MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
Course Web Site
http://www.papademas.net/occ/math114
Textbook Companion Web Site
http://wps.prenhall.com/chet_cleaves_cmupdate_7
Welcome to Week 13
This Week’s Agenda
•
•
•
Review Chapter 13 in the Course Textbook ( Factoring )
Commence completion of Lab Assignments 25 and 26
( download from the course Web site )
Commence completion of Homework Assignment(s) 13
( download from the course Web site )
Chapter Sections
Ch 13: Factoring
Section 1: The Distributive Property and Common Factors
Section 2: Factoring Special Products
Section 3: Factoring General Trinomials
Chapter 12 Review ( Rational Exponents, Roots and Radicals )
Here is a review of Rational Exponents, Roots and Radicals.
[ Example ]
Evaluate √ ( 324 ) .
[ Solution ]
We first ask if the number 324 a perfect square. Yes, since 18 × 18 is 324 .
Is it not true that ( − 18 ) × ( − 18 ) also equals 324 .
Hence √ ( 324 ) = ± 18 .
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1
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
[ Example ]
Evaluate 2 √ ( 121 ) .
[ Solution ]
The small 2 is referred to as a root index. If a principal square root symbol
does not have an index then a square root is understood.
Our answer here is ± 11 , i.e. we have two solutions here!
[ Example ]
Evaluate 4 √ ( 256 ) .
[ Solution ]
This is equivalent to 256
(1/4)
which evaluates to ± 4 , since
256 = ( 4 ) ( 4 ) ( 4 ) ( 4 )
or
256 = ( − 4 ) ( − 4 ) ( − 4 ) ( − 4 )
[ Example ]
Evaluate the following expression:
2 ( 3 √ ( 729 ) ) 4
[ Solution ]
For the radical expression 2 ( 3 √ ( 729 ) )
4
the 2 is a factor
the 3 is the root index
the 729 is the radicand
the 4 is an exponent
2 ( 3 √ ( 729 ) ) 4 is equivalent to
2 [ ( 729 ) ( 1 / 3 ) ] 4 is equivalent to
2(9)4 =
2 ( 6561 ) =
13,122
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2
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
[ Example ]
Which of these is greater in quantity?
(a)
10
√(1)
(b)
2
√(4)
(c)
3
√ ( 27 )
(d)
8
√(4)
[ Solution ]
(a)
(c)
10
√(1) = 1
3
√ ( 27 ) = 3
(b)
(d)
2
8
√(4) = 2
√ ( 4 ) = 1.189207115
[ Example ]
Subtract the radicals, simplifying if possible: 6 √ ( 5 ) − 8 √ ( 5 )
[ Solution ]
Here we combine like terms: − 2 √ ( 5 )
6 √ ( 5 ) − 8 √ ( 5 ) = −2 √ ( 5 )
[ Example ]
Add the radicals, simplifying if possible.
11 √ ( 75 ) + 8 √ ( 147 )
[ Solution ]
If possible, rewrite each radicand as the product of a perfect square
and some other number.
11 √ ( 75 ) + 8 √ ( 147 ) =
11 √ ( 25 ) √ ( 3 ) + 8 √ ( 49 ) √ ( 3 ) =
11 × 5 √ ( 3 ) + 8 × 7 √ ( 3 ) =
55 √ ( 3 ) + 56 √ ( 3 ) =
111 √ ( 3 )
[ Example ]
Perform the indicated operation. Simplify your answer.
√(3)[4√(6) + 3√(6)]
[ Solution ]
4√(3)√(6) + 3√(3)√(6) =
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3
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
4√(9)√(2) + 3√(9)√(2) =
12 √ ( 2 ) + 9 √ ( 2 ) =
21 √ ( 2 ) =
Mathematica Screen Snapshot ( Exact Value )
Mathematica Screen Snapshot ( Approximate Value )
[ Example ]
Perform the indicated operation and leave your answer in the form a + b i ,
where i is equivalent to the square root of − 1 .
(4 + 6i ) − (6 − 3i )
[ Solution ]
Apply the distributive property.
(4 + 6i ) − (6 − 3i )
= 4 + 6i − 6 + 3i
= −2 + 9i
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apply distributive property
combine like terms
4
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
[ Example ]
Perform the indicated operation and leave your answer in the form a + b i .
(2 − 3i )(4 + i )
[ Solution ]
Use FOIL here! FOIL means First OUTER INNER Last
(2 − 3i )(4 + i ) =
8
2i
− 12 i
F
O
I
− 3 i2
L
8 + 2 i − 12 i − 3 ( − 1 )
11 − 10 i
Chapter 13 Review ( Factoring )
Factoring an expression refers to writing in its component form.
Section 1: The Distributive Property and Common Factors
What is distributive property? The distributive property together with the
associative and commutative properties comprise three properties of
real numbers.
The distributive property states that for three given real numbers, a , b and c ,
a ( b + c ) equals a b + a c
The reverse process to the distributive property is called factoring, i.e.
a b + a c equals a ( b + c )
Factoring means considering the factors or multipliers or terms that comprise
an expression.
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5
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
[ Example ]
Your gross pay has two factors: hourly wage and hours worked. Hence a
gross pay of $ 500 has factors of $ 50 and 10 if you are paid $ 50 a hour to
work for 10 hours. Also a gross pay of $ 500 has factors of $ 10 and 50 if you
are paid $ 10 a hour to work for 50 hours. Both factors yield a product of
$ 500 , as shown in the MS Excel snapshot given below.
[ Example ]
What are the factors of 3 a 2 b ?
[ Solution ]
The factors here are 3 , a 2 and b .
[ Example ]
What are the factors of 5 ( a + 2 b ) ?
[ Solution ]
The factors here are 5 and ( a + 2 b ) .
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6
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
[ Example ]
How else can we factor the expression 6 x y
2
?
[ Solution ]
One way is 3 ( 2 ) ( x ) ( y ) ( y ) .
Another way is 6 · ( x ) ( y ) ( y ) .
Another way is 3 · 2 · x · y 2 .
[ Example ]
What are the factors of 1,776 ?
[ Solution ]
Perform successive division by 2 until a prime quotient is obtained.
1,776 ÷ 2 = 888
888 ÷ 2 = 444
444 ÷ 2 = 222
222 ÷ 2 = 111
111 ÷ 3 = 37 ( is a prime number )
Notice that the factoring process ends with a prime number!
Hence 1,776 = 2 4 × 3 × 37
( a prime factorization since 2 , 3 and 37 are all prime numbers )
[ Example ]
24 x
Factor the expression.
4
− 15 x
3
+ 33 x
2
[ Solution ]
24 x
4
− 15 x
Here 3 x
2
3
+ 33 x
2
= 3x
2
( 8x
2
− 5 x + 11 )
is referred to as the GCF .
GCF = Greatest Common Factor
The general rule of factoring is to first find a common factor.
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7
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
[ Example ]
Factor completely.
3x2 − 6x
[ Solution ]
3x2 − 6x = 3x(x −2)
Here the factor ( 3 x ) is our GCF .
[ Example ( Applications ) ]
The product of two consecutive odd integers is 1,023 . What expressions can
we use for the unknown factors?
[ Solution ]
The unknown factors are ( x ) and ( x + 2 ) .
The actual values are 31 × 33 = 1,023 .
These number can be found by solving a quadratic equation.
( x ) ( x + 2 ) = 1,023
or
x 2 + 2 x = 1,023
or
x 2 + 2 x − 1,023 = 0
from which
( x + 33 ) ( x − 31 ) = 0
Hence, by the zero product rule, the factors are 31 and 33 .
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8
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
[ Example ( Applications ) ]
The area of rectangular rug having an area of 750 square feet. The length of
rug is 30 feet and its width is expressed as x + 2 . Find the value of x .
[ Solution ]
The factors of an area are the length and the width or:
Area = Length × Width
750 = 30 ( x + 2 )
from which
750 = 30 x + 60
or
30 x + 60 = 750
and
30 x = 750 − 60
30 x = 690
x = 690 ÷ 30 or 23 feet
Checking our answer: 23 × 25 = 750
Section 2: Factoring Special Products
Some special types of factoring are:
Difference of Two Squares
a2− b2=(a+b)(a−b)
Sum of Two Squares
a 2 + b 2 ( is not factorable )
Perfect Square Trinomial
a2+2ab + b2 = (a+b)(a+b)
or
a2−2ab + b2 = (a−b)(a−b)
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9
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
Sum of Two Cubes
a3+ b3 = (a+b)(a2 − ab + b2)
Difference of Two Cubes
a3− b3 = (a−b)(a2 + ab + b2)
Factoring a Difference of Two Squares
[ Example ]
Factor completely.
a2 − 72
[ Solution ]
a2 − 72 = (a + 7)(a − 7)
Here the signs on the right are the different.
[ Example ]
Factor completely.
92 – b2
[ Solution ]
92 – b2 = (9 + b)(9 − b)
[ Example ]
Factor completely.
36 x
2
− 0.25
[ Solution ]
36 x 2 − 0.25 = ( 6 x + 0.50 ) ( 6 x − 0.50 )
[ Example ]
What are the factors of x 2 − 25 ?
[ Solution ]
The factors of x 2 − 25 are ( x + 5 ) and ( x − 5 ) .
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10
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
Factoring a Sum of Two Squares
The sum of two squares a 2 + b 2 cannot be factored.
[ Example ]
Factor completely.
8a2+ 2b2
[ Solution ]
8a2+ 2b2 = 2(4a2+ b2)
The expression ( 4 a 2 + b 2 ) is a sum of two squares and hence not
factorable.
Perfect Square Trinomial
x2 + 2xy + y2 = (x+y)(x+y)
Here the signs on the right are the same.
x2 − 2xy + y2 = (x − y)(x − y)
Here the signs on the right are the same.
[ Example ]
Factor Completely.
x
2
+ 14 x y + 49 y 2
[ Solution ]
x
2
+ 14 x y + 49 y 2 = ( x + 7 y ) ( x + 7 y )
This is a perfect square trinomial.
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11
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
[ Example ( Sum of Two Cubes ) ]
Factor Completely.
x
3
+ 512
[ Solution ]
To Factor a Sum of Two Cubes:
Step 1
Determine the Base Numbers
3
x + 512 = ( x ) 3 + ( 8 ) 3
Step 2
x
3
+ 512 = ( x + 8 ) (
Step 3
x
3
3
)
Square the Base Numbers
+ 512 = ( x + 8 ) ( x
Step 4
x
Write the First Factor
2
2
8
)
Multiple the Base Numbers and Change the Sign of the Product
+ 512 = ( x + 8 ) ( x
2
− 8 x + 64 )
Section 3: Factoring General Trinomials
[ Example ( Trinomial Whose Leading Coefficient is 1 ) ]
Factor the trinomial completely.
x
2
− 11 x + 24
[ Solution ]
The factors of 24 that have a sum or difference of 11 are 3 and 8 .
x
2
− 11 x + 24 = ( x − 3 ) ( x − 8 )
[ Example ( Trinomial Whose Leading Coefficient is Different From 1 ) ]
Factor the trinomial completely.
2x
2
− x − 15
[ Solution ]
The factors of 15 that have a sum or difference of − 1 , when the leading
coefficient is 2 , are 3 and 5 .
2x
2
− x − 15 = ( 2 x + 5 ) ( x − 3 )
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12
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
[ Example ( Using Factoring to Solve a Quadratic Equation ) ]
Use factoring to solve the quadratic equation.
5x
2
= 8x
[ Solution ]
Rearrange the equation 5 x
equation.
5x
2
2
= 8 x into the general form of a quadratic
− 8x = 0
Factor the left hand side.
x ( 5x − 8 ) = 0
Apply the zero product rule.
x = 0 or ( 5 x − 8 ) = 0
x = 0 or x = 8 / 5
[ Example ( Using Factoring to Solve a Quadratic Equation ) ]
Use factoring to solve the quadratic equation.
(x + 9)(x + 2) = 0
[ Solution ]
By the zero product rule, ( x + 9 ) ( x + 2 ) = 0 means that
( x + 9 ) = 0 or ( x + 2 ) = 0 .
From which x = − 9 or x = − 2 .
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13
MATH 114
Applied Mathematics I
Week 13
Factoring
Lecture Notes
Student Name
Factoring By Grouping
[ Example ( Factoring by Grouping ) ]
Factor completely.
6x
2
− 3x + 4xy − 2y
[ Solution ]
First factor each pair of terms.
6x
2
− 3x + 4xy − 2y =
3x(2x − 1)+ 2y(2x − 1) =
(3x + 2y )(2x − 1)
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14