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Warm Up

Warm Up. Solve the following equations/inequalities. Mini Quiz. Solve the following equations/inequalities. Solving Higher Degree Polynomials. Let’s Review Synthetic Substitution. To use synthetic substitution/ division:.

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Warm Up

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  1. Warm Up • Solvethe following equations/inequalities

  2. Mini Quiz • Solvethe following equations/inequalities

  3. Solving Higher Degree Polynomials

  4. Let’s Review Synthetic Substitution

  5. To use synthetic substitution/ division: • Write the polynomial in standard form (Insert zeros for missing terms) • List coefficients • If you have x = c, divide by c • If you have (x + c) or (x – c), use the opposite of c • Drop down the first coefficient • Multiply, add down, and repeat

  6. Using Synthetic Substitution Use synthetic division to evaluate f(x) = 2x4+ -8x2+ 5x- 7 when x = 3. Ex. EQ: How do we use Long Division and Synthetic Division?

  7. Using Synthetic Substitution Polynomial in standard form 3 Coefficients x-value SOLUTION 2x4 + 0x3 + (–8x2) + 5x + (–7) Polynomial in standard form 2 0 –8 5 –7 3• Coefficients 6 18 30 105 35 10 98 2 6 The value of f(3) is the last number you write, In the bottom right-hand corner. EQ: How do we use Long Division and Synthetic Division?

  8. Ex. 1 Evaluate 4x3+ x + 7 when x = 2 2

  9. Ex. 1 Evaluate 4x3+ x + 7 when x = 2 add down multiply by the box number 2 41

  10. Synthetic Division • We will use synthetic division to divide one polynomial by another. • Ex.  ****When we divide we are finding FACTORS!!

  11. To determine if a given binomial is a factor • Divide synthetically • If your remainder is 0, then the binonial is a factor • Ex. Is x – 3 a factor for

  12. Ex. 1 (4x3 + x + 7)  (x – 2) Set = to zero & solve for x 2

  13. Ex. 1 (4x3 + x + 7)  (x – 2) add down multiply by the box number 2 41 Remainder Quotient

  14. Ex. 2 You try… (x3 + 6x2 – x – 30)  (x – 2) 30

  15. Ex. 3 (2x3 + 7x2 – 5)  (x + 3)

  16. Classwork Do Now • Worksheet • Homework pg. 87 # 7 - 12

  17. Classwork: Complete (10 min) • Divide synthetically

  18. Higher Degree Polynomial Functions

  19. An example of a polynomial function… f(x) = 6x4 + x3 – 21x2 – 15x + 36 Leading coefficient Degree (highest exponent) (in front)

  20. 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. “root” “zero” and “solution” all mean the same thing

  21. From Last week… 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

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

  23. How many roots? How many factors? f(x) = x3 + 2x2 – 3x + 12 f(x) = x3 + 2x2 – 3x + 12 3 3 f(x) = x5 + 14x4 – 4 f(x) = x5 + 14x4 – 4 5 5 The degree tells you the number of roots or factors the polynomial will have.

  24. Repeated roots… Ex. 1 P(x) = x2 – 10x + 25 P(x) = (x - 5)(x - 5) x = 5 x = 5 5 has a multiplicity of two If a root occurs k times, it has a multiplicity of k.

  25. Repeated roots… f(x) = x2 f(x) = (x+3)(x+3)(x+3) x = 0, 0 x = -3, -3, -3 0 has a multiplicity of two -3 has a multiplicity of three

  26. Conjugate pairs -If is a root, then is also a root. Imaginary roots always come in pairs. -If a + biis a root, then a – bi is also a root. Irrational roots always come in pairs.

  27. Our goal is to be able to find all of the roots of a polynomial.

  28. Finding Roots

  29. Fundamental Theorem of Algebra -Every polynomial will have at least one linear factor

  30. Sometimes you can find all of the roots of a polynomial without doing much work.

  31. Suppose a polynomial of degree 3 has 3 – 4i and 9 as roots. Find all of the roots. 3 Ex. 2 Roots Because the degree is 3… 9 we should have 3 roots. 3 – 4i Because one root is 3 – 4i… 3 + 4i then 3 + 4i is also a root. Imaginary roots always come in pairs.

  32. Ex. 3 You try… Suppose a polynomial of degree 6 has -2 + 5i, -i, and as roots. Find all of the roots. 6 Because the degree is 6… Roots we should have 6 roots. -2 + 5i -2 – 5i -i Imaginary roots always come in pairs. i Irrational roots always come in pairs.

  33. Ex. 4 Find all of the roots. f(x) = x2 – 2x - 8 If quadratic: (x – 4)(x + 2) factor OR quadratic formula

  34. What if it is a higher degree polynomial? • If given a root or a factor, use synthetic division with the given root/factor • Then use factoring or the quadratic formula to solve the remaining quadratic

  35. Ex. 5 Given a zero, find all the zeros P(x) = x3 – 6x2 + 13x – 20;4 We should get zero for a remainder… 20 4 -8 why? 1 -2 5 Now we have a quadratic, what can we do? Factor OR quadratic formula

  36. Ex. 5 Given a zero, find all the zeros P(x) = x3 – 6x2 + 13x – 20;4 , 12i 20 4 -8 1 -2 5

  37. Warm Up • Find ALL zeros

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