1 / 78

College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson

College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson. Polynomial and Rational Functions. 4. Real Zeros of Polynomials. 4.4. Real Zeros of Polynomials.

addison
Download Presentation

College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. College Algebra Fifth Edition James StewartLothar RedlinSaleem Watson

  2. Polynomial and Rational Functions 4

  3. Real Zeros of Polynomials 4.4

  4. Real Zeros of Polynomials • The Factor Theorem tells us that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. • In this section, we study some algebraic methods that help us find the real zeros of a polynomial, and thereby factor the polynomial.

  5. Rational Zeros of Polynomials

  6. Rational Zeros of Polynomials • To help us understand the upcoming theorem, let’s consider the polynomial • From the factored form, we see that the zeros of P are 2, 3, and –4. • When the polynomial is expanded, the constant 24 is obtained by multiplying (–2) x (–3) x 4.

  7. Rational Zeros of Polynomials • This means that the zeros of the polynomial are all factors of the constant term. • The following generalizes this observation.

  8. Rational Zeros Theorem • If the polynomial has integer coefficients, then every rational zero of P is of the form where: • p is a factor of the constant coefficient a0. • q is a factor of the leading coefficient an.

  9. Rational Zeros Theorem—Proof • If p/q is a rational zero, in lowest terms, of the polynomial P, we have:

  10. Rational Zeros Theorem—Proof • Now, p is a factor of the left side, so it must be a factor of the right as well. • As p/q is in lowest terms, p and q have no factor in common; so, p must be a factor of a0. • A similar proof shows that q is a factor of an.

  11. Rational Zeros Theorem • From the theorem, we see that: • If the leading coefficient is 1 or –1, the rational zeros must be factors of the constant term.

  12. E.g. 1—Using the Rational Zeros Theorem • Find the rational zeros of P(x) = x3 – 3x + 2 • Since the leading coefficient is 1, any rational zero must be a divisor of the constant term 2. • So, the possible rational zeros are ±1 and ±2. • We test each of these possibilities.

  13. E.g. 1—Using the Rational Zeros Theorem • The rational zeros of P are 1 and –2.

  14. E.g. 2—Finding Rational Zeros • Factor the polynomial P(x) = 2x3 + x2 – 13x + 6 • By the Rational Zeros Theorem, the rational zeros of P are of the form • The constant term is 6 and the leading coefficient is 2. • Thus,

  15. E.g. 2—Finding Rational Zeros • The factors of 6 are: ±1, ±2, ±3, ±6 • The factors of 2 are: ±1, ±2 • Thus, the possible rational zeros of P are:

  16. E.g. 2—Finding Rational Zeros • Simplifying the fractions and eliminating duplicates, we get the following list of possible rational zeros:

  17. E.g. 2—Finding Rational Zeros • To check which of these possiblezeros actually arezeros, we need to evaluate P at each of these numbers. • An efficient way to do this is to use synthetic division.

  18. E.g. 2—Finding Rational Zeros • Test if 1 is a zero: • Remainder is not 0. • So, 1 is nota zero.

  19. E.g. 2—Finding Rational Zeros • Test if 2 is a zero: • Remainder is0. • So, 2 isa zero.

  20. E.g. 2—Finding Rational Zeros • From the last synthetic division, we see that 2 is a zero of P and that P factors as:

  21. Finding the Rational Zeros of a Polynomial • These steps explain how we use the Rational Zeros Theorem with synthetic division to factor a polynomial. • List possible zeros. • Divide. • Repeat.

  22. Step 1 to Finding the Rational Zeros of a Polynomial • List possible zeros. • List all possible rational zeros using the Rational Zeros Theorem.

  23. Step 2 to Finding the Rational Zeros of a Polynomial • Divide. • Use synthetic division to evaluate the polynomial at each of the candidates for rational zeros that you found in Step 1. • When the remainder is 0, note the quotient you have obtained.

  24. Step 3 to Finding the Rational Zeros of a Polynomial • Repeat. • Repeat Steps 1 and 2 for the quotient. • Stop when you reach a quotient that is quadratic or factors easily. • Use the quadratic formula or factor to find the remaining zeros.

  25. E.g. 3—Using the Theorem and the Quadratic Formula • Let P(x) = x4 – 5x3 – 5x2 + 23x + 10. • Find the zeros of P. • Sketch the graph of P.

  26. Example (a) E.g. 3—Theorem & Quad. Formula • The leading coefficient of P is 1. • So, all the rational zeros are integers. • They are divisors of the constant term 10. • Thus, the possible candidates are:±1, ±2, ±5, ±10

  27. Example (a) E.g. 3—Theorem & Quad. Formula • Using synthetic division, we find that 1 and 2 are not zeros.

  28. Example (a) E.g. 3—Theorem & Quad. Formula • However, 5 is a zero.

  29. Example (a) E.g. 3—Theorem & Quad. Formula • Also, P factors as:

  30. Example (a) E.g. 3—Theorem & Quad. Formula • We now try to factor the quotient x3 – 5x – 2 • Its possible zeros are the divisors of –2, namely, ±1, ±2 • We already know that 1 and 2 are not zeros of the original polynomial P. • So, we don’t need to try them again.

  31. Example (a) E.g. 3—Theorem & Quad. Formula • Checking the remaining candidates –1 and –2, we see that –2 is a zero.

  32. Example (a) E.g. 3—Theorem & Quad. Formula • Also, P factors as:

  33. Example (a) E.g. 3—Theorem & Quad. Formula • Now, we use the quadratic formula to obtain the two remaining zeros of P: • The zeros of P are: 5, –2 , ,

  34. Example (b) E.g. 3—Theorem & Quad. Formula • Now that we know the zeros of P, we can use the methods of Section 4.2 to sketch the graph. • If we want to use a graphing calculator instead, knowing the zeros allows us to choose an appropriate viewing rectangle. • It should be wide enough to contain all the x-intercepts of P.

  35. Example (b) E.g. 3—Theorem & Quad. Formula • Numerical approximations to the zeros of P are: 5, –2, 2.4, –0.4

  36. Example (b) E.g. 3—Theorem & Quad. Formula • So, in this case, we choose the rectangle [–3, 6] by [–50, 50] and draw the graph.

  37. Descartes’ Rule of Signs and Upper and Lower Bounds for Roots

  38. Descartes’ Rule of Signs • In some cases, the following rule is helpful in eliminating candidates from lengthy lists of possible rational roots. • It was discovered by the French philosopher and mathematician René Descartes around 1637.

  39. Variation in Sign • To describe this rule, we need the concept of variation in sign. • Suppose P(x) is a polynomial with real coefficients, written with descending powers of x (and omitting powers with coefficient 0). • A variation in signoccurs whenever adjacent coefficients have opposite signs.

  40. Variation in Sign • For example, P(x) = 5x7 – 3x5 – x4 + 2x2 + x – 3 has three variations in sign.

  41. Descartes’ Rule of Signs • Let P be a polynomial with real coefficients. • The number of positive real zeros of P(x) is either equal to the number of variations in sign in P(x) or is less than that by an even whole number. • The number of negative real zeros of P(x) is either equal to the number of variations in sign in P(-x) or is less than that by an even whole number.

  42. E.g. 4—Using Descartes’ Rule • Use Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros of the polynomialP(x) = 3x6 + 4x5 + 3x3 – x – 3 • The polynomial has one variation in sign. • So, it has one positive zero.

  43. E.g. 4—Using Descartes’ Rule • Now,P(–x) = 3(–x)6 + 4(–x)5 +3(–x)3 – (–x) – 3 = 3x6 – 4x5 – 3x3 + x – 3 • Thus, P(–x) has three variations in sign. • So, P(x) has either three or one negative zero(s), making a total of either two or four real zeros.

  44. Upper and Lower Bounds for Roots • We say that a is a lower boundand b is an upper boundfor the zeros of a polynomial if every real zero c of the polynomial satisfies a ≤c ≤b. • The next theorem helps us find such bounds for the zeros of a polynomial.

  45. The Upper and Lower Bounds Theorem • Let P be a polynomial with real coefficients. • If we divide P(x) by x –b (with b > 0) using synthetic division, and if the row that contains the quotient and remainder has no negative entry, then b is an upper bound for the real zeros of P. • If we divide P(x) by x –a (with a < 0) using synthetic division, and if the row that contains the quotient and remainder has entries that are alternately nonpositive and nonnegative, then a is a lower bound for the real zeros of P.

  46. Upper and Lower Bounds Theorem • A proof of this theorem is suggested in Exercise 95. • The phrase “alternately nonpositive and nonnegative” simply means that: • The signs of the numbers alternate, with 0 considered to be positive or negative as required.

  47. E.g. 5—Upper & Lower Bounds for Zeros of Polynomial • Show that all the real zeros of the polynomialP(x) = x4 – 3x2 + 2x – 5 • lie between –3 and 2.

  48. E.g. 5—Upper & Lower Bounds for Zeros of Polynomial • We divide P(x) by x – 2 and x + 3 using synthetic division. • All entries are positive.

  49. E.g. 5—Upper & Lower Bounds for Zeros of Polynomial • Entries alternate in sign.

  50. E.g. 5—Upper & Lower Bounds for Zeros of Polynomial • By the Upper and Lower Bounds Theorem, –3 is a lower bound and 2 is an upper bound for the zeros. • Neither –3 nor 2 is a zero (the remainders are not 0 in the division table). • So, all the real zeros lie between these numbers.

More Related