College Algebra K /DC Monday, 24 February 2014

1. College Algebra K/DCMonday, 24 February 2014 • OBJECTIVETSW (1) factor polynomials into linear factors, and (2) find all zeros of polynomials. • ASSIGNMENT DUE WEDNESDAY(not tomorrow) • Sec. 3.2: p. 327 (37-40 all, 53-58 all) • QUIZ ON WEDNESDAY • Sec. 3.2 only. • Assignments for sec. 3.3 will be due on Monday, 03 March 2014 (TEST day). • PI Day will be observed on Friday, 14 March 2014. • Be thinking about what “pie” you will bring.

2. Deciding Whether a Number is a Zero (From sec. 3.2) • Let . Is k = 1 + 3i a zero? Use synthetic division with k = 1 + 3i. YES, f(1 + 3i) = 0, sok = 1 + 3iis a zero.

3. Forms of Answers (From sec. 3.2) • Directions:Divide; write your answer as a polynomial. • Ex: • Directions:Express each polynomial of the formf(x) = (x – k) • q(x) + r for the given value of k. • Ex: Ex:

4. Forms of Answers (From sec. 3.2) • Directions:Use the remainder theorem and synthetic division to find f(k). • Ex: • Directions:Use synthetic division to decide whether the given number is a zero of the given polynomial. Give a reason for your answer. • Ex: NO, f(5) = 320, so k = 5 is not a zero.

5. Zeros of Polynomial Functions 3.3 Factor Theorem ▪ Number of Zeros

6. Factoring into Linear Factors vs. Finding Zeros Different directions lead to different forms for your answer. Ex:Factor the given polynomial into linear factors. Ex:Find the zeros of the given polynomial.

7. Factoring a Polynomial Given a Zero • Factor f(x) = 2x3 – 3x2 – 5x + 6 into linear factors, given that 1 is a zero of f(x).

8. Factoring a Polynomial Given a Zero • Factor into linear factors, given that 5 is a zero of f(x). Since 5 is a zero of f, x – 5 is a factor. Divide f(x) by x – 5. The quotient is .

9. Finding Zeros of a Polynomial • For the polynomial f(x) = x3 + 4x2 – 11x – 30, one zero is –2. Find all others. The zeros of f (x) are −2, −5, and 3.

10. Finding Zeros of a Polynomial • For the polynomial f (x) = 2x 3 – x 2 – 25x – 12, one zero is −3. Find all others. The zeros of f (x) are −3, −½, and 4. Now factor 2x 2 – 7x – 4. 2x 2 – 7x – 4 = (2x + 1)(x – 4) Set each of these equal to zero and solve. 2x + 1 = 0 ⇒ x = −½x – 4 = 0 ⇒ x = 4

11. Conjugate Zeros Theorem • If f(x) defines a polynomial function having only real coefficients and if (a + bi) is a zero of f(x), where a and b are real numbers, then (a– bi) is also a zero of f(x). • This is used ONLY WITH i’s!!!!!

12. Finding Zeros of a Polynomial • For the polynomial f(x) = x4 + x3 – x2 + x – 2, one zero is i. Find all others. The zeros are i, −i, −2, and 1.

13. Finding All Zeros of a Polynomial Function Given One Zero Find all zeros ofgiven that 2 + i is a zero. Since f(x) has only real coefficients, and 2 + i is a zero, then 2 –i is also a zero. Use synthetic division to divide f(x), using k = 2 + i.

14. Finding All Zeros of a Polynomial Function Given One Zero Now use synthetic division to dividethe quotient polynomial by k = 2 –i. Now set this equal to zero, factor, and solve for x. The zeros of f(x) are 2 + i, 2 – i, –5, and 2.

15. Multiplicity • For the polynomial • the linear factorization is • or • (x + 1) is a factor of multiplicity 3.

16. Assignment • Sec. 3.3: pp. 337-338 (17-33 odd) • Write the problem and solve. • Due on Monday, 03 March 2014 (TEST day).

17. Class Problems [02/24/2014]Due before you leave today. • Factor into linear factors given that k is a zero of f (x). • 1)f (x) = 2x3 – 9x2 – 11x + 30; k = 5 • 2)f (x) = 2x3 + 19x2 + 25x + 8; k = −1 • 3)f (x) = 3x3 – 5x2 – 16x + 12; k = 2/3 • For each polynomial function, one zero is given. Find all others. • 4)f (x) = 3x3 + 2x2 – 7x + 2; 1/3 • 5)f (x) = x4 – 3x3 + 3x2 – 3x + 2; −i • 6)f (x) = x4 – 3x3 + 2x2 + 2x – 4; 1 + i (Your answer will be of the form f(x) = (x – k)(x – c2)(x – c3).) (Your answer will be: The zeros are ….)

18. Assignment: Sec. 3.3: pp. 337-338 (17-33 odd)Due on Monday, 03 March 2014. • Factor f(x) into linear factors given that k is a zero of f(x). • For each polynomial function, one zero is given. Find all others.

19. College Algebra K/DCTuesday, 25 February 2014 • OBJECTIVETSW use the Rational Zeros Theorem to find zeros of polynomials. • TEST ON MONDAY • Sec. 3.1 – 3.3. • ASSIGNMENT DUE TOMORROW • Sec. 3.2: p. 327 (37-40 all, 53-58 all) • QUIZ: Sec. 3.2 is tomorrow. • REMINDERS • PI Day: Friday, 14 March 2014

20. Zeros of Polynomial Functions 3.3 Rational Zeros Theorem

21. Rational Zeros Theorem • If is a rational number written in lowest terms, • and if is a zero of f (a polynomial function with • integer coefficients), then p is a factor of the • constant term and q is a factor of the leading • coefficient.

22. Using the Rational Zeros Theorem Possible rational zeros, :(always reduced) • For the given polynomial function, (a) list all possible rational zeros, (b) find all zeros, and(c) factor f(x). • (a) List all possible rational zeros. p must be a factor of a0 = 4:(constant) ±1,±2, ±4 q must be a factor of a4 = 8:(leading coefficient) ±1,±2, ±4, ±8

23. Using the Rational Zeros Theorem (b) Find all zeros. • Use trial and error to find the first zero. Is 1 a zero? 1 is not a zero. Is –1 a zero? –1 is a zero.

24. Using the Rational Zeros Theorem The rational zeros are . • Use the quotient polynomial to find the next factor. Is 4 a zero? 4is a zero. (c) Factor f(x) into linear factors.

25. Assignment • Sec. 3.3: p. 338 (35-42 all) • Write the problem. • DIRECTIONS: • List all possible rational zeros; • Use synthetic division (trial and error) to find all zeros; • Factor f (x) into linear factors. • Due on Monday, 03 March 2014 (TEST day).

26. Sec. 3.3: In-Class Problems [02/25/2014] Due before you leave today. • For each problem, (a)list all possible rational zeros, (b)find all zeros (real and complex) of each polynomial function, and (c) factor f (x).

27. Assignment: p. 338 (35-42 all)Due on Monday, 03 March 2014 (TEST day) • For each polynomial function, (a) list all possible rational zeros, (b) find all zeros, and (c) factor f(x).

28. College Algebra K/DCWednesday, 26 February 2014 • OBJECTIVE TSW (1) quiz over sec. 3.2, and (2) complete WS REVIEW Sec. 3.1 – 3.3. • TEST ON MONDAY • Sec. 3.1 – 3.3. • ASSIGNMENT DUE RIGHT NOW • Sec. 3.2: p. 327 (37-40 all, 53-58 all) wire basket • ASSIGNMENTS DUE MONDAY • WS Sec. 3.3  will be given on Friday • Sec. 3.3: pp. 337-338 (17-33 odd) • Sec. 3.3: p. 338 (35-42 all) • REMINDERPI Day: Friday, 14 March 2014

29. College Algebra K/DCFriday, 28 February 2014 • OBJECTIVE TSW (1) finish exploring sec. 3.3, and (2) review for Monday’s test covering sec. 3.1 – 3.3. • TEST ON MONDAY • Sec. 3.1 – 3.3. • ASSIGNMENTS DUE MONDAY • WS Sec. 3.3 • Sec. 3.3: pp. 337-338 (17-33 odd) • Sec. 3.3: p. 338 (35-42 all) • REMINDERPI Day: Friday, 14 March 2014

30. Finding Functions That Meet Conditions • Suppose you knew that a polynomial function of degree 3 had zeros of 3, −1, and −2, and that f(1) = −2. • What is the function? • First, multiply the factors: • f(x) = a(x – 3)(x + 1)(x + 2) • f(x) = a(x2 – 2x – 3)(x + 2) • f(x) = a(x3 – 7x – 6)

31. Finding Functions That Meet Conditions • Suppose you knew that a polynomial function of degree 3 had zeros of 3, −1, and −2, and that f(1) = −2. • f(x) = a(x3 – 7x – 6) • Since f(1) = −2, • −2 = a(13 – 7(1) – 6) • −2 = a(−12) • a = 1/6 • So the function is f(x) = 1/6x3 – 7/6x – 1

32. Finding Functions That Meet Conditions • Zeros of 2 with multiplicity 2 and zero of 1 multiplicity 1, f(5) = 12. Find a polynomial of degree 3 meeting these conditions.

33. Finding Functions That Meet Conditions • Zeros of 3i and 1, degree 3, f(0) = −4.

34. Assignment • ASSIGNMENTS DUE MONDAY • WS Sec. 3.3 • Sec. 3.3: pp. 337-338 (17-33 odd) • Sec. 3.3: p. 338 (35-42 all)