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# Zeros of p(x)

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# Zeros of p(x)

Finding the zeros of Polynomial Function - Math 4 Topics

Finding the zeros of Polynomial Function - Math 4 Topics

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### Zeros of p(x)

1. 1. Finding the Zeros of a Polynomial Function Reynaldo B. Pantino, T2
2. 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. 3. Recapitulations What is remainder theorem? What is synthetic division? What is factoring? What is zero of a function?
4. 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. 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. 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. 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. 8. Discussions UNLOCKING OF DIFFICULTIES The degree of a polynomial function corresponds to the number of zeros of the polynomial.
9. 9. Discussions UNLOCKING OF DIFFICULTIES A depressed equation of P is an equation which has a degree less that of P.
10. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 21. Assignments On page 103, answers numbers 6, 12, 18,19, & 20. Ref. Advanced Algebra, Trigonometry & Statistics What is rational Zero Theorm? Pp. 105