Download Chapter 3: Exponents and Polynomials

Chapter 12
Exponents and
Polynomials
Chapter Sections
12.1 – Exponents
12.2 – Negative Exponents and Scientific Notation
12.3 – Introduction to Polynomials
12.4 – Adding and Subtracting Polynomials
12.5 – Multiplying Polynomials
12.6 – Special Products
12.7 – Dividing Polynomials
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§ 12.1
Exponents
Exponents
Exponents that are natural numbers are
shorthand notation for repeating factors.
34 = 3 • 3 • 3 • 3
3 is the base
4 is the exponent (also called power)
Note by the order of operations that exponents
are calculated before other operations.
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Evaluating Exponential Expressions
Example
Evaluate each of the following expressions.
34 = 3 • 3 • 3 • 3 = 81
(–5)2 = (– 5)(–5) = 25
–62 = – (6)(6) = –36
(2 • 4)3 = (2 • 4)(2 • 4)(2 • 4) = 8 • 8 • 8 = 512
3 • 42 = 3 • 4 • 4 = 48
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Evaluating Exponential Expressions
Example
Evaluate each of the following expressions.
a.) Find 3x2 when x = 5.
3x2 = 3(5)2 = 3(5 · 5) = 3 · 25 = 75
b.) Find –2x2 when x = –1.
–2x2 = –2(–1)2 = –2(–1)(–1) = –2(1) = –2
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The Product Rule
Product Rule (applies to common bases only)
am · an = am+n
Example
Simplify each of the following expressions.
32 · 34 = 32+4 = 36 = 3 · 3 · 3 · 3 · 3 · 3= 729
x4 · x5 = x4+5 = x9
z3 · z2 · z5= z3+2+5 = z10
(3y2)(– 4y4) = 3 · y2 (– 4) · y4 = 3(– 4)(y2 · y4) = – 12y6
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The Power Rule
Power Rule
(am)n = amn
Example
Simplify each of the following expressions.
(23)3 = 23·3 = 29 = 512
(x4)2 = x4·2 = x8
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The Power of a Product Rule
Power of a Product Rule
(ab)n = an · bn
Example
Simplify (5x2y)3.
(5x2y)3 = 53 · (x2)3 · y3 = 125x6 y3
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The Power of a Quotient Rule
Power of a Quotient Rule

a
b
Example
Simplify
n
an
 n
b

a
b
n
an
 n
b
(5x2y)3 = 53 · (x2)3 · y3 = 125x6 y3
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The Power of a Quotient Rule
Power of a Quotient Rule

a
b
n
an
 n
b
Example
Simplify the following expression.
4
 
 
 p 
p
 3  
3
3
r
3r


2
2 4
4

p

3 r 
2 4
4
3 4
(Power of
product
rule)
p8
 12
81r
(Power
rule)
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The Quotient Rule
Quotient Rule (applies to common bases only)
am
mn

a
an
a0
Example
Simplify the following expression.
9a 4b 7   9  a 4  b 7 
41
72
3 5



3
(
a
)(
b
)

3
a
b
2




2
 3  a  b 
3ab
Group common
bases together
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Zero Exponent
Zero exponent
a0 = 1, a  0
Note: 00 is undefined.
Example
Simplify each of the following expressions.
50 = 1
(xyz3)0 = x0 · y0 · (z3)0 = 1 · 1 · 1 = 1
–x0 = –(x0) = – 1
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§ 12.2
Negative Exponents and
Scientific Notation
Negative Exponents
Using the quotient rule from section 3.1,
4
x
46
2
x x
6
x
x0
But what does x -2 mean?
x
x x x x
1
1


 2
6
x
x x x x x x x x x
4
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Negative Exponents
So, in order to extend the quotient rule to
cases where the difference of the exponents
would give us a negative number we define
negative exponents as follows.
If a  0, and n is an integer, then
a
n
1
 n
a
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Simplifying Expressions
Example
Simplify by writing each of the following expressions with
positive exponents or calculating.
1 1
3  2 
3
9
1
7
x  7
x
2
2
2x  4
x
4
Remember that without parentheses, x
is the base for the exponent –4, not 2x
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Simplifying Expressions
Example
Simplify by writing each of the following expressions with
positive exponents or calculating.
x
3
1
 3
x
1
1
3   2  
3
9
2
(3) 2 
1
1

2
(3)
9
Notice the difference in results when the
parentheses are included around 3.
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Simplifying Expressions
Example
Simplify by writing each of the following expressions
with positive exponents.
1)
2)
3
1
x
1
3



x
3
1
x
1
3
x
1
(Note that to convert a power with a negative
2
4
2
x
y exponent to one with a positive exponent, you
x

 2 simply switch the power from a numerator to a
4
1
y
x denominator, or vice versa, and switch the
exponent to its positive value.)
4
y
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Summary of Exponent Rules
If m and n are integers and a and b are real numbers, then:
Product Rule for exponents am · an = am+n
Power Rule for exponents (am)n = amn
Power of a Product (ab)n = an · bn
n
an
a
Power of a Quotient    n , b  0
b
b
Quotient Rule for exponents
am
mn

a
, a0
n
a
Zero exponent a0 = 1, a  0
Negative exponent
a n 
1
, a0
n
a
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Simplifying Expressions
Simplify by writing the following expression with
positive exponents or calculating.
 3 ab 
  4 7 3 
3 a b 
2
3
2

3
3
2
4
3
ab
7
ab


3  a  b 

 3  a  b 
2
3  2
Power of a quotient rule
 2 2
4 2
3 2
7 2
2
3  2
Power of a product rule
8
8
34 a 6b 2
34 a14b 2
a
a
 8 14 6  8 6 6  348 a146b 26  34 a 8b 4  4 4 
3b
3a b
3ab
81b 4
Power rule for exponents
Quotient rule for exponents
Negative exponents
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Negative exponents
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Scientific Notation
In many fields of science we encounter very large or
very small numbers. Scientific notation is a
convenient shorthand for expressing these types of
numbers.
A positive number is written in scientific notation if
it is written as a product of a number a, where 1  a
< 10, and an integer power r of 10.
a  10r
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Scientific Notation
To Write a Number in Scientific Notation
1) Move the decimal point in the original number to the
left or right, so that the new number has a value
between 1 and 10.
2) Count the number of decimal places the decimal point
is moved in Step 1.
• If the original number is 10 or greater, the count is
positive.
• If the original number is less than 1, the count is
negative.
3) Multiply the new number in Step 1 by 10 raised to an
exponent equal to the count found in Step 2.
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Scientific Notation
Example
Write each of the following in scientific notation.
Have to move the decimal 3 places to the left, so that the
1) 4700
new number has a value between 1 and 10.
Since we moved the decimal 3 places, and the original
number was > 10, our count is positive 3.
4700 = 4.7  103
2)
0.00047
Have to move the decimal 4 places to the right, so that
the new number has a value between 1 and 10.
Since we moved the decimal 4 places, and the original
number was < 1, our count is negative 4.
0.00047 = 4.7  10-4
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Scientific Notation
To Write a Scientific Notation Number in
Standard Form
•
Move the decimal point the same number of
spaces as the exponent on 10.
• If the exponent is positive, move the
decimal point to the right.
• If the exponent is negative, move the
decimal point to the left.
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Scientific Notation
Example
Write each of the following in standard notation.
1) 5.2738  103
Since the exponent is a positive 3, we move the decimal 3
places to the right.
5.2738  103 = 5273.8
2)
6.45  10-5
Since the exponent is a negative 5, we move the decimal
5 places to the left.
00006.45  10-5 = 0.0000645
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Operations with Scientific Notation
Multiplying and dividing with numbers written in scientific
notation involves using properties of exponents.
Example
Perform the following operations.
1) (7.3  10-2)(8.1  105) = (7.3 · 8.1)  (10-2 · 105)
= 59.13  103
= 59,130
1.2 10 4 1.2 10 4
5

0
.
3

10


 0.000003
2)
9
9
4 10
4 10
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§ 12.3
Introduction to
Polynomials
Polynomial Vocabulary
Term – a number or a product of a number and
variables raised to powers
Coefficient – numerical factor of a term
Constant – term which is only a number
Polynomial is a sum of terms involving
variables raised to a whole number exponent,
with no variables appearing in any
denominator.
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Polynomial Vocabulary
In the polynomial 7x5 + x2y2 – 4xy + 7
There are 4 terms: 7x5, x2y2, -4xy and 7.
The coefficient of term 7x5 is 7,
of term x2y2 is 1,
of term –4xy is –4 and
of term 7 is 7.
7 is a constant term.
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Types of Polynomials
Monomial is a polynomial with 1 term.
Binomial is a polynomial with 2 terms.
Trinomial is a polynomial with 3 terms.
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Degrees
Degree of a term
To find the degree, take the sum of the exponents
on the variables contained in the term.
Degree of a constant is 0.
Degree of the term 5a4b3c is 8 (remember that c
can be written as c1).
Degree of a polynomial
To find the degree, take the largest degree of any
term of the polynomial.
Degree of 9x3 – 4x2 + 7 is 3.
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Evaluating Polynomials
Evaluating a polynomial for a particular value involves
replacing the value for the variable(s) involved.
Example
Find the value of 2x3 – 3x + 4 when x = 2.
2x3 – 3x + 4 = 2( 2)3 – 3( 2) + 4
= 2( 8) + 6 + 4
=6
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Combining Like Terms
Like terms are terms that contain exactly the same variables
raised to exactly the same powers.
Warning!
Only like terms can be combined through addition and
subtraction.
Example
Combine like terms to simplify.
x2y + xy – y + 10x2y – 2y + xy
= x2y + 10x2y + xy + xy – y – 2y
(Like terms are grouped together)
= (1 + 10)x2y + (1 + 1)xy + (– 1 – 2)y = 11x2y + 2xy – 3y
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§ 12.4
Adding and Subtracting
Polynomials
Adding and Subtracting Polynomials
Adding Polynomials
Combine all the like terms.
Subtracting Polynomials
Change the signs of the terms of the polynomial
being subtracted, and then combine all the like
terms.
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Adding and Subtracting Polynomials
Example
Add or subtract each of the following, as indicated.
1) (3x – 8) + (4x2 – 3x +3) = 3x – 8 + 4x2 – 3x + 3
= 4x2 + 3x – 3x – 8 + 3
= 4x2 – 5
2) 4 – (– y – 4) = 4 + y + 4 = y + 4 + 4 = y + 8
3) (– a2 + 1) – (a2 – 3) + (5a2 – 6a + 7)
= – a2 + 1 – a2 + 3 + 5a2 – 6a + 7
= – a2 – a2 + 5a2 – 6a + 1 + 3 + 7 = 3a2 – 6a + 11
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Adding and Subtracting Polynomials
In the previous examples, after discarding the
parentheses, we would rearrange the terms so
that like terms were next to each other in the
expression.
You can also use a vertical format in
arranging your problem, so that like terms are
aligned with each other vertically.
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§ 12.5
Multiplying
Polynomials
Multiplying Polynomials
Multiplying polynomials
• If all of the polynomials are monomials, use the
associative and commutative properties.
• If any of the polynomials are not monomials,
use the distributive property before the
associative and commutative properties. Then
combine like terms.
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Multiplying Polynomials
Example
Multiply each of the following.
1) (3x2)(– 2x) = (3)(– 2)(x2 · x) = – 6x3
2) (4x2)(3x2 – 2x + 5)
= (4x2)(3x2) – (4x2)(2x) + (4x2)(5)
= 12x4 – 8x3 + 20x2
(Distributive property)
(Multiply the monomials)
3) (2x – 4)(7x + 5) = 2x(7x + 5) – 4(7x + 5)
= 14x2 + 10x – 28x – 20
= 14x2 – 18x – 20
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Multiplying Polynomials
Example
Multiply (3x + 4)2
Remember that a2 = a · a, so (3x + 4)2 = (3x + 4)(3x + 4).
(3x + 4)2 = (3x + 4)(3x + 4) = (3x)(3x + 4) + 4(3x + 4)
=
9x2 + 12x + 12x + 16
=
9x2 + 24x + 16
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Multiplying Polynomials
Example
Multiply (a + 2)(a3 – 3a2 + 7).
(a + 2)(a3 – 3a2 + 7) = a(a3 – 3a2 + 7) + 2(a3 – 3a2 + 7)
a4 – 3a3 + 7a + 2a3 – 6a2 +
14
= a4 – a3 – 6a2 + 7a + 14
=
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Multiplying Polynomials
Example
Multiply (3x – 7y)(7x + 2y)
(3x – 7y)(7x + 2y) = (3x)(7x + 2y) – 7y(7x + 2y)
= 21x2 + 6xy – 49xy + 14y2
= 21x2 – 43xy + 14y2
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Multiplying Polynomials
Example
Multiply (5x – 2z)2
(5x – 2z)2 = (5x – 2z)(5x – 2z) = (5x)(5x – 2z) – 2z(5x – 2z)
= 25x2 – 10xz – 10xz + 4z2
= 25x2 – 20xz + 4z2
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Multiplying Polynomials
Example
Multiply (2x2 + x – 1)(x2 + 3x + 4)
(2x2 + x – 1)(x2 + 3x + 4)
= (2x2)(x2 + 3x + 4) + x(x2 + 3x + 4) – 1(x2 + 3x + 4)
=
2x4 + 6x3 + 8x2 + x3 + 3x2 + 4x – x2 – 3x – 4
=
2x4 + 7x3 + 10x2 + x – 4
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Multiplying Polynomials
You can also use a vertical format in arranging
the polynomials to be multiplied.
In this case, as each term of one polynomial is
multiplied by a term of the other polynomial,
the partial products are aligned so that like
terms are together.
This can make it easier to find and combine like
terms.
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§ 12.6
Special Products
The FOIL Method
When multiplying 2 binomials, the distributive
property can be easily remembered as the FOIL
method.
F – product of First terms
O – product of Outside terms
I – product of Inside terms
L – product of Last terms
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Using the FOIL Method
Example
Multiply (y – 12)(y + 4)
(y – 12)(y + 4)
Product of First terms is y2
(y – 12)(y + 4)
Product of Outside terms is 4y
(y – 12)(y + 4)
Product of Inside terms is -12y
(y – 12)(y + 4)
Product of Last terms is -48
F O
I
L
(y – 12)(y + 4) = y2 + 4y – 12y – 48
= y2 – 8y – 48
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Using the FOIL Method
Example
Multiply (2x – 4)(7x + 5)
L
F
F
O
I
L
(2x – 4)(7x + 5) = 2x(7x) + 2x(5) – 4(7x) – 4(5)
I
O
= 14x2 + 10x – 28x – 20
= 14x2 – 18x – 20
We multiplied these same two binomials together in the
previous section, using a different technique, but arrived at the
same product.
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Special Products
In the process of using the FOIL method on products
of certain types of binomials, we see specific patterns
that lead to special products.
Squaring a Binomial
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
Multiplying the Sum and Difference of Two Terms
(a + b)(a – b) = a2 – b2
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Special Products
Although you will arrive at the same results
for the special products by using the
techniques of this section or last section,
memorizing these products can save you some
time in multiplying polynomials.
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§ 12.7
Dividing Polynomials
Dividing Polynomials
Dividing a polynomial by a monomial
Divide each term of the polynomial separately by
the monomial.
Example
 12a 3  36a  15
 12a 3 36a 15



3a
3a
3a 3a
5
2
 4a  12 
a
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Dividing Polynomials
Dividing a polynomial by a polynomial other
than a monomial uses a “long division”
technique that is similar to the process known
as long division in dividing two numbers,
which is reviewed on the next slide.
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Dividing Polynomials
168
43 7256
43
295
258
37 6
344
32
Divide 43 into 72.
Multiply 1 times 43.
Subtract 43 from 72.
Bring down 5.
Divide 43 into 295.
Multiply 6 times 43.
Subtract 258 from 295.
Bring down 6.
Divide 43 into 376.
Multiply 8 times 43.
Subtract 344 from 376.
Nothing to bring down.
We then write our result as
32
168 .
43
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Dividing Polynomials
As you can see from the previous example, there is
a pattern in the long division technique.
Divide
Multiply
Subtract
Bring down
Then repeat these steps until you can’t bring
down or divide any longer.
We will incorporate this same repeated technique
with dividing polynomials.
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Dividing Polynomials
4x  5
2
7 x  3 28x  23x  15
2
28 x  12 x
 35 x  15
 35x 15
Divide 7x into 28x2.
Multiply 4x times 7x+3.
Subtract 28x2 + 12x from 28x2 – 23x.
Bring down – 15.
Divide 7x into –35x.
Multiply – 5 times 7x+3.
Subtract –35x–15 from –35x–15.
Nothing to bring down.
So our answer is 4x – 5.
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Dividing Polynomials
2 x  10
2
2 x  7 4x  6 x  8
2
4 x  14 x
20x  8
20x  70
78
Divide 2x into 4x2.
Multiply 2x times 2x+7.
Subtract 4x2 + 14x from 4x2 – 6x.
Bring down 8.
Divide 2x into –20x.
Multiply -10 times 2x+7.
Subtract –20x–70 from –20x+8.
Nothing to bring down.
We write our final answer as 2 x  10 
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( 2 x  7)
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