1 / 29

Chapter 6

Chapter 6. Polynomials and Polynomial Functions. 6.1 Using Properties of Exponents. Properties of Exponents. Let a and b be real numbers and let m and n be integers. PRODUCT OF POWERS PROPERTY a m • a n = a m+n POWER OF A POWER PROPERTY ( a m ) n = a mn

kellsie
Télécharger la présentation

Chapter 6

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. Chapter 6 Polynomials and Polynomial Functions

  2. 6.1 Using Properties of Exponents

  3. Properties of Exponents Let a and b be real numbers and let m and n be integers. PRODUCT OF POWERS PROPERTY am• an= am+n POWER OF A POWER PROPERTY (am)n= amn POWER OF A PRODUCT PROPERTY (ab)m= ambm NEGATIVE EXPONENT PROPERTY ZERO EXPONENT PROPERTY a0 = 1, a ≠ 0 QUOTIENT OF POWERS PROPERTY POWER OF A QUOTIENT PROPERTY

  4. A number is expressed in Scientific Notation if it is in the form c X 10n where 1 ≤ c < 10 and n is an integer. Decimal to scientific Notation: 5,284,000 = 5.284 X 106 0.0000075 = 7.5 X 10-6 Scientific Notation to decimal: 3.45670123 X 108 = 345,670,123 6.546 X 10-5 = 0.00006546

  5. Classwork: Pg 326 – 328 (1-15,48,52,53) Homework: Pg 326 – 328 (16-46 every 3rd, 47,49-51,54-56)

  6. 6.2 Evaluating and Graphing Polynomial Functions

  7. A Polynomial Function : Leading Coefficient: Constant Term: a0 Degree: n Standard Form: terms are written in descending order of exponents from left to right

  8. A Polynomial Function in Standard Form: Leading Coefficient: Constant Term: 5 Degree: 3 Standard Form: terms are written in descending order of exponents from left to right

  9. Polynomial Classifications

  10. X 3 Synthetic Substitution Use synthetic Substitution to evaluate f(x)=2x4 - 8x2 + 5x – 7 when x = 3. Write the value of x and the coefficients of ƒ(x) as shown. Polynomial in standard form 2x4 + 0x3 - 8x2 + 5x – 7 x-value Coefficients 3 2 0 -8 5 -7 add 6 18 30 105 f(3) = 98 The value of f(x) is the last number written in the bottom right hand corner 2 6 10 35 98

  11. END BEHAVIOR FOR POLYNOMIAL FUNCTIONS For an>0 and n is even, both ends point up For an>0 and n is odd, left end points down and right end points up For an<0 and n is even, both ends point down For an<0 and n is odd, left end points up and right end points down See chart on pg 331

  12. Graphing Polynomial Functions To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior. ƒ(x) = x3 + x2 - 4x – 1

  13. Classwork: Pg 333 – 336 (1-13, 65,68) Homework: Pg 333 – 336 (15,16,20,21,22,25,27,28,36,37,41,45,47,49-52,53,56,59,62,66,70, 80, 81)

  14. 6.3 Adding, Subtracting, and Multiplying Polynomials

  15. To add or subtract polynomials, add or subtract the coefficients of like terms. You can use a vertical or horizontal format. 3x3+ 2x2 – x - 7 + x3 - 10x2 - x + 8 (9x3 - 2x + 1) + (5x2 + 12x - 4) = 9x3 + 5x2 - 2x + 12x + 1 – 4 = 9x3 + 5x2 + 10x - 3 4x3 - 8x2 - x + 1 8x3 - 3x2 - 2x + 9 -2x3 - 6x2 + x - 1 6x3 – 9x2 – x + 8 8x3 - 3x2 - 2x + 9 -(2x3 + 6x2 -x + 1) Add the opposite. (2x2 + 3x) -(3x2 + x - 4)= 2x2 + 3x - 3x2 - x + 4 = - x2 + 2x + 4 Add the opposite.

  16. Multiplying Polynomials To Multiply polynomials, use the distributive property regardless if you use the vertical or horizontal method. Horizontally (x - 3)(3x2 - 2x - 4)=x(3x2)+x(-2x)+x(-4)-3(3x2)-3(-2x)-3(-4) =3x3 – 2x2- 4x – 9x2 + 6x + 12 =3x3 – 11x2 + 2x +12 • Vertically • - x2 + 2x + 4 • X x - 3 • 3x2 - 6x - 12 • x3 + 2x2 +4x • x + 5x2 – 2x - 12 Multiply -x2 + 2x + 4 by -3. Multiply -x2 + 2x + 4 by x. Combine like terms.

  17. Special Product Patterns SUM AND DIFFERENCE Example (a + b)(a - b) = a2 - b2 (x + 3)(x - 3) = x2 - 9 SQUARE OF A BINOMIAL (a + b)2 = a2 + 2ab + b2 (y + 4)2 = y2 + 8y + 16 (a - b)2 = a2 - 2ab + b2 (3t2 - 2)2 = 9t4 - 12t2 + 4 CUBE OF A BINOMIAL (a + b)3 = a3 + 3a2b + 3ab2 + b3 (x + 1)3 = x3 + 3x2 + 3x + 1 (a - b)3 = a3 - 3a2b + 3ab2 - b3 (p - 2)3 = p3 - 6p2 + 12p - 8

  18. Classwork: Pg 341 – 343 (1-12) Homework: Pg 341 – 343 (13 – 60 every 3rd, 62,63)

  19. 6.4 Factoring and Solving Polynomial Equations

  20. In Chapter 5 you learned how to factor the following types of quadratic expressions. TYPE EXAMPLE General trinomial 2x2 - 5x - 12 = (2x + 3)(x - 4) Perfect square trinomial x2 + 10x + 25 = (x + 5)2 Difference of two squares 4x2 - 9 = (2x + 3)(2x - 3) Common monomial factor 6x2 + 15x = 3x(2x + 5) In this lesson you will learn how to factor other types of polynomials.

  21. SPECIAL FACTORING PATTERNS SUM OF TWO CUBES Example a3 + b3 = (a + b)(a2 - ab + b2) x3 + 8 = (x + 2)(x2 - 2x + 4) DIFFERENCE OF TWO CUBES a3 - b3 = (a - b)(a2 + ab + b2) 8x3 - 1 = (2x - 1)(4x2 + 2x + 1) x3 + 27 = x3 + 33Sum of two cubes = (x + 3)(x2 - 3x + 9) 16u5 - 250u2 = 2u2(8u3 - 125) Factor common monomial. = 2u2[(2u)3 – 53] Difference of two cubes = 2u2(2u - 5)(4u2 + 10u + 25)

  22. Factor By Grouping The pattern for this is as follows. ra + rb + sa + sb = r(a + b) + s(a + b) = (r + s)(a + b) x3 - 2x2 - 9x + 18 = x2(x - 2) - 9(x - 2) Factor by grouping. = (x2 - 9)(x - 2) = (x + 3)(x - 3)(x - 2) Difference of squares

  23. In Chapter 5 you learned how to use the zero product property to solve factorable quadratic equations. You can extend this technique to solve some higher-degree polynomial equations. Solve 2x5 + 24x = 14x3. 2x5 + 24x = 14x3Write original equation. 2x5 - 14x3 + 24x = 0 Rewrite in standard form. 2x(x4 - 7x2 + 12) = 0 Factor common monomial. 2x(x2 - 3)(x2 - 4) = 0 Factor trinomial. 2x(x2 - 3)(x + 2)(x - 2) = 0 Factor difference of squares. x = 0, x = , x = - , x = -2, or x = 2 Zero product property The solutions are 0, , , -2, and 2. Check these in the original equation.

  24. Classwork: Pg 348 – 350 (1-3,5-16,88) Homework: Pg 348 – 350 (18,21,24,27-32,33,37,41,45,49,50,51,68,71,77,80,89)

  25. 6.5 The Remainder and Factor Theorems If a polynomial ƒ(x) is divided by x -k, then the remainder is r = ƒ(k). REMAINDER THEOREM A polynomial ƒ(x) has a factor x - k if and only if ƒ(k) = 0. FACTOR THEOREM

  26. Polynomial Long Division Divide 2x4 + 3x3 + 5x - 1 by x2 - 2x + 2. Subtract 2x2(x2 - 2x+2) Subtract 7x(x2 - 2x+2) Subtract 10(x2 - 2x+2) Remainder Answer

  27. Synthetic Division Divide x3 + 2x2 - 6x - 9 by x - 2 Solve for x: x-2=0 X=2 2 1 2 -6 -9 2 8 4 1 4 2 -5 When writing solution, reduce the degree of the polynomial by 1. Add the remainder divided by the divisor at the end of the answer Answer

  28. If the remainder is 0 then the divisor is a solution or a zero or a factor. You can use this fact to help factor completely.

  29. Classwork: Pg 356-358 (4-13,39,47) Homework: P 356 – 358 (15-54 every 3rd)

More Related