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UNIT 2, LESSON 5 DESCARTES LAW OF SIGNS. FINDING POLYNOMIALS WITH GIVEN ZEROS If we are given the zeros of a polynomial, we can generate the polynomial.

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Presentation on theme: "UNIT 2, LESSON 5 DESCARTES LAW OF SIGNS. FINDING POLYNOMIALS WITH GIVEN ZEROS If we are given the zeros of a polynomial, we can generate the polynomial."— Presentation transcript:

1 UNIT 2, LESSON 5 DESCARTES LAW OF SIGNS

2 FINDING POLYNOMIALS WITH GIVEN ZEROS If we are given the zeros of a polynomial, we can generate the polynomial by first creating the factors of the polynomial.

3 FINDING POLYNOMIALS WITH GIVEN ZEROS If we are given the zeros of a polynomial, we can generate the polynomial by first creating the factors of the polynomial. For example, if we have the zeros 1 and 3, we can begin by using those zeros to write the factors of our polynomial:

4 FINDING POLYNOMIALS WITH GIVEN ZEROS If we are given the zeros of a polynomial, we can generate the polynomial by first creating the factors of the polynomial. For example, if we have the zeros 1 and 3, we can begin by using those zeros to write the factors of our polynomial:

5 FINDING POLYNOMIALS WITH GIVEN ZEROS The coefficient is there to remind us that there are infinite polynomials with these zeros. For now, let’s assume the coefficient is 1.

6 COMPLEX ROOTS Find a function having zeros 1, 3i, and -3i.

7 COMPLEX ROOTS Find a function having zeros 1, 3i, and -3i.

8 COMPLEX ROOTS Find a function having zeros 1, 3i, and -3i.

9 COMPLEX ROOTS Find a function having zeros 1, 3i, and -3i.

10 COMPLEX ROOTS Find a function having zeros 1, 3i, and -3i.

11 MULTIPLICITY Find a polynomial with -1 as a zero of multiplicity 3, 4 as a zero of multiplicity 1, and 0 as a zero of multiplicity 1.

12 MULTIPLICITY Find a polynomial with -1 as a zero of multiplicity 3, 4 as a zero of multiplicity 1, and 0 as a zero of multiplicity 1. Set up the factors:

13 MULTIPLICITY Find a polynomial with -1 as a zero of multiplicity 3, 4 as a zero of multiplicity 1, and 0 as a zero of multiplicity 1. Set up the factors:

14 MULTIPLICITY Find a polynomial with -1 as a zero of multiplicity 3, 4 as a zero of multiplicity 1, and 0 as a zero of multiplicity 1. Set up the factors:

15 MORE ABOUT COMPLEX ROOTS If a complex number is a zero of a polynomial, then its conjugate is also a zero of the polynomial.

16 MORE ABOUT COMPLEX ROOTS If a complex number is a zero of a polynomial, then its conjugate is also a zero of the polynomial. This is also true of rational zeros.

17 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

18 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

19 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

20 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

21 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

22 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

23 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

24 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

25 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

26 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

27 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

28 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

29 MORE ABOUT COMPLEX ROOTS Find a polynomial function with zeros and.

30 RATIONAL ZEROS THEOREM Let where all coefficients are integers. If p/q is a zero of P(x), then p is a factor of a 0 and q is a factor of a n.

31 RATIONAL ZEROS THEOREM Factor into linear factors.

32 RATIONAL ZEROS THEOREM Factor into linear factors.

33 RATIONAL ZEROS THEOREM Factor into linear factors.

34 DESCARTES' RULE OF SIGNS The number of positive roots of a polynomial with real coefficients is equal to the number of "changes of sign" in the list of coefficients, or is less than this number by a multiple of 2. http://www.cut-the-knot.org/fta/ROS2.shtml

35 HOW MANY OF THE ROOTS ARE POSITIVE? http://www.mathsisfun.com/algebra/polynomials-rule-signs.html First, rewrite the polynomial from highest to lowest exponent (ignore any "zero" terms, so it does not matter that x 4 and x 3 are missing): -3x 5 + x 2 + 4x – 2 Then, count how many times there is a change of sign (from plus to minus, or minus to plus):

36 HOW MANY OF THE ROOTS ARE POSITIVE? http://www.mathsisfun.com/algebra/polynomials-rule-signs.html The number of sign changes is the maximum number of positive roots There are 2 changes in sign, so there are at most 2 positive roots (maybe less).

37 HOW MANY OF THE ROOTS ARE NEGATIVE? http://www.mathsisfun.com/algebra/polynomials-rule-signs.html By doing a similar calculation we can find out how many roots are negative...... but first we need to put "-x" in place of "x", like this:

38 HOW MANY OF THE ROOTS ARE NEGATIVE? http://www.mathsisfun.com/algebra/polynomials-rule-signs.html One change only, so there is 1 negative root.

39 HOW MANY ROOTS IN TOTAL? http://www.mathsisfun.com/algebra/polynomials-rule-signs.html The Fundamental Theorem of Algebra states that a polynomial will have exactly as many roots as its degree. OK, we have gathered lots of info. We know all this: positive roots: 2, or 0 negative roots: 1 total number of roots: 5 So, after a little thought, the overall result is: 5 roots: 2 positive, 1 negative, 2 complex (one pair), or 5 roots: 0 positive, 1 negative, 4 complex (two pairs)

40 Consider the polynomial 3x 6 - 5x 4 + 2x 3 - 7x 2 - x + 2. http://www.mathopolis.com/questions/q.php?id=1131

41 Consider the polynomial 3x 6 - 5x 4 + 2x 3 - 7x 2 - x + 2. 6 total roots http://www.mathopolis.com/questions/q.php?id=1131

42 Consider the polynomial 3x 6 - 5x 4 + 2x 3 - 7x 2 - x + 2. 6 total roots 4 sign changes, so 4, 2, or 0 positive roots http://www.mathopolis.com/questions/q.php?id=1131

43 Consider the polynomial 3x 6 - 5x 4 + 2x 3 - 7x 2 - x + 2. 6 total roots 4 sign changes, so 4, 2, or 0 positive roots Now replace x by -x: ⇒ 3(-x) 6 - 5(-x) 4 + 2(-x) 3 - 7(-x) 2 - (-x) + 2 = 3x 6 - 5x 4 - 2x 3 - 7x 2 + x + 2 http://www.mathopolis.com/questions/q.php?id=1131

44 Consider the polynomial 3x 6 - 5x 4 + 2x 3 - 7x 2 - x + 2. 6 total roots 4 sign changes, so 4, 2, or 0 positive roots Now replace x by -x: ⇒ 3(-x) 6 - 5(-x) 4 + 2(-x) 3 - 7(-x) 2 - (-x) + 2 = 3x 6 - 5x 4 - 2x 3 - 7x 2 + x + 2 2 sign changes, so 2 or 0 negative roots http://www.mathopolis.com/questions/q.php?id=1131

45 Consider the polynomial 2x + 5x 4 - 3x 2 - 7x 6 http://www.mathopolis.com/questions/q.php?id=1131

46 Consider the polynomial 2x + 5x 4 - 3x 2 - 7x 6 First write the terms in the correct order: 2x + 5x 4 - 3x 2 - 7x 6 = -7x 6 + 5x 4 - 3x 2 + 2x http://www.mathopolis.com/questions/q.php?id=1131

47 Consider the polynomial 2x + 5x 4 - 3x 2 - 7x 6 First write the terms in the correct order: 2x + 5x 4 - 3x 2 - 7x 6 = -7x 6 + 5x 4 - 3x 2 + 2x Next factor out x : x(-7x 5 + 5x 3 - 3x + 2) This means x = 0 is one of the roots. http://www.mathopolis.com/questions/q.php?id=1131

48 Consider the polynomial 2x + 5x 4 - 3x 2 - 7x 6 First write the terms in the correct order: 2x + 5x 4 - 3x 2 - 7x 6 = -7x 6 + 5x 4 - 3x 2 + 2x Next factor out x : x(-7x 5 + 5x 3 - 3x + 2) This means x = 0 is one of the roots. There are 3 changes of sign. So the number of positive roots is either 3 or 1. http://www.mathopolis.com/questions/q.php?id=1131

49 Consider the polynomial 2x + 5x 4 - 3x 2 - 7x 6 Now replace x by -x: ⇒ -7(-x) 5 + 5(-x) 3 - 3(-x) + 2 = 7x 5 - 5x 3 + 3x + 2 This has 2 changes of sign. Therefore the number of negative roots is either 2 or 0. http://www.mathopolis.com/questions/q.php?id=1131

50 THINK! How many complex roots does the polynomial 2x 5 + 8x 3 + 3x - 7 have? http://www.mathopolis.com/questions/q.php?id=1131

51 THINK! How many complex roots does the polynomial 2x 5 + 8x 3 + 3x - 7 have? 1 change of sign, so 1 positive root. http://www.mathopolis.com/questions/q.php?id=1131

52 THINK! How many complex roots does the polynomial 2x 5 + 8x 3 + 3x - 7 have? 1 change of sign, so 1 positive root. -2x 5 - 8x 3 - 3x - 7 0 changes of sign, so 0 negative roots. http://www.mathopolis.com/questions/q.php?id=1131

53 THINK! How many complex roots does the polynomial 2x 5 + 8x 3 + 3x - 7 have? There are 5 roots altogether, only 1 of which is real. So, there are 4 complex roots. http://www.mathopolis.com/questions/q.php?id=1131

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