1. Copyright © 2010 Pearson Education, Inc
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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
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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
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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.)
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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
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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
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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.)
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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)
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9. Example continued c) d) Solution c) d)
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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.)
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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
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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.)
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13. Example Simplify: a) (x3)4 b) (42)6 Solution a) (x3)4 = x34 = x12
= 416
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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.)
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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
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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.)
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17. Example Simplify: a) b) c) Solution a) b)
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18. continued Example Simplify: c) Solution c)
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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:
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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
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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.)
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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 =
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23. continued Example c) (4)2 = d) xy1 =
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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
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25. continued Example c) (3a2b4)3 = 33(a2)3(b4) = 27 a6b12 = d) e) f)
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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.)
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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.
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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.)
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29. Example Simplify: Solution
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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.
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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 =
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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
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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
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34. Example Simplify. (6.2  109)  (8.0  108) Solution
(6.2  109)  (8.0  108) =
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35. 4.3 Polynomials Terms Types of Polynomials Degree and Coefficients
Combining Like Terms
Evaluating Polynomials and Applications
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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
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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.
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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
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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
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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
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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.
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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
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43. Example Evaluate x3 + 4x + 7 for x = 3. Solution For x = 3, we have
= (27) + 4(3) + 7
= 27 + (12) + 7
= 22
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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.
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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.
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46. Addition and Subtraction of Polynomials
4.4
Addition and Subtraction of Polynomials
Addition of Polynomials
Opposites of Polynomials
Subtraction of Polynomials
Problem Solving
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47. Example Add. (6x3 + 7x  2) + (5x3 + 4x2 + 3) Solution
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48. Example Add: (3  4x + 2x2) + (6 + 8x  4x2 + 2x3) Solution
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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
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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.
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51. Example Simplify: (8x4  x3 + 9x2  2x + 72) Solution
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52. Subtraction of Polynomials
We can now subtract one polynomial from another by adding the opposite of the polynomial being subtracted.
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53. Example Subtract: (10x5 + 2x3  3x2 + 5)  (3x5 + 2x4  5x3  4x2)
Solution
= 10x5 + 2x3  3x x5  2x4 + 5x3 + 4x2
= 13x5  2x4 + 7x3 + x2 + 5
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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
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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
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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
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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.
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58. Multiplication of Polynomials
4.5
Multiplication of Polynomials
Multiplying Monomials
Multiplying a Monomial and a Polynomial
Multiplying Any Two Polynomials
Checking by Evaluating
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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.
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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
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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)
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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.
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63. Example Solution Multiply: 5x2(x3  4x2 + 3x  5)
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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
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65. continued Solution b) (3x  2)(x  1) = (3x  2)x  (3x  2)1
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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.
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67. Example Multiply: (5x3 + x2 + 4x)(x2 + 3x) Solution 5x3 + x2 + 4x
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68. Example Multiply: (3x2  4)(2x2  3x + 1) Solution 2x2  3x + 1
6x4 + 9x3  11x2 + 12x  4
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69. 4.6 Special Products Products of Two Binomials
Multiplying Sums and Differences of Two Terms
Squaring Binomials
Multiplications of Various Types
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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
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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.
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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
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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.
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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.
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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
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76. continued Example Solution b) (6 + 5w) (6  5w) = 62  (5w)2
c) (4t3  3)(4t3 + 3) = (4t3)2  32
= 16t6  9
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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.
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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
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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
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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.
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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
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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
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83. Example Multiply continued f) (2x + 1)2 = 4x2 + 2(2x)(1) + 1
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84. Polynomials in Several Variables
4.7
Polynomials in Several Variables
Evaluating Polynomials
Like Terms and Degree
Addition and Subtraction
Multiplication
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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
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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
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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).
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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
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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.
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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
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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
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92. Example
Subtract:
(5x2y + 2x3y2 + 4x2y3 + 7y)  (5x2y  7x3y2 + x2y2  6y)
Solution
= (5x2y + 2x3y2 + 4x2y3 + 7y)  5x2y + 7x3y2  x2y2 + 6y
= 9x3y2 + 4x2y3  x2y2 + 13y
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93. Multiplication Example Multiply: (4x2y  3xy + 4y)(xy + 3y) Solution
4x3y2  3x2y2 + 4xy2
4x3y2 + 9x2y2  5xy2 + 12y2
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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
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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
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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
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97. Division of Polynomials
4.8
Division of Polynomials
Dividing by a Monomial
Dividing by a Binomial
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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
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99. Example Divide. Solution
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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.
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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
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102. Solution continued
Multiply 4 by the divisor, x + 3, using the distributive law
Subtract
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103. Example Divide 15x2  22x + 14 by 3x  2. Solution
The answer is 5x  4 with R6. Another way to write the answer is
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104. Example Divide x5  3x4  4x2 + 10x by x  3 . Solution The answer is
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