Transcript bei05_ppt_04

Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 1- 1
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
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
The Product Rule
For any number a and any positive
integers m and n,
a m  a n  a m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
Slide 4- 4
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 = x3+5
= x8
Adding exponents
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 5
Example continued
b) 62  67  63
d) (w3z4)(w3z7)
Solution
b) 62  67  63 = 62+7+3
= 612
c) (x + y)6(x + y)9
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
Slide 4- 6
The Quotient Rule
For any nonzero number a and any
positive integers m and n for which m > n,
m
a
mn
a .
n
a
(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
Slide 4- 7
Example
Divide and simplify each of the following. (Here “simplify”
means express the product as one base to a power whenever
possible.)
14
9
7
7 9
(6
y
)
x
8
6
r
t
a)
b)
c)
d)
x3
83
(6 y ) 6
4r 3 t
Solution
9
x
a)
 x 9 3
x3
 x6
b) 87
7 3

8
83
 84
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 8
Example continued
c)
(6 y )14
(6 y ) 6
d)
6r 7 t 9
4r 3 t
Solution
(6 y)14
14 6
8

(6
y
)

(6
y
)
c)
(6 y)6
6r 7 t 9 6 r 7 t 9
d)
  3
3
4r t 4 r t
6 7 3 91 3 4 8
  r t  r t
4
2
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 9
The Exponent Zero
For any real number a, with a ≠ 0,
a  1.
0
(Any nonzero number raised to the 0
power is 1.)
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 10
Example
Simplify: a) 12450
d) (1)80
e) 80.
b) (3)0
c) (4w)0
Solution
a) 12450 = 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
Slide 4- 11
The Power Rule
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
Slide 4- 12
Example
Simplify: a)(x3)4
b) (42)6
Solution
a) (x3)4 = x34
= x12
b) (42)8 = 428
= 416
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 13
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
Slide 4- 14
Example
Simplify: a)(3x)4
Solution
a) (3x)4 = 34x4 = 81x4
b) (2x3)2
c) (a2b3)7(a4b5)
b) (2x3)2 = (2)2(x3)2
= 4x6
c) (a2b3)7(a4b5) = (a2)7(b3)7a4b5
= a14b21a4b5
= a18b26
Multiplying exponents
Adding exponents
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 15
Raising a Quotient to a Power
For any real numbers a and b, b ≠ 0,
and any whole number n,
n
n
a
a
 
   n.
b
b
(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
Slide 4- 16
Example
w
Simplify: a)  
4
3
 3 
b)  5 
b 
4
 2a 
c)  4 
 b 
5
2
Solution
3
3
3
w
w
w


a)
   3 
4
4
64
4
3
34

b)  5   5 4
(b )
b 
81
81
 54  20
b
b
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 17
Example continued
Simplify: c)  2a4 
 b 
5
2
Solution
5 2
5 2


2
a
(2
a
)
c) 


4
b


(b 4 ) 2
22 (a5 )2 4a10

 8
4 2
b
b
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 18
Definitions and Properties of Exponents
For any whole numbers m and n,
1 as an exponent:
0 as an exponent:
The Product Rule:
The Quotient Rule:
The Power Rule:
Raising a product to a power:
Raising a quotient to a power:
a1 = a
a0 = 1
a m  a n  a mn
am
mn

a
an
(am)n = amn
(ab)n = anbn
n
an
a
   n
b
b
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 19
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
Negative Exponents
For any real number a that is nonzero and any
integer n,
a
n
1
 n.
a
(The numbers a-n and an are reciprocals of each
other.)
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 21
Example
Express using positive exponents, and, if possible,
simplify.
a) m5
b) 52
c) (4)2
d) xy1
Solution
1
5
a) m = 5
m
b)
52
1
1
= 2 
25
5
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 22
Example continued
c)
(4)2 =
d)
xy1
1
1
1


2
(4)(4) 16
(4)
 1
1 x
= x 1   x  
 y y
y 
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 23
Example
Simplify. Do not use negative exponents in the answer.
a) w5  w3
b) (x4)3
c) (3a2b4)3
a 5
d)
6
a
1
e)
9
b
7
f)
w
z 6
Solution
a) w5  w3  w5 ( 3)  w2
b) (x4)3 = x(4)(3) = x12
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 24
Example continued
c) (3a2b4)3 = 33(a2)3(b4)
6
27a
= 27 a6b12 =
12
b
5
a
d) 6  a 5( 6)  a1  a
a
1
 ( 9 )
9
e)

b

b
9
b
6
w7
1
1
z
f) 6  w7  6  7  z 6  7
z
z
w
w
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 25
Factors and Negative Exponents
For any nonzero real numbers a and b and
any integers m and n,
a  n bm
 n.
m
b
a
(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
Slide 4- 26
Example
6

20
x
Simplify.
4 y 3 z 4
Solution
We can move the negative factors to the other side of
the fraction bar if we change the sign of each exponent.
6
20 x
5 z
 3 6
3 4
4y z
y z
4
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 27
Reciprocals and Negative Exponents
For any nonzero real numbers a and b and any
integer n,
a
 
b
n
n
b
  .
a
(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
Slide 4- 28
Example
Simplify:  a 
 
 3b 
4
Solution
2
2
a 
 3b 
   4 
a 
 3b 
4
2
(3b) 2
 4 2
(a )
2
2
2
3 b
9b
 8  8
a
a
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 29
Scientific Notation
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
Slide 4- 30
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.
3.842000. 3.842  106 = 3,842,000
b) Since the exponent is negative, the decimal
point moves left 7 places.
0.0000005.3 5.3  107 = 0.00000053
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 31
Example
Write in scientific notation:
a) 94,000
b) 0.0423
Solution
a) We need to find m such that 94,000
= 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 0.0423 we move the
decimal point 2 places to the left.
0.0423 = 4.23  102
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 32
Multiplying and Dividing Using Scientific
Notation
Products and quotients of numbers written in scientific
notation are found using the rules for exponents.
8)(2.2  105)
Simplify:
(1.7

10
Example
Solution (1.7  108)(2.2  105)
= 1.7  2.2  108  105
= 3.74  108 +(5)
= 3.74  103
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 33
Example
Simplify. (6.2  109)  (8.0  108)
Solution
9
9
6.2

10
6.2
10

 8
(6.2  109)  (8.0  108) =
8
8.0  10
8.0
10
 0.775  10 17
 7.75 101 1017
 7.75  1018
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 34
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
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, 24 x3y
A polynomial is a monomial or a sum of monomials.
Examples of polynomials:
5w + 8, 3x2 + x + 4, x, 0, 75y6
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 36
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
Slide 4- 37
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
Slide 4- 38
The degree of a term of a polynomial is the number
of variable factors in that term.
Example
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.
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 39
The part of a term that is a constant factor is the
coefficient of that term. The coefficient of 4y is 4.
Example
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.
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 40
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
4x2, 9x3, 6x4, 8x, and 7.
The coefficients are
4,
9, 6, 8 and 7.
The degree of each term is 2, 3, 4, 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
Slide 4- 41
Example Combine like terms.
a) 4y4  9y4
b) 7x5 + 9 + 3x2 + 6x2  13  6x5
c) 9w5  7w3 + 11w5 + 2w3
Solution
a) 4y4  9y4 = (4  9)y4 = 5y4
b) 7x5 + 9 + 3x2 + 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
Slide 4- 42
Example
Evaluate x3 + 4x + 7 for x = 3.
Solution For x = 3, we have
x3 + 4x + 7 = (3)3 + 4(3) + 7
= (27) + 4(3) + 7
= 27 + (12) + 7
= 22
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 43
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
Slide 4- 44
Example
The average number of accidents per day involving drivers of
age r can be approximated by the polynomial 0.4r2  40r + 1039.
Find the average number of accidents per day involving 25-yearold drivers.
Solution 0.4r2  40r + 1039 = 0.4(25)2  40(25) + 1039
= 0.4(625)  1000 + 1039
= 250  1000 + 1039
= 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
Slide 4- 45
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
Example
Add. (6x3 + 7x  2) + (5x3 + 4x2 + 3)
Solution
(6x3 + 7x  2) + (5x3 + 4x2 + 3)
= (6 + 5)x3 + 4x2 + 7x + (2 + 3)
= x3 + 4x2 + 7x + 1
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 47
Example
Add: (3  4x + 2x2) + (6 + 8x  4x2 + 2x3)
Solution
(3  4x + 2x2) + (6 + 8x  4x2 + 2x3)
= (3  6) + (4 + 8)x + (2  4)x2 + 2x3
= 3 + 4x  2x2 + 2x3
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 48
Example
Add: 10x5  3x3 + 7x2 + 4 and 6x4  8x2 + 7 and
4x6  6x5 + 2x2 + 6
Solution
10x5  3x3 + 7x2 + 4
6x4
 8x2 + 7
4x6  6x5
+ 2x2 + 6
4x6 + 4x5 + 6x4  3x3 + x2 + 17
The answer is 4x6 + 4x5 + 6x4  3x3 + x2 + 17.
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 49
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
Slide 4- 50
Example
Simplify: (8x4  34 x3 + 9x2  2x + 72)
Solution
(8x4  34 x3 + 9x2  2x + 72)
=
8x4
+
3 3
4x
 9x2 + 2x  72
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 51
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
Slide 4- 52
Example Subtract:
(10x5 + 2x3  3x2 + 5)  (3x5 + 2x4  5x3  4x2)
Solution
(10x5 + 2x3  3x2 + 5)  (3x5 + 2x4  5x3  4x2)
= 10x5 + 2x3  3x2 + 5 + 3x5  2x4 + 5x3 + 4x2
= 13x5  2x4 + 7x3 + x2 + 5
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 53
Example
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
Slide 4- 54
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
Slide 4- 55
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
5 ft
7 ft
w ft
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 56
Example continued
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
Slide 4- 57
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
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
Slide 4- 59
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
Slide 4- 60
Example
Multiply: a) x and x + 7
b) 6x(x2  4x + 5)
Solution
a) x(x + 7) = x  x + x  7
= x2 + 7x
b) 6x(x2  4x + 5) = (6x)(x2)  (6x)(4x) + (6x)(5)
= 6x3  24x2 + 30x
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 61
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
Slide 4- 62
Example
Multiply: 5x2(x3  4x2 + 3x  5)
Solution
5x2(x3  4x2 + 3x  5) = 5x5  20x4 + 15x3  25x2
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 63
Example
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3+5x+53
= x2 + 3x + 5x + 15
= x2 + 8x + 15
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 64
continued
Solution
b) (3x  2)(x  1) = (3x  2)x  (3x  2)1
= x(3x  2)  1(3x  2)
= x  3x  x  2  1  3x  1(2)
= 3x2  2x  3x + 2
= 3x2  5x + 2
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 65
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
Slide 4- 66
Example
Multiply: (5x3 + x2 + 4x)(x2 + 3x)
Solution
5x3 + x2 + 4x
x2 + 3x
15x4 + 3x3 + 12x2
5x5 + x4 + 4x3
5x5 + 16x4 + 7x3 + 12x2
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 67
Example
Multiply: (3x2  4)(2x2  3x + 1)
Solution
2x2  3x + 1
3x2
4
 8x2 + 12x  4
6x4 + 9x3  3x2
6x4 + 9x3  11x2 + 12x  4
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 68
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
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
L
F
(A + B)(C + D)
I
O
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 70
Example
Multiply: (x + 4)(x2 + 3)
Solution
F
O I
L
(x + 4)(x2 + 3) = x3 + 3x + 4x2 + 12
I
3 + 4x2 + 3x + 12
=
x
O
F
L
The terms are rearranged in descending
order for the final answer.
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 71
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
Slide 4- 72
Example continued
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
Slide 4- 73
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
Slide 4- 74
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
Slide 4- 75
Example continued
Solution
b) (6 + 5w) (6  5w) = 62  (5w)2
= 36  25w2
c) (4t3  3)(4t3 + 3) = (4t3)2  32
= 16t6  9
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 76
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; These are called perfect-square
(A – B)2 = A2 – 2AB + B2; trinomials.*
*In some books, these are called trinomial squares.
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 77
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
Slide 4- 78
Example continued (A + B)2 = A2 2AB + B2
Solution
b) (y  7)2 = y2  2  y  7 + 72
= 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
Slide 4- 79
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
Slide 4- 80
Example Multiply.
a) (x + 5)(x  5)
c) (x + 9)(x + 9)
e) (p + 2)(p2 + 3p  2)
b) (w  7)(w + 4)
d) 3x2(4x2 + x  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
Slide 4- 81
Example Multiply continued
c) (x + 9)(x + 9) = x2 + 18x + 81
d) 3x2(4x2 + x  2) = 12x4 + 3x3  6x2
p2 + 3p  2
p+2
2p2 + 6p  4
p3 + 3p2  2p
p3 + 5p2 + 4p  4
e)
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 82
Example Multiply continued
f) (2x + 1)2 = 4x2 + 2(2x)(1) + 1
= 4x2 + 4x + 1
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 83
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
Example
Evaluate the polynomial 5 + 4x + xy2 + 9x3y2
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
Slide 4- 85
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.
r
Approximate its surface area.
Solution
We evaluate the polynomial for
h
h = 50 ft and r = 9 ft. If 3.14 is
used to approximate , we have
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 86
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 ft2 + 508.68 ft2
 3334.68 ft2
Note that the unit in the answer (square feet) is a
unit of area. The surface area is about 3334.7 ft2
(square feet).
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 87
Recall that the degree of a monomial is the number of
variable factors in the term.
Example Identify the coefficient and the degree of
each term and the degree of the polynomial
10x3y2  15xy3z4 + yz + 5y + 3x2 + 9.
Term
Coefficient
Degree
10x3y2
10
15xy3z4
yz
5y
15
1
5
5
8
2
1
3
9
2
0
3x2
9
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Degree of the
Polynomial
8
Slide 4- 88
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
Slide 4- 89
Example Combine like terms.
a) 10x2y + 4xy3  6x2y  2xy3
b) 8st  6st2 + 4st2 + 7s3 + 10st  12s3 + t  2
Solution
a) 10x2y + 4xy3  6x2y  2xy3
= (10 6)x2y + (42)xy3
= 4x2y + 2xy3
b) 8st  6st2 + 4st2 + 7s3 + 10st  12s3 + t  2
= 5s3  2st2 + 18st + t  2
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 90
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
Slide 4- 91
Example
Subtract:
(5x2y + 2x3y2 + 4x2y3 + 7y)  (5x2y  7x3y2 + x2y2  6y)
Solution
(5x2y + 2x3y2 + 4x2y3 + 7y)  (5x2y  7x3y2 + x2y2  6y)
= (5x2y + 2x3y2 + 4x2y3 + 7y)  5x2y + 7x3y2  x2y2 + 6y
= 9x3y2 + 4x2y3  x2y2 + 13y
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 92
Multiplication
Example
Multiply: (4x2y  3xy + 4y)(xy + 3y)
Solution
4x2y  3xy + 4y
xy + 3y
12x2y2  9xy2 + 12y2
4x3y2  3x2y2 + 4xy2
4x3y2 + 9x2y2  5xy2 + 12y2
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 93
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)2
d) (7a2b + 3b)(7a2b  3b)
e) (3x3y2 + 7t)(3x3y2 + 7t)
f) (3x + 1  4y)(3x + 1 + 4y)
Solution
a) (x + 6y)(2x  3y) = 2x2  3xy + 12xy  18y2
= 2x2 + 9xy  18y2
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
FOIL
Slide 4- 94
Solution continued
b) (5x + 7y)2 = (5x)2 + 2(5x)(7y) + (7y)2
= 25x2 + 70xy + 49y2
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
Slide 4- 95
Solution continued
e) (3x3y2 + 7t)(3x3y2 + 7t)
= (7t  3x3y2)(7t + 3x3y2)
= (7t)2  (3x3y2)2
= 49t2  9x6y4
f) (3x + 1  4y)(3x + 1 + 4y)
= (3x + 1)2  (4y)2
= 9x2 + 6x + 1  16y2
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 96
4.8
Division of Polynomials

Dividing by a Monomial

Dividing by a Binomial
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Dividing by a Monomial
To divide a polynomial by a monomial, we divide each
term by the monomial.
5 + 24x4  12x3 by 6x
Divide.
x
Example
5
4
3
5
4
3
x

24
x

12
x
x
24
x
12
x
Solution



6x
6x
6x
6x
1 51 24 41 12 31
 x 
x  x
6
6
6
1 4
 x  4 x3  2 x 2
6
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 98
Example
Divide. 21a b  14a b  7a b  7a b
5
4
3 2
2
2
Solution
21a5b4  14a3b2  7a 2b 21a5b4 14a3b 2 7a 2b



2
2
2
7a b
7a b 7a b 7a 2b
21 5 2 41  14  32 2 1  7 
  a b  a b  
7
7
 7 
 3a b  2ab  1
2 3
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 99
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
Slide 4- 100
Example
Divide x2 + 7x + 12 by x + 3.
Solution
x
2
x  3 x  7 x  12
( x 2  3 x )
4x
Multiply x + 3 by x, using the distributive
law
Subtract by changing signs and adding
Now we “bring down” the next term.
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 101
Solution continued
x+4
x  3 x 2  7 x  12
( x 2  3 x )
4 x  12
(4 x  12)
0
Multiply 4 by the divisor, x + 3, using the
distributive law
Subtract
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 102
Example
Divide 15x2  22x + 14 by 3x  2.
Solution
5x  4
3 x  2 15x 2  22 x  14
(15 x 2  10 x)
 12 x  14
( 12 x  8)
6
The answer is 5x  4 with R6. Another way to write the answer is
6
5x  4 
.
3x  2
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 103
Example
Divide x5  3x4  4x2 + 10x by x  3 .
Solution
x4
 4x  2
x 3
x5  3x 4  0 x3  4 x 2  10 x  0
( x 5  3 x 4 )
0 x3  4 x 2  10 x
  4 x 2  12 x 
 2x  0
( 2 x  6)
6
The answer is x  4 x  2 
.
x3
4
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
6
Slide 4- 104