Presentation is loading. Please wait.

Presentation is loading. Please wait.

Copyright © 2010 Pearson Education, Inc

Similar presentations


Presentation on theme: "Copyright © 2010 Pearson Education, Inc"— Presentation transcript:

1 Copyright © 2010 Pearson Education, Inc
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 1- 1

2 4 Polynomials 4.1 Exponents and Their Properties
4.2 Negative Exponents and Scientific Notation 4.3 Polynomials 4.4 Addition and Subtraction of Polynomials 4.5 Multiplication of Polynomials 4.6 Special Products 4.7 Polynomials in Several Variables 4.8 Division of Polynomials Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

3 Exponents and Their Products
4.1 Exponents and Their Products Multiplying Powers with Like Bases Dividing Powers with Like Bases Zero as an Exponent Raising a Power to a Power Raising a Product or a Quotient to a Power Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

4 The Product Rule For any number a and any positive integers m and n, (To multiply powers with the same base, keep the base and add the exponents.) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

5 Example Multiply and simplify each of the following. (Here “simplify” means express the product as one base to a power whenever possible.) a) x3  x5 b) 62  67  63 c) (x + y)6(x + y)9 d) (w3z4)(w3z7) Solution a) x3  x5 = x Adding exponents = x8 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

6 continued Example b) 62  67  63 c) (x + y)6(x + y)9 d) (w3z4)(w3z7)
Solution b) 62  67  63 = = 612 c) (x + y)6(x + y)9 = (x + y)6+9 = (x + y)15 d) (w3z4)(w3z7) = w3z4w3z7 = w3w3z4z7 = w6z11 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

7 The Quotient Rule For any nonzero number a and any positive integers m and n for which m > n, (To divide powers with the same base, subtract the exponent of the denominator from the exponent of the numerator.) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

8 Example Divide and simplify each of the following. (Here “simplify” means express the product as one base to a power whenever possible.) a) b) c) d) Solution a) b) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

9 Example continued c) d) Solution c) d)
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

10 The Exponent Zero (Any nonzero number raised to the 0 power is 1.)
For any real number a, with a ≠ 0, (Any nonzero number raised to the 0 power is 1.) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

11 Example Simplify: a) 12450 b) (3)0 c) (4w)0 d) (1)80 e) 80.
Solution a) = 1 b) (3)0 = 1 c) (4w)0 = 1, for any w  0. d) (1)80 = (1)1 = 1 e) 80 is read “the opposite of 90” and is equivalent to (1)90: 90 = (1)90 = (1)1 = 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

12 The Power Rule (am)n = amn.
For any number a and any whole numbers m and n, (am)n = amn. (To raise a power to a power, multiply the exponents and leave the base unchanged.) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

13 Example Simplify: a) (x3)4 b) (42)6 Solution a) (x3)4 = x34 = x12
= 416 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

14 Raising a Product to a Power
For any numbers a and b and any whole number n, (ab)n = anbn. (To raise a product to a power, raise each factor to that power.) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

15 Example Simplify: a) (3x)4 b) (2x3)2 c) (a2b3)7(a4b5) Solution
a) (3x)4 = 34x4 = 81x4 b) (2x3)2 = (2)2(x3)2 = 4x6 c) (a2b3)7(a4b5) = (a2)7(b3)7a4b5 = a14b21a4b Multiplying exponents = a18b Adding exponents Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

16 Raising a Quotient to a Power
For any real numbers a and b, b ≠ 0, and any whole number n, (To raise a quotient to a power, raise the numerator to the power and divide by the denominator to the power.) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

17 Example Simplify: a) b) c) Solution a) b)
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

18 continued Example Simplify: c) Solution c)
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

19 Definitions and Properties of Exponents For any whole numbers m and n,
1 as an exponent: a1 = a 0 as an exponent: a0 = 1 The Product Rule: The Quotient Rule: The Power Rule: (am)n = amn Raising a product to a power: (ab)n = anbn Raising a quotient to a power: Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

20 Negative Exponents and Scientific Notation
4.2 Negative Exponents and Scientific Notation Negative Integers as Exponents Scientific Notation Multiplying and Dividing Using Scientific Notation Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

21 For any real number a that is nonzero and any integer n,
Negative Exponents For any real number a that is nonzero and any integer n, (The numbers a-n and an are reciprocals of each other.) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

22 Example Express using positive exponents, and, if possible, simplify.
a) m5 b) 52 c) (4)2 d) xy1 Solution a) m5 = b) 52 = Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

23 continued Example c) (4)2 = d) xy1 =
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

24 Example Simplify. Do not use negative exponents in the answer.
a) b) (x4)3 c) (3a2b4)3 d) e) f) Solution a) b) (x4)3 = x(4)(3) = x12 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

25 continued Example c) (3a2b4)3 = 33(a2)3(b4) = 27 a6b12 = d) e) f)
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

26 Factors and Negative Exponents
For any nonzero real numbers a and b and any integers m and n, (A factor can be moved to the other side of the fraction bar if the sign of the exponent is changed.) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

27 Example Simplify. Solution
We can move the negative factors to the other side of the fraction bar if we change the sign of each exponent. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

28 Reciprocals and Negative Exponents
For any nonzero real numbers a and b and any integer n, (Any base to a power is equal to the reciprocal of the base raised to the opposite power.) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

29 Example Simplify: Solution
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

30 Scientific Notation N × 10m,
Scientific notation for a number is an expression of the type N × 10m, where N is at least 1 but less than 10 (that is, 1 ≤ N < 10), N is expressed in decimal notation, and m is an integer. Note that when m is positive the decimal point moves right m places in decimal notation. When m is negative, the decimal point moves left |m| places. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

31 Example Convert to decimal notation: a) 3.842  106 b) 5.3  107
Solution a) Since the exponent is positive, the decimal point moves right 6 places.  106 = 3,842,000 b) Since the exponent is negative, the decimal point moves left 7 places.  107 = Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

32 Example Write in scientific notation: a) 94,000 b) 0.0423 Solution
a) We need to find m such that 94, = 9.4  10m. This requires moving the decimal point 4 places to the right. 94,000 = 9.4  104 b) To change 4.23 to we move the decimal point 2 places to the left. = 4.23  102 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

33 Multiplying and Dividing Using Scientific Notation
Products and quotients of numbers written in scientific notation are found using the rules for exponents. Simplify: (1.7  108)(2.2  105) Solution (1.7  108)(2.2  105) = 1.7  2.2  108  105 = 3.74  108 +(5) = 3.74  103 Example Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

34 Example Simplify. (6.2  109)  (8.0  108) Solution
(6.2  109)  (8.0  108) = Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

35 4.3 Polynomials Terms Types of Polynomials Degree and Coefficients
Combining Like Terms Evaluating Polynomials and Applications Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

36 Terms A term can be a number, a variable, a product of numbers and/or variables, or a quotient of numbers and/or variables. A term that is a product of constants and/or variables is called a monomial. Examples of monomials: 8, w, x3y A polynomial is a monomial or a sum of monomials. Examples of polynomials: 5w + 8, 3x2 + x + 4, x, 0, y6 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

37 Example Identify the terms of the polynomial 7p5  3p3 + 3. Solution
The terms are 7p5, 3p3, and 3. We can see this by rewriting all subtractions as additions of opposites: 7p5  3p3 + 3 = 7p5 + (3p3) + 3 These are the terms of the polynomial. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

38 A polynomial that is composed of two terms is called a binomial, whereas those composed of three terms are called trinomials. Polynomials with four or more terms have no special name. Monomials Binomials Trinomials No Special Name 5x2 3x + 4 3x2 + 5x + 9 5x3  6x2 + 2xy  9 8 4a5 + 7bc 7x7  9z3 + 5 a4 + 2a3  a2 + 7a  2 8a23b3 10x3  7 6x2  4x  ½ 6x6  4x5 + 2x4  x3 + 3x  2 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

39 The degree of a term of a polynomial is the number of variable factors in that term.
Determine the degree of each term: a) 9x5 b) 6y c) 9 Solution a) The degree of 9x5 is 5. b) The degree of 6y is 1. c) The degree of 9 is 0. Example Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

40 The part of a term that is a constant factor is the coefficient of that term. The coefficient of 4y is 4. Identify the coefficient of each term in the polynomial. 5x4  8x2y + y  9 Solution The coefficient of 5x4 is 5. The coefficient of 8x2y is 8. The coefficient of y is 1, since y = 1y. The coefficient of 9 is simply 9. Example Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

41 The terms are 4x2, 9x3, 6x4, 8x, and 7.
The leading term of a polynomial is the term of highest degree. Its coefficient is called the leading coefficient and its degree is referred to as the degree of the polynomial. Consider this polynomial 4x2  9x3 + 6x4 + 8x  7. The terms are x2, 9x3, 6x4, 8x, and 7. The coefficients are , 9, 6, and 7. The degree of each term is 2, 3, , 1, and 0. The leading term is 6x4 and the leading coefficient is 6. The degree of the polynomial is 4. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

42 Example Combine like terms. a) 4y4  9y4
b) 7x x2 + 6x2  13  6x5 c) 9w5  7w3 + 11w5 + 2w3 Solution a) 4y4  9y4 = (4  9)y4 = 5y4 b) 7x x2 + 6x2  13  6x5 = 7x5  6x5 + 3x2 + 6x2 + 9  13 = x5 + 9x2  4 c) 9w5  7w3 + 11w5 + 2w3 = 9w5 + 11w5  7w3 + 2w3 = 20w5  5w3 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

43 Example Evaluate x3 + 4x + 7 for x = 3. Solution For x = 3, we have
= (27) + 4(3) + 7 = 27 + (12) + 7 = 22 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

44 Example In a sports league of n teams in which each team plays every other team twice, the total number of games to be played is given by the polynomial n2  n. A boys’ soccer league has 12 teams. How many games are played if each team plays every other team twice? Solution We evaluate the polynomial for n = 12: n2  n = 122  12 = 144  12 = 132. The league plays 132 games. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

45 Example Solution 0.4r2  40r + 1039 = 0.4(25)2  40(25) + 1039
The average number of accidents per day involving drivers of age r can be approximated by the polynomial 0.4r2  40r Find the average number of accidents per day involving 25-year-old drivers. Solution 0.4r2  40r = 0.4(25)2  40(25) = 0.4(625)  = 250  = 289 There are, on average, approximately 289 accidents each day involving 25-year-old drivers. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

46 Addition and Subtraction of Polynomials
4.4 Addition and Subtraction of Polynomials Addition of Polynomials Opposites of Polynomials Subtraction of Polynomials Problem Solving Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

47 Example Add. (6x3 + 7x  2) + (5x3 + 4x2 + 3) Solution
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

48 Example Add: (3  4x + 2x2) + (6 + 8x  4x2 + 2x3) Solution
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

49 Example Add: 10x5  3x3 + 7x2 + 4 and 6x4  8x2 + 7 and 4x6  6x5 + 2x2 + 6 Solution 10x  3x3 + 7x2 + 4 6x  8x2 + 7 4x6  6x x2 + 6 4x6 + 4x5 + 6x4  3x3 + x2 + 17 The answer is 4x6 + 4x5 + 6x4  3x3 + x Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

50 The Opposite of a Polynomial
To find an equivalent polynomial for the opposite, or additive inverse, of a polynomial, change the sign of every term. This is the same as multiplying the polynomial by –1. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

51 Example Simplify: (8x4  x3 + 9x2  2x + 72) Solution
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

52 Subtraction of Polynomials
We can now subtract one polynomial from another by adding the opposite of the polynomial being subtracted. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

53 Example Subtract: (10x5 + 2x3  3x2 + 5)  (3x5 + 2x4  5x3  4x2)
Solution = 10x5 + 2x3  3x x5  2x4 + 5x3 + 4x2 = 13x5  2x4 + 7x3 + x2 + 5 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

54 Example Solution (8x5 + 2x3  10x)  (4x5  5x3 + 6)
Subtract: (8x5 + 2x3  10x)  (4x5  5x3 + 6) Solution (8x5 + 2x3  10x)  (4x5  5x3 + 6) = 8x5 + 2x3  10x + (4x5) + 5x3  6 = 4x5 + 7x3  10x  6 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

55 Example Write in columns and subtract:
(6x2  4x + 7)  (10x2  6x  4) Solution 6x2  4x + 7 (10x2  6x  4) 4x2 + 2x + 11 Remember to change the signs Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

56 Example A 6-ft by 5-ft hot tub is installed on an outdoor deck measuring w ft by w ft. Find a polynomial for the remaining area of the deck. Solution 1. Familiarize. We make a drawing of the situation as follows. w ft 7 ft 5 ft Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

57 continued Example 2. Translate. Rewording: Area of Area of Area
deck  tub = left over Translating: w ft  w ft  5 ft  7 ft = Area left over 3. Carry out. w ft2  35 ft2 = Area left over. 4. Check. As a partial check, note that the units in the answer are square feet, a measure of area, as expected. 5. State. The remaining area in the yard is (x2  35)ft2. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

58 Multiplication of Polynomials
4.5 Multiplication of Polynomials Multiplying Monomials Multiplying a Monomial and a Polynomial Multiplying Any Two Polynomials Checking by Evaluating Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

59 To Multiply Monomials To find an equivalent expression for the product of two monomials, multiply the coefficients and then multiply the variables using the product rule for exponents. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

60 Example Multiply: a) (6x)(7x) b) (5a)(a) c) (8x6)(3x4) Solution
a) (6x)(7x) = (6  7) (x  x) = 42x2 b) (5a)(a) = (5a)(1a) = (5)(1)(a  a) = 5a2 c) (8x6)(3x4) = (8  3) (x6  x4) = 24x6 + 4 = 24x10 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

61 Example Solution Multiply: a) x and x + 7 b) 6x(x2  4x + 5)
a) x(x + 7) = x2 + 7x b) 6x(x2  4x + 5) = 6x3  24x2 + 30x = x  x + x  7 = (6x)(x2)  (6x)(4x) + (6x)(5) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

62 The Product of a Monomial and a Polynomial
To multiply a monomial and a polynomial, multiply each term of the polynomial by the monomial. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

63 Example Solution Multiply: 5x2(x3  4x2 + 3x  5)
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

64 Example a) x + 3 and x + 5 b) 3x  2 and x  1 Solution
Multiply each of the following. a) x + 3 and x + 5 b) 3x  2 and x  1 Solution a) (x + 3)(x + 5) = (x + 3)x + (x + 5)3 = x(x + 3) + 5(x + 5) = x  x + x   x + 5  3 = x2 + 3x + 5x + 15 = x2 + 8x + 15 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

65 continued Solution b) (3x  2)(x  1) = (3x  2)x  (3x  2)1
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

66 The Product of Two Polynomials
To multiply two polynomials P and Q, select one of the polynomials, say P. Then multiply each term of P by every term of Q and combine like terms. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

67 Example Multiply: (5x3 + x2 + 4x)(x2 + 3x) Solution 5x3 + x2 + 4x
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

68 Example Multiply: (3x2  4)(2x2  3x + 1) Solution 2x2  3x + 1
6x4 + 9x3  11x2 + 12x  4 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

69 4.6 Special Products Products of Two Binomials
Multiplying Sums and Differences of Two Terms Squaring Binomials Multiplications of Various Types Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

70 The FOIL Method To multiply two binomials, A + B and C + D, multiply the First terms AC, the Outer terms AD, the Inner terms BC, and then the Last terms BD. Then combine like terms, if possible. (A + B)(C + D) = AC + AD + BC + BD Multiply First terms: AC. Multiply Outer terms: AD. Multiply Inner terms: BC Multiply Last terms: BD FOIL (A + B)(C + D) O I F L Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

71 Example Multiply: (x + 4)(x2 + 3) Solution F O I L
(x + 4)(x2 + 3) = x3 + 3x + 4x2 + 12 = x3 + 4x2 + 3x + 12 O I F L The terms are rearranged in descending order for the final answer. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

72 Example Multiply. a) (x + 8)(x + 5) b) (y + 4) (y  3)
c) (5t3 + 4t)(2t2  1) d) (4  3x)(8  5x3) Solution a) (x + 8)(x + 5) = x2 + 5x + 8x + 40 = x2 + 13x + 40 b) (y + 4) (y  3) = y2  3y + 4y  12 = y2 + y  12 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

73 continued Example Solution
c) (5t3 + 4t)(2t2  1) = 10t5  5t3 + 8t3  4t = 10t5 + 3t3  4t d) (4  3x)(8  5x3) = 32  20x3  24x + 15x4 = 32  24x  20x3 + 15x4 In general, if the original binomials are written in ascending order, the answer is also written that way. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

74 The Product of a Sum and Difference
The product of the sum and difference of the same two terms is the square of the first term minus the square of the second term. (A + B)(A – B) = A2 – B2. This is called a difference of squares. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

75 Example Multiply. a) (x + 8)(x  8) b) (6 + 5w) (6  5w)
c) (4t3  3)(4t3 + 3) Solution (A + B)(A  B) = A2  B2 a) (x + 8)(x  8) = x2  82 = x2  64 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

76 continued Example Solution b) (6 + 5w) (6  5w) = 62  (5w)2
c) (4t3  3)(4t3 + 3) = (4t3)2  32 = 16t6  9 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

77 The Square of a Binomial
The square of a binomial is the square of the first term, plus twice the product of the two terms, plus the square of the last term. (A + B)2 = A2 + 2AB + B2; (A – B)2 = A2 – 2AB + B2; These are called perfect-square trinomials.* *In some books, these are called trinomial squares. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

78 Example Multiply. a) (x + 8)2 b) (y  7)2 c) (4x  3x5)2 Solution
(A + B)2 = A2+2AB + B2 a) (x + 8)2 = x2 + 2x8 + 82 = x2 + 16x + 64 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

79 continued (A + B)2 = A2 2AB + B2 Example
Solution b) (y  7)2 = y2  2  y  = y2  14y + 49 c) (4x  3x5)2 = (4x)2  2  4x  3x5 + (3x5)2 = 16x2  24x6 + 9x10 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

80 Multiplying Two Polynomials
1. Is the multiplication the product of a monomial and a polynomial? If so, multiply each term of the polynomial by the monomial. 2. Is the multiplication the product of two binomials? If so: a) Is the product of the sum and difference of the same two terms? If so, use the pattern (A + B)(A  B) = (A  B)2. b) Is the product the square of a binomial? If so, us the pattern (A + B)2 = A2 + 2AB + B2, or (A – B)2 = A2 – 2AB + B2. c) If neither (a) nor (b) applies, use FOIL. 3. Is the multiplication the product of two polynomials other than those above? If so, multiply each term of one by every term of the other. Use columns if you wish. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

81 Example Multiply. a) (x + 5)(x  5) b) (w  7)(w + 4)
c) (x + 9)(x + 9) d) 3x2(4x2 + x  2) e) (p + 2)(p2 + 3p  2) f) (2x + 1)2 Solution a) (x + 5)(x  5) = x2  25 b) (w  7)(w + 4) = w2 + 4w  7w  28 = w2  3w  28 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

82 Example Multiply continued c) (x + 9)(x + 9) = x2 + 18x + 81
d) 3x2(4x2 + x  2) = 12x4 + 3x3  6x2 e) p2 + 3p  2 p + 2 2p2 + 6p  4 p3 + 3p2  2p p3 + 5p2 + 4p  4 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

83 Example Multiply continued f) (2x + 1)2 = 4x2 + 2(2x)(1) + 1
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

84 Polynomials in Several Variables
4.7 Polynomials in Several Variables Evaluating Polynomials Like Terms and Degree Addition and Subtraction Multiplication Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

85 Example Evaluate the polynomial x + xy2 + 9x3y for x = 3 and y = 4. Solution We substitute 3 for x and 4 for y: 5 + 4x + xy2 + 9x3y2 = 5 + 4(3) + (3)(42) + 9(3)3(4)2 = 5  12  48  3888 = 3943 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

86 Example The surface area of a right circular cylinder is given by the polynomial 2rh + 2r2 where h is the height and r is the radius of the base. A barn silo has a height of 50 feet and a radius of 9 feet. Approximate its surface area. Solution We evaluate the polynomial for h = 50 ft and r = 9 ft. If 3.14 is used to approximate , we have h r Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

87 continued h = 50 ft and r = 9 ft
2rh + 2r2  2(3.14)(9 ft)(50 ft) + 2(3.14)(9 ft)2  2(3.14)(9 ft)(50 ft) + 2(3.14)(81 ft2)  2826 ft ft2  ft2 Note that the unit in the answer (square feet) is a unit of area. The surface area is about ft2 (square feet). Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

88 Recall that the degree of a monomial is the number of variable factors in the term.
Identify the coefficient and the degree of each term and the degree of the polynomial 10x3y2  15xy3z4 + yz + 5y + 3x2 + 9. Example Term Coefficient Degree Degree of the Polynomial 10x3y2 10 5 8 15xy3z4 15 yz 1 2 5y 3x2 3 9 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

89 Like Terms Like, or similar terms either have exactly the same variables with exactly the same exponents or are constants. For example, 9w5y4 and 15w5y4 are like terms and 12 and 14 are like terms, but 6x2y and 9xy3 are not like terms. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

90 Combine like terms. Example a) 10x2y + 4xy3  6x2y  2xy3
b) 8st  6st2 + 4st2 + 7s3 + 10st  12s3 + t  2 Solution = (10 6)x2y + (42)xy3 = 4x2y + 2xy3 = 5s3  2st2 + 18st + t  2 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

91 Addition and Subtraction
Example Add: (6x3 + 4y  6y2) + (7x3 + 5x2 + 8y2) Solution (6x3 + 4y  6y2) + (7x3 + 5x2 + 8y2) = (6 + 7)x3 + 5x2 + 4y + (6 + 8)y2 = x3 + 5x2 + 4y + 2y2 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

92 Example Subtract: (5x2y + 2x3y2 + 4x2y3 + 7y)  (5x2y  7x3y2 + x2y2  6y) Solution = (5x2y + 2x3y2 + 4x2y3 + 7y)  5x2y + 7x3y2  x2y2 + 6y = 9x3y2 + 4x2y3  x2y2 + 13y Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

93 Multiplication Example Multiply: (4x2y  3xy + 4y)(xy + 3y) Solution
4x3y2  3x2y2 + 4xy2 4x3y2 + 9x2y2  5xy2 + 12y2 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

94 The special products discussed in Section 4.5 can speed up your work.
Example Multiply. a) (x + 6y)(2x  3y) b) (5x + 7y)2 c) (a4  5a2b2) d) (7a2b + 3b)(7a2b  3b) e) (3x3y2 + 7t)(3x3y2 + 7t) f) (3x + 1  4y)(3x y) Solution a) (x + 6y)(2x  3y) = 2x2  3xy + 12xy  18y2 = 2x2 + 9xy  18y FOIL Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

95 Solution continued b) (5x + 7y)2 = (5x)2 + 2(5x)(7y) + (7y)2
c) (a4  5a2b2)2 = (a4)2  2(a4)(5a2b2) + (5a2b2)2 = a8  10a6b2 + 25a4b4 d) (7a2b + 3b)(7a2b  3b) = (7a2b)2  (3b)2 = 49a4b2  9b2 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

96 Solution continued e) (3x3y2 + 7t)(3x3y2 + 7t)
= (7t  3x3y2)(7t + 3x3y2) = (7t)2  (3x3y2)2 = 49t2  9x6y4 f) (3x + 1  4y)(3x y) = (3x + 1)2  (4y)2 = 9x2 + 6x + 1  16y2 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

97 Division of Polynomials
4.8 Division of Polynomials Dividing by a Monomial Dividing by a Binomial Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

98 Dividing by a Monomial Example
To divide a polynomial by a monomial, we divide each term by the monomial. Divide. x5 + 24x4  12x3 by 6x Solution Example Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

99 Example Divide. Solution
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

100 Dividing by a Binomial For divisors with more than one term, we use long division, much as we do in arithmetic. Polynomial are written in descending order and any missing terms in the dividend are written in, using 0 for the coefficients. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

101 Example Divide x2 + 7x + 12 by x + 3. Solution
Now we “bring down” the next term. Multiply x + 3 by x, using the distributive law Subtract by changing signs and adding Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

102 Solution continued Multiply 4 by the divisor, x + 3, using the distributive law Subtract Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

103 Example Divide 15x2  22x + 14 by 3x  2. Solution
The answer is 5x  4 with R6. Another way to write the answer is Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

104 Example Divide x5  3x4  4x2 + 10x by x  3 . Solution The answer is
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley


Download ppt "Copyright © 2010 Pearson Education, Inc"

Similar presentations


Ads by Google