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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.

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

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  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.

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