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Agenda

Agenda. Fri Oct. 2 Remainder and Factor Theorems. Mon Oct. 5 Find Rational Zeros. Warm – Up Tests not graded 2.5 – Remainder and Factor Theorems 2.5 – Find zeros, solutions or roots of functions HW. Tues Oct. 6 Fundamental Theorem of Algebra. WEB Oct. 7 Graphs of Polynomial Functions.

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Agenda

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  1. Agenda Fri Oct. 2 Remainder and Factor Theorems Mon Oct. 5 Find Rational Zeros • Warm – Up • Tests not graded • 2.5 – Remainder and Factor Theorems • 2.5 – Find zeros, solutions or roots of functions • HW Tues Oct. 6 Fundamental Theorem of Algebra WEB Oct. 7 Graphs of Polynomial Functions

  2. An example of a polynomial function… f(x) = 6x4 + x3 – 21x2 – 15x + 36 When we solve a polynomial function, we are looking for the numbers that make the equation equal to zero OR where the graph crosses the x-axis (x=0). “root” “zero” and “solution” all mean the same thing

  3. On a much smaller scale… f(x) = x2 + x – 6 To solve this equation… x2 + x – 6 = 0 Set equal to zero (x + 3)(x – 2) = 0 Factor Set each factor equal to zero x + 3 = 0 x – 2 = 0 x = - 3 x = 2 solve

  4. f(x) = 6x4 + x3 – 21x2 – 15x + 36 • With larger polynomials, solving by factoring is impractical. • Later in the chapter we’ll learn new methods for solving a polynomial.

  5. How do we determine if a number is a root/zero/solution? Solutions: f(x) = x2 + x – 6 x = - 3 x = 2 f(-3) = (-3)2 + (-3) – 6 = 9 – 3 – 6 = 0 f(2) = (2)2 + (2) – 6 = 4 + 2 – 6 = 0 An x value is a “zero” if you get zero after evaluating the function at that value for x.

  6. You try… 1. Determine if -5 is a root of f(x) = x3 – 5x2 + x – 5 f(-5) = (-5)3 – 5(-5)2 + (-5) – 5 = -125 – 125 – 5 – 5 = -260 No, -5 is not a root

  7. 2. Determine if -i is a root of p(x) = x3 – 5x2 + x – 5 f(-i) = (-i)3 – 5(-i)2 + (-i) – 5 i = i i2 = -1 i3 = -i i4 = 1 = - i3 – 5i2 – i – 5 = -(-i) – 5(-1) – i – 5 = i + 5 – i – 5 = 0 Yes, -i is a root.

  8. You try… 3. Determine if 2i is a root of g(x) = x3 – 3x2 + x – 3 f(2i) = (2i)3 – 3(2i)2 + (2i) – 3 i = i i2 = -1 i3 = -i i4 = 1 = 8i3 – 3(4i2) + 2i – 3 = -8i + 12 +2i – 3 = 9 – 6i NO

  9. Sometimes we write a polynomial as the product of its factors… f(x) = 6x4 + x3 – 21x2 – 15x + 36 f(x) = (x + 1)(x + 2)(2x – 3)(3x – 4) OR f(x) = x2 + x – 6 f(x) = (x + 3)(x – 2)

  10. One way to determine if something is a factor is long division…

  11. 4. Determine if x-2 is a factor… Because the remainder is zero… Yes, it’s a factor.

  12. 5. Determine if x2+3 is a factor of f(x) = x3 –x2 + 5x – 4 – 1 x x3 – x2 + 5x – 4 x2 + 0x + 3 x3 + 0x2 + 3x + 2x - x2 – 4 Because the remainder is not zero… - x2 + 0x – 3 2x – 1 No, it’s not a factor.

  13. 11.2 Remainder and Factor Theorems

  14. Remainder Theorem This Theorem says that if you divide a polynomial function f(x) by a linear binomial {(x – k)}, then the remainder is the value of the function at k. In other words f(k) = r, where r is the remainder after dividing f(x) by (x – k)

  15. You can use Long Division or Synthetic Division to evaluate a function at a given value of the variable.

  16. Long division can be a real pain…

  17. 1. Synthetic Division (6x3 – 19x2 + 16x – 4)  (x – 2)  List coefficients (with spaces) 2 6 -19 16 -4  Place root in corner box (opposite of factor) 12 -14 4  bring down 6 2 -7 0  multiply by box remainder  add down Because the remainder is zero… (x - 2) is a factor 2 is a root

  18. 1. Synthetic Division (6x3 – 19x2 + 16x – 4)  (x – 2) 2 6 -19 16 -4 12 -14 4 6 2 -7 0 remainder (6x3 – 19x2 + 16x – 4)= (x – 2)(6x2 – 7x + 2)

  19. 2. Synthetic Division (x5 – 64)  (x + 1) -1 1 0 0 0 0 -64 -1 1 -1 1 -1 1 -1 1 -1 1 -65 remainder Because the remainder is not zero… (x + 1) is not a factor -1 is not a root

  20. 2. Synthetic Division (x5 – 64)  (x + 1) -1 1 0 0 0 0 -64 -1 1 -1 1 -1 1 -1 1 -1 1 -65 remainder (x5 – 64) = (x + 1)(x4 – x3 + x2 – x + 1) – 65

  21. Compare the previous example to… f(x)= x5 – 64 given f(-1)= find Using Substitution we get….. f(-1)= -1 - 64 f(-1)= -65 Is this the same thing we got with SD

  22. Factor given that Because we know that or is a factor of Use Synthetic Division to find the other factors.

  23. -3 2 11 18 9 -6 -15 -9 2 5 3 0 remainder We now have……

  24. Now that we have a linear binomial factor and a quadratic trinomial factor we can continue factoring to find the other solutions, zeros, roots. Can be factored into 2 binomial Factored Completely we have……

  25. One zero of f(x) = x3 + 6x2 + 3x – 10 is x = -5. Find the other zeros of the function. Use synthetic division -5 1 6 3 -10 -5 -5 10 remainder 1 1 -2 0 Write the new factors To be continued……………….

  26. Can the trinomial be factored? By the Factor Theorem, the zeros of f are 2, -3, and 3.

  27. Homework • Work Sheet

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