. 1.) P(x) = x3 + 5x2 - 2x 24 x = 2 2.) P(x) = x3 - x2 - 7x + 3 x = 3 3.) P(x) = x3 8x2 + 20x 16x = 2 4.) P(x) = x3 + 5x2 - 9x 45 x = -5 5.) P(x) = x3 + 3x2 + 3x + 1 x = -1
  • 21. Assignments On page 103, answers numbers 6, 12, 18,19, & 20. Ref. Advanced Algebra, Trigonometry & Statistics What is rational Zero Theorm? Pp. 105