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CHAPTER
6
Factoring
6.1
6.2
6.3
6.4
6.5
6.6
6.7
Greatest Common Factor and Factoring by
Grouping
Factoring Trinomials of the Form x2 + bx + c
Factoring Trinomials of the Form ax2 + bx + c,
where a  1
Factoring Special Products
Strategies for Factoring
Solving Quadratic Equations by Factoring
Graphs of Quadratic Equations and Functions
Copyright © 2011 Pearson Education, Inc.
6.1
Greatest Common Factor and Factoring
by Grouping
1. List all possible factors for a given number.
2. Find the greatest common factor of a set of numbers or
monomials.
3. Write a polynomial as a product of a monomial GCF and
a polynomial.
4. Factor by grouping.
Copyright © 2011 Pearson Education, Inc.
Factored form: A number or an expression written as a
product of factors.
Following are some examples of factored form:
An integer written in factored form with integer factors:
28 = 2 • 14
A monomial written in factored form with monomial
factors: 8x5 = 4x2 • 2x3
A polynomial written in factored form with a monomial
factor and a polynomial factor: 2x + 8 = 2(x + 4)
A polynomial written in factored form with two
polynomial factors: x2 + 5x + 6 = (x + 2)(x + 3)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 3
Example 1
List all natural number factors of 36.
Solution:
To list all the natural number factors, we can divide
36 by 1, 2, 3, and so on, writing each divisor and
quotient pair as a product until we have all possible
combinations.
1 • 36
2 • 18
3 • 12
The natural number factors of 36
4•9
are 1, 2, 3, 4, 6, 9, 12, 18, 36.
6•6
Copyright © 2011 Pearson Education, Inc.
Slide 6- 4
Greatest common factor (GCF): The largest natural
number that divides all given numbers with no
remainder.
Listing Method for Finding GCF
To find the GCF of a set of numbers by listing,
1. List all possible factors for each given number.
2. Search the lists for the largest factor common to all
lists.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 5
Example 2
Find the GCF of 48 and 54.
Solution:
Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54
The GCF of 48 and 54 is 6.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 6
Prime Factorization Method for Finding GCF
To find the GCF of a given set of numbers by prime
factorization,
1. Write the prime factorization of each given number
in exponential form.
2. Create a factorization for the GCF that includes
only those prime factors common to all
factorizations, each raised to its smallest exponent
in the factorizations.
3. Multiply the factors in the factorization created in
Step 2.
Note: If there are no common prime factors, then
the GCF is 1.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 7
Example 3a
Find the GCF of 45 and 72.
Solution:
Write the prime factorization 45 and 72 in factored
form.
45 = 32 • 5
72 = 23 • 32
The common prime factor is 3.
GCF = 32 = 9
Copyright © 2011 Pearson Education, Inc.
Slide 6- 8
Example 3b
Find the GCF of 3240 and 8316.
Solution:
Write the prime factorization of each number.
3240 = 23 • 34 • 5
8316 = 22 • 33 • 7 • 11
The common prime factors are 2 and 3. Take the
term with the smallest exponent.
GCF = 22 • 33 = 108
Copyright © 2011 Pearson Education, Inc.
Slide 6- 9
Example 4a
Find the GCF of a4b4c5 and a3b7.
Solution:
Each monomial is already written as a prime
factorization. The common primes are a and b.
The smallest exponent for a is 3.
The smallest exponent for b is 4.
GCF = a3b4
Copyright © 2011 Pearson Education, Inc.
Slide 6- 10
Example 4b
Find the GCF of 45a3b and 30a2.
Solution:
Write the prime factorization of each monomial,
treating the variables like prime factors.
45a3b = 32 • 5 • a3 • b
30a2 = 2 • 3 • 5 • a2
The common prime factors are 3, 5, and a.
GCF = 3 • 5 • a2 = 15a2
Copyright © 2011 Pearson Education, Inc.
Slide 6- 11
Factoring a Monomial GCF Out of a Polynomial
To factor a monomial GCF out of a given
polynomial,
1. Find the GCF of the terms in the polynomial.
2. Rewrite the given polynomial as a product of the
GCF and the quotient of the polynomial and the
GCF.
 Given polynomial 
polynomial = GCF 

GCF


Copyright © 2011 Pearson Education, Inc.
Slide 6- 12
Example 5
Factor 18 x 2  24 x.
Solution
1. Find the GCF of 18x2 and 24x.
The GCF is 6x.
2. Write the given polynomial as a product of
the GCF and the quotient of the polynomial
and the GCF.
2

18 x  24 x 
2
18 x  24 x  6 x 

6x


Copyright © 2011 Pearson Education, Inc.
Slide 6- 13
continued
Factor 18 x 2  24 x.
Solution
2

18
x
 24 x 
2
18 x  24 x  6 x 

6x


 18 x 2 24 x 
 6x 


6x 
 6x
 6 x(3 x  4)
Check We can check by multiplying the
factored form using the distributive property.
This is left to you.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 14
Example 6
Factor 25x3 y 2  35x 4 .
Solution
1. Find the GCF of both terms.
The GCF is 5x3
2. Write the given polynomial as a product of
the GCF and the quotient of the polynomial
and the GCF.
3 2
4

3 2
4
3 25 x y  35 x 
25 x y  35 x  5 x 

3
5x


Copyright © 2011 Pearson Education, Inc.
Slide 6- 15
continued
Factor 25x3 y 2  35x 4
Solution
3 2
4


25
x
y

35
x
3 2
4
3
25 x y  35 x  5 x 

3
5
x


3 2
4

35 x 
3 25 x y
 5x 
 3 
3
5x 
 5x
3
2
 5 x (5 y  7 x)
Check We can check by multiplying the
factored form using the distributive property.
This is left to you.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 16
Example 7
Factor. 9 x yz  15x y  18x y
2
3
4
2
Solution
2
3
4 2
1. Find the GCF of 9 x yz, 15 x y, and  18 x y .
Because the first term in the polynomial is
negative, we will factor out the negative of
the GCF to avoid a negative first term inside
the parentheses. We will factor out 3x2 y.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 17
continued
2. Write the given polynomial as the product of the
GCF and the parentheses containing the quotient of
the given polynomial and the GCF.
2
3
4 2


9
x
yz

15
x
y

18
x
y 
2
2
3
4 2
9 x yz  15x y  18x y  3x y 

2

3
x
y


2
3
4 2


9
x
yz
15
x
y
18
x
y 
2
 3x y 



2
2
2

3
x
y

3
x
y

3
x
y


 3x 2 y  3z  5 x  6 x 2 y 
Copyright © 2011 Pearson Education, Inc.
Slide 6- 18
Example 8
Factor. a  b  5  8 b  5
Solution: Notice that this expression is a sum of
two products, a and (b + 5), and 8 and (b + 5).
Further, note that (b + 5) is the GCF of the two
products.
 a  b  5  8  b  5 
a  b  5  8  b  5   b  5  

b5


 a  b  5 8  b  5 
  b  5 


b5 
 b5
  b  5 a  8
Copyright © 2011 Pearson Education, Inc.
Slide 6- 19
Factoring by Grouping
To factor a four-term polynomial by grouping,
1. Factor out any monomial GCF (other than 1) that is
common to all four terms.
2. Group together pairs of terms and factor the GCF
out of each pair or group.
3. If there is a common binomial factor, then factor it
out.
4. If there is no common binomial factor, then
interchange the middle two terms and repeat the
process. If there is still no common binomial factor,
then the polynomial cannot be factored by
grouping.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 20
Example 9a
Factor. 8 p3  24 p 2  3 pq  9q
Solution First we look for a monomial GCF (other
than 1). This polynomial does not have one.
Because the polynomial has four terms, we now
try to factor by grouping.
8 p3  24 p 2  3 pq  9q
  8 p 3  24 p 2    3 pq  9q 
 8 p2  p  3  3q  p  3
  p  3  8 p 2  3q 
Copyright © 2011 Pearson Education, Inc.
Slide 6- 21
Example 9b
Factor. 6a 2b  18a 2  14ab  42a
Solution There is a monomial GCF, 2a; factor this
from all four terms.
6a 2b  18a 2  14ab  42a  2a(3ab  9a  7b  21)
 2a  3ab  9a    7b  21 
 2a 3a  b  3  7  b  3 
 2a  b  3 3a  7 
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Slide 6- 22
Factor by factoring out the GCF.
56 x 4 y 2  32 xy 2  72 x3 y
3
2
4
xy
14
x
y

8
y

18
x
a)


b) 4 xy 14 x 2 y  8 y  18 xy 
c) 8 xy  7 x 2 y  4 xy  9 x 2 
d) 8 xy  7 x3 y  4 y  9 x 2 
6.1
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Slide 6- 23
Factor by factoring out the GCF.
56 x 4 y 2  32 xy 2  72 x3 y
3
2
4
xy
14
x
y

8
y

18
x
a)


b) 4 xy 14 x 2 y  8 y  18 xy 
c) 8 xy  7 x 2 y  4 xy  9 x 2 
d) 8 xy  7 x3 y  4 y  9 x 2 
6.1
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Slide 6- 24
Factor by grouping.
b  3bc  7b  21c
2
a)
b  c  7  3
b)  b  3c  b  7 
c)
 c  3b  7
d)  b  3 c  7 
6.1
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Slide 6- 25
Factor by grouping.
b  3bc  7b  21c
2
a)
b  c  7  3
b)  b  3c  b  7 
c)
 c  3b  7
d)  b  3 c  7 
6.1
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Slide 6- 26
6.2
Factoring Trinomials of the
Form x2 + bx + c
1. Factor trinomials of the form x2 + bx + c.
2. Factor out a monomial GCF, then factor the trinomial of
the form x2 + bx + c.
Copyright © 2011 Pearson Education, Inc.
Following are some examples of trinomials of the
form x2 + bx + c.
x2 + 5x + 6 or x2 –7x + 12 or x2 – 5x – 24
Products in the form x2 + bx + c are the result of the
product of two binomials.
When we factor a trinomial of the form x2 + bx + c,
we reverse the FOIL process, using the fact that b is
the sum of the last terms in the binomials and c is the
product of the last terms in the binomials.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 28
Factoring x2 + bx + c
To factor a trinomial of the form x2 + bx + c,
1. Find two numbers with a product equal to c and a
sum equal to b.
2. The factored trinomial will have the form:
(x + first number) (x + second number).
Note: The signs in the binomial factors can be
minus signs, depending on the signs of b and c.
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Slide 6- 29
Example 1a
Factor. x2 – 6x + 8
Solution: We must find a pair of numbers whose
product is 8 and whose sum is –6. If two
numbers have a positive product and negative
sum, they must both be negative. Following is a
table listing the products and sums:
Product
Sum
(–1)(–8) = 8
–1 + (–8) = –9
(–2)(–4) = 8
–2 + (–4) = –6
Copyright © 2011 Pearson Education, Inc.
This is the
correct
combination.
Slide 6- 30
continued
Answer
x2 – 6x + 8 = (x – 2)(x – 4)
Check We can check by multiplying the binomial
factors to see if their product is the
original polynomial.
(x – 2)(x – 4) = x2 – 4x – 2x + 8
= x2 – 6x + 8
Copyright © 2011 Pearson Education, Inc.
Multiply the factors using
FOIL.
The product is the original
polynomial.
Slide 6- 31
Example 1b
Factor. x2 + 2x – 24
Solution: We must find a pair of numbers whose
product is –24 and whose sum is 2. Because the
product is negative, the two numbers have
different signs.
Product
Sum
(–2)(12) = 24
–2 + 12 = 10
(–4)(6) = 24
–4 + 6 = 2
(–8)(3) = –24
–8 + 3 = –5
Copyright © 2011 Pearson Education, Inc.
This is the
correct
combination.
Slide 6- 32
continued
Answer x2 + 2x – 24 = (x – 4)(x + 6)
Check We can check by multiplying the binomial
factors to see if their product is the
original polynomial.
(x – 4)(x + 6) = x2 + 6x – 4x – 24
= x2 + 2x – 24
Copyright © 2011 Pearson Education, Inc.
Multiply the factors using
FOIL.
The product is the original
polynomial.
Slide 6- 33
Example 1c
Factor. x2 – 6x – 27
Solution: We must find a pair of numbers whose
product is –27 and whose sum is 6. Because the
product is negative, the two numbers have
different signs.
Product
Sum
(–1)(27) = 27
–2 + 27 = 25
(–9)(3) = 27
–9 + 3 = 6
Copyright © 2011 Pearson Education, Inc.
This is the
correct
combination.
Slide 6- 34
continued
Answer x2 – 6x – 27 = (x – 9)(x + 3)
Check We can check by multiplying the binomial
factors to see if their product is the
original polynomial.
(x – 9)(x + 3) = x2 + 3x – 9x – 27
= x2 – 6x – 27
Copyright © 2011 Pearson Education, Inc.
Multiply the factors using
FOIL.
The product is the original
polynomial.
Slide 6- 35
Example 1d
Factor. x2 + 2x + 5
Solution: We must find a pair of numbers whose
product is 5 and whose sum is 2. If two numbers
have a positive product and a positive sum, both
must be positive
Product
Sum
(1)(5) = 5
1+5=6
There is no
combination of
factors whose
product is 5 and
sum is 2.
The polynomial cannot be factored. The
polynomial is prime.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 36
Example 2a
Factor. a2 – ab – 20b2
Solution: We must find a pair of terms whose
product is 20b2 and whose sum is –1b. These
terms would have to be –5b and 4b.
Answer a2 – ab – 20b2 = (a – 5b)(a + 4b)
Check (a – 5b)(a + 4b) = a2 + 4ab – 5ab – 20b2
= a2 – ab – 20b2
Copyright © 2011 Pearson Education, Inc.
Slide 6- 37
Example 3a
Factor. 4xy3 + 12xy2 – 72xy
Solution
First, we look for a monomial GCF (other than 1).
Notice that the GCF of the terms is 4xy.
Factoring out the monomial, we have
4xy3 + 12xy2 – 72xy = 4xy(y2 + 3y – 18)
Now try to factor the trinomial to two binomials.
We must find a pair of numbers whose product is
–18 and whose sum is 3.
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Slide 6- 38
continued
Product
Sum
(–1)(18) = –18
–1 + 18 = 17
(–2)(9) = – 18
–2 + 9 = 7
(–3)(6) = – 18
–3 + 6 = 3
This is the correct
combination.
Answer
4xy3 + 12xy2 – 72xy = 4xy(y – 3)(y + 6)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 39
Example 3b
Factor. x4 + 2x3 + 5x2
Solution
First, factor out the GCF, x2.
x2 ( x2 + 2x + 5)
Now try to factor the trinomial to two binomials.
We must find a pair of numbers whose product is
5 and whose sum is 2.
From a previous example, we learned this
polynomial is prime. The final factored form is
x2 ( x2 + 2x + 5)
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Slide 6- 40
Factor. x2 + 5x – 36
a) (x + 3)(x – 12)
b) (x – 3)(x + 12)
c) (x + 9)(x – 4)
d) (x – 9)(x + 4)
6.2
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Slide 6- 41
Factor. x2 + 5x – 36
a) (x + 3)(x – 12)
b) (x – 3)(x + 12)
c) (x + 9)(x – 4)
d) (x – 9)(x + 4)
6.2
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Slide 6- 42
Factor completely. 5rs3 – 10rs2 – 40rs
a) 5rs(s2 – 2s – 8)
b) 5rs(s2 + 2s – 8)
c) 5rs(s + 2)(s – 4)
d) 5rs(s – 2)(s + 4)
6.2
Copyright © 2011 Pearson Education, Inc.
Slide 6- 43
Factor completely. 5rs3 – 10rs2 – 40rs
a) 5rs(s2 – 2s – 8)
b) 5rs(s2 + 2s – 8)
c) 5rs(s + 2)(s – 4)
d) 5rs(s – 2)(s + 4)
6.2
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Slide 6- 44
6.3
Factoring Trinomials of the Form
ax2 + bx + c, where a  1
1. Factor trinomials of the form ax2 + bx + c,
where a  1, by trial.
2. Factor trinomials of the form ax2 + bx + c,
where a  1, by grouping.
Copyright © 2011 Pearson Education, Inc.
We will focus on factoring trinomials in which the
coefficient of the squared term is other than 1,
such as the following:
3x2 + 17x + 10
8x2 + 29x – 12
In general, like trinomials of the form x2 + bx + c,
trinomials of the form ax2 + bx + c, where a  1,
also have two binomial factors.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 46
Factoring by Trial and Error
To factor a trinomial of the form ax2 + bx + c, where
a ≠ 1, by trial and error,
1. Look for a monomial GCF in all the terms. If there
is one, factor it out.
2. Write a pair of first terms whose product is ax2.
ax2





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Slide 6- 47
3. Write a pair of last terms whose product is c.
c





4. Verify that the sum of the inner and outer products
is bx (the middle term of the trinomial).





Inner
+ Outer
bx
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Slide 6- 48
If the sum of the inner and outer products is not
bx, try the following:
a. Exchange the last terms of the binomials from
step 3, then repeat step 4.
b. For each additional pair of last terms, repeat steps
3 and 4.
c. For each additional pair of first terms, repeat
steps 2 - 4.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 49
Example 1
Factor. 6 x 2  13x  5
Solution
The first terms must multiply to equal 6x2.
These could be x and 6x, or 2x and 3x.
6 x 2  13x  5  
+



The last terms must multiply to equal –5.
Because –5 is negative, the last terms in the
binomials must have different signs. This factor
pair must be 1 and 5.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 50
continued
Now we multiply binomials with various
combinations of these first and last terms until we find
a combination whose inner and outer products
combine to equal 13x.
 x  5 6x 1  6x22  x  30x  5  6x22  29x  5
 x  1 6x  5  6x  5x  6x  5  6x  x  5
3x  1 2x  5  6x2 15x  2x  5  6x2 13x  5 Incorrect
combinations.
2
2
3
x

5
2
x

1

6
x

3
x

10
x

5

6
x

7
x

5



2
2
2
x

1
3
x

5

6
x

10
x

3
x

5

6
x
 7x  5



2
2
2
x

5
3
x

1

6
x

2
x

15
x

5

6
x
 13x  5 Correct



combination.
Answer 6x2  13x  5   2x  53x 1
Copyright © 2011 Pearson Education, Inc.
Slide 6- 51
Example 2
3
2
21
x

60
x
 9x
Factor.
Solution First, we factor out the monomial GCF, 3x.
21x3  60 x 2  9 x  3x  7 x 2  20 x  3
Now we factor the trinomial within the parentheses.
The first terms must multiply to equal 7x2.
These could be x and 7x.
3x  7 x 2  20 x  3  3x 


+

The last terms must multiply to equal 3.
Because 3 is a prime number, its factors are 1
and 3.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 52
continued
Now we multiply binomials with various
combinations of these first and last terms until we find
a combination whose inner and outer products
combine to equal –20x.
3x  x  1 7 x  3  3x  7 x 2  3x  7 x  3  3x  7 x 2  4 x  3
3x  x  1 7 x  3  3x  7 x 2  3x  7 x  3  3x  7 x 2  4 x  3
3x  x  3 7 x  1  3x  7 x 2  x  21x  3  3x  7 x 2  20 x  3
3x  x  3 7 x  1  3x  7 x 2  x  21x  3  3x  7 x 2  20 x  3
Correct combination.
Answer 21x3  60x2  9x  3x  x  3 7 x  1
Copyright © 2011 Pearson Education, Inc.
Slide 6- 53
Objective 2
Factor trinomials of the form
ax2 + bx + c, where a  1, by
grouping.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 54
Factoring ax2 + bx + c, where a ≠ 1, by Grouping
To factor a trinomial of the form ax2 + bx + c, where
a ≠ 1, by grouping,
1. Look for a monomial GCF in all the terms. If there
is one, factor it out.
2. Multiply a and c.
3. Find two factors of this product whose sum is b.
4. Write a four-term polynomial in which bx is written
as the sum of two like terms whose coefficients are
the two factors you found in step 3.
5. Factor by grouping.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 55
Example 3a
Factor. 2 x 2  15 x  7
Solution
Notice that for this trinomial, a = 2, b = –15, and
c = 7. We begin my multiplying a and c: (2)(7) = 14.
Now we find two factors of 14 whose sum is –15.
Notice that these two factors must both be negative.
Factors of ac
Sum of Factors of ac
(–2)(–7) = 14
–2 + (–7) = –9
(–1)(–14) = 14
–1 + (– 14) = –15
Copyright © 2011 Pearson Education, Inc.
Correct
Slide 6- 56
continued
2 x 2 –15x  7  2 x 2 –x – 14x  7
 x  2 x  1  7  2 x 1
  2 x 1   x  7 
Copyright © 2011 Pearson Education, Inc.
Slide 6- 57
Example 3b
Factor. 4 x 2  17 x  4
Solution
Notice that for this trinomial, a = 4, b = 17, and
c = 4. We begin my multiplying a and c: (4)(4) = 16.
Now we find two factors of 16 whose sum is 17. Notice
that these two factors must both be positive.
Factors of ac
Sum of Factors of ac
(1)(16) = 16
1 + 16 = 17
(2)(8) = 16
2 + 8 = 10
(4)(4) = 16
4+4=8
Copyright © 2011 Pearson Education, Inc.
Correct
Slide 6- 58
continued
4 x 2 +17x  4  4 x 2 + x + 16x  4
 x  4 x  1  4  4 x  1
  4 x  1   x  4
Copyright © 2011 Pearson Education, Inc.
Slide 6- 59
Example 3c
Factor.12 x 4  14 x3  6 x 2
Solution
Notice that there is a GCF, 2x2, that we can factor out.
2x2(6x2 + 7x – 3)
Now we factor the trinomial within the parentheses.
a = 6, c = 3; ac = (6)(3) = 18
Find two factors of 18 whose sum is 7. Because the
product is negative, the two factors will have different
signs.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 60
Example 3c
Factor.12 x 4  14 x3  6 x 2
Solution
Factors of ac
Sum of Factors of ac
(1)(18) = 18
1 + 18 = 17
(2)(9) = 18
2 + 9 = 7
(3)(6) = 18
3 + 6 = 3
Correct
2x2(6x2 + 7x – 3)
Now write 7x as 2x + 9x and then factor by
grouping.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 61
continued
2 x 2 (6 x 2 +7x  3)  2 x 2 (6 x 2 2x + 9x  3)
 2x2  2x(3x 1)  3(3x 1)
 2 x2  2 x  3 3x 1
Copyright © 2011 Pearson Education, Inc.
Slide 6- 62
Factor completely. 6x2 –33x – 63
a) 3(2x + 7)(x – 3)
b) 3(2x + 3)(x – 7)
c) 3(2x – 3)(x + 7)
d) 3(2x – 7)(x + 3)
6.3
Copyright © 2011 Pearson Education, Inc.
Slide 6- 63
Factor completely. 6x2 –33x – 63
a) 3(2x + 7)(x – 3)
b) 3(2x + 3)(x – 7)
c) 3(2x – 3)(x + 7)
d) 3(2x – 7)(x + 3)
6.3
Copyright © 2011 Pearson Education, Inc.
Slide 6- 64
Factor completely. 2x2 +3x – 20
a) (2x + 2)(x – 10)
b) (2x + 4)(x – 5)
c) (2x – 5)(x + 4)
d) (2x – 10)(x + 2)
6.3
Copyright © 2011 Pearson Education, Inc.
Slide 6- 65
Factor completely. 2x2 +3x – 20
a) (2x + 2)(x – 10)
b) (2x + 4)(x – 5)
c) (2x – 5)(x + 4)
d) (2x – 10)(x + 2)
6.3
Copyright © 2011 Pearson Education, Inc.
Slide 6- 66
6.4
1.
2.
3.
4.
Factoring Special Products
Factor perfect square trinomials.
Factor a difference of squares.
Factor a difference of cubes.
Factor a sum of cubes.
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Factoring Perfect Square Trinomials
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
Copyright © 2011 Pearson Education, Inc.
Slide 6- 68
Example 1a
Factor. 9a2 + 6a + 1
Solution
This trinomial is a perfect square because the first
and the last terms are perfect squares and twice the
product of their roots is the middle term.
9a2 + 6a + 1
The square root of 9a2 is 3a.
The square root of 1 is 1.
Twice the product of 3a and 1 is (2)(3a)(1) = 6a,
which is the middle term.
Answer 9a2 + 6a + 1 = (3a + 1)2
Copyright © 2011 Pearson Education, Inc.
Slide 6- 69
Example 1b
Factor. 16x2 – 56x + 49
Solution
This trinomial is a perfect square.
16x2 – 56x + 49
The square root of 16x2 is 4x.
The square root of 49 is 7.
Twice the product of 4x and 7 is (2)(4x)(7) = 56x,
which is the middle term.
Answer 16x2 – 56x + 49 = (4x – 7)2
Copyright © 2011 Pearson Education, Inc.
Use a2 – 2ab + b2 = (a – b)2,
where a = 4x and b = 7.
Slide 6- 70
Example 2a
Factor. 9x2 – 6xy + y2
Solution
9x2 – 6xy + y2 = (3x – y)2
Use a2 – 2ab + b2 = (a – b)2,
where a = 3x and b = y.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 71
Example 2b
Factor. 9a3b  42a 2b  49ab
Solution
9a3b  42a 2b  49ab  ab(9a 2  42a  49)
 ab(3a  7)
2
Copyright © 2011 Pearson Education, Inc.
Factor out the
monomial GCF, ab.
Use a2 – 2ab + b2 = (a – b)2,
where a = 3a and b = 7.
Slide 6- 72
Factoring a Difference of Squares
a2 – b2 = (a + b)(a – b)
Warning: A sum of squares a2 + b2 is
prime and cannot be factored.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 73
Example 3a
Factor. 9x2 – 16y2
Solution
This binomial is a difference of squares because
9x2 – 16y2 = (3x)2 – (4y)2 .
To factor it, we use the rule a2 – b2 = (a + b)(a – b).
a2 – b2 = (a + b)(a – b)
9x2 – 16y2 = (3x)2 – (4y)2 = (3x + 4y)(3x – 4y)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 74
Example 3b
Factor. 16n4 – 25
Solution
This binomial is a difference of squares, where
a = 4n2 and b = 5.
16n4 – 25 = (4n2 + 5)(4n2 – 5)
Copyright © 2011 Pearson Education, Inc.
Use a2 – b2 = (a + b)(a – b).
Slide 6- 75
Example 3c
Factor. 72a 9  98a 7
Solution
The terms in this binomial have a monomial GCF, 2a7.
72a 9  98a 7  2a7 (36a 2  49)
 2a 7 (6a  7)(6a  7)
Copyright © 2011 Pearson Education, Inc.
Factor out the GCF.
Factor 36a2 – 49, using
a2 – b2 = (a + b)(a – b)
with a = 6a and b = 7.
Slide 6- 76
Example 4
Factor. x 4  625
Solution
The binomial is a difference of squares, where a = x2
and b = 25.
x 4  625  ( x 2  25)( x 2  25)
 ( x 2  25)( x  5)( x  5)
Copyright © 2011 Pearson Education, Inc.
Factor x2 – 25, using
a2 – b2 = (a + b)(a – b)
with a = x and b = 5.
Slide 6- 77
Factoring a Difference of Cubes
a3 – b3 = (a – b)(a2 + ab + b2)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 78
Example 5a
Factor. 125x3 – 27
Solution
This binomial is a difference of cubes.
a3 – b3 = (a – b) (a2 + a b + b2)
125x3 – 27 = (5x)3 – (3)3 = (5x – 3)((5x)2 + (5x)(3) + (3)2)
= (5x – 3)(25x2 + 15x + 9)
Note: The trinomial may seem like a perfect
square. However, to be a perfect square, the
middle term should be 2ab. In this trinomial,
we only have ab, so it cannot be factored.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 79
Example 5b
Factor. 64a 4  27ab3
Solution
The binomial has a GCF, a.
3
3

a
(64
a

27
b
)
64a  27ab
4
3
 a(4a  3b)  (4a) 2  (4a)(3b)  (3b) 2 
 a(4a  3b)(16a 2  12ab  9b2 )
Copyright © 2011 Pearson Education, Inc.
Slide 6- 80
Factoring a Sum of Cubes
a3 + b3 = (a + b)(a2 – ab + b2)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 81
Example 6a
Factor. 125x3 + 64
Solution
This binomial is a difference of cubes.
a3 + b3 = (a + b) (a2  a b + b2)
125x3 + 64 = (5x)3 + (4)3 = (5x + 4)((5x)2  (5x)(4) + (4)2)
= (5x + 4)(25x2  20x + 16)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 82
Example 6b
Factor. 6x +162xy3
Solution
The terms in this binomial have a monomial GCF, 6x.
6x +162xy3 = 6x(1 + 27y3)
= 6x(1 + 3y)((1)2 – (1)(3y) + (3y)2)
= 6x(1 + 3y)(1 – 3y + 9y2)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 83
Factor completely. 4a2 – 20a + 25
a) (2a + 5)2
b) (2a – 5)2
c) (4a + 5)2
d) (4a – 5)2
6.4
Copyright © 2011 Pearson Education, Inc.
Slide 6- 84
Factor completely. 4a2 – 20a + 25
a) (2a + 5)2
b) (2a – 5)2
c) (4a + 5)2
d) (4a – 5)2
6.4
Copyright © 2011 Pearson Education, Inc.
Slide 6- 85
Factor completely. 9x2 – 49
a) (3x + 5)2
b) (3x + 7)(3x – 7)
c) (3x – 7)2
d) (7x + 3)(7x – 3)
6.4
Copyright © 2011 Pearson Education, Inc.
Slide 6- 86
Factor completely. 9x2 – 49
a) (3x + 5)2
b) (3x + 7)(3x – 7)
c) (3x – 7)2
d) (7x + 3)(7x – 3)
6.4
Copyright © 2011 Pearson Education, Inc.
Slide 6- 87
Factor completely. 2n2 + 24n + 72
a) 2(n + 6)2
b) 2(n + 6)(n – 6)
c) 2(n – 6)2
d) (2n + 6)(2n – 6)
6.4
Copyright © 2011 Pearson Education, Inc.
Slide 6- 88
Factor completely. 2n2 + 24n + 72
a) 2(n + 6)2
b) 2(n + 6)(n – 6)
c) 2(n – 6)2
d) (2n + 6)(2n – 6)
6.4
Copyright © 2011 Pearson Education, Inc.
Slide 6- 89
6.5
Strategies for Factoring
1. Factor polynomials.
Copyright © 2011 Pearson Education, Inc.
Factoring a Polynomial
To factor a polynomial, first factor out any monomial GCF, then
consider the number of terms in the polynomial. If the polynomial
has
I. Four terms, try to factor by grouping.
II. Three terms, determine whether the trinomial is a perfect square.
A. If the trinomial is a perfect square, then
consider its form.
1. If in form a2 + 2ab + b2, the factored form is (a + b)2.
2. If in form a2  2ab + b2, the factored form is (a  b)2.
B. If the trinomial is not a perfect square, consider its form.
1. If in form x2 + bx + c, find two factors of c whose
sum is b, and write the factored form as
(x + first number)(x + second number).
Copyright © 2011 Pearson Education, Inc.
Slide 6- 91
Factoring a Polynomial continued
2. If in form ax2 + bx + c, where a  1, then use trial and
error. Or, find two factors of ac whose sum is b; write these
factors as coefficients of two like terms that, when combined,
equal bx; and then factor by grouping.
III. Two terms, determine if the binomial is a difference of squares,
sum of cubes, or difference of cubes.
A. If given a binomial that is a difference of squares, a2 – b2,
the factors are conjugates and the factored form is
(a + b)(a – b). Note that a sum of squares cannot be factored.
B. If given a binomial that is a sum of cubes, a3 + b3, the
factored form is (a + b)(a2 – ab + b2).
C. If given a binomial that is a difference of cubes, a3 – b3, the
factored form is (a – b)(a2 + ab + b2).
Note: Always check to see if any of the factors can be factored.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 92
Example 1a
Factor. 12x2 – 8x – 15
Solution
There is no GCF.
Not a perfect square, since the first and last terms
are not perfect squares.
Use trial and error or grouping.
(x – 3)(12x + 5) = 12x2 + 5x – 36x – 15 No
(6x – 3)(2x + 3) = 12x2 + 18x – 6x – 9 No
(6x + 5)(2x – 3) = 12x2 – 18x + 10x – 15
12x2 – 8x – 15 Correct
Copyright © 2011 Pearson Education, Inc.
Slide 6- 93
Example 1b
Factor. 5x3 – 10x2 – 120x
Solution
5x(x2 – 2x – 24)
Factored out the monomial GCF, 5x.
Look for two numbers whose product is –24 and
whose sum is 2.
Product
Sum
(1)(24) = 24
4(6) = 24
1 + 24 = 23
4 + (6) = 2
Correct combination.
5x(x + 4)(x – 6)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 94
Example 1c
Factor. 8a4 – 72n2
Solution
8a4 – 72n2 = 8(a4 – 9n2)
Factor out the monomial GCF, 8.
a4 – 9n2 is a difference of squares
= 8(a2 – 3n)(a2 + 3n)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 95
Example 1d
Factor. 12y5 + 84y3
Solution
12y3(y2 + 7)
Factor out the monomial GCF, 12y3.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 96
Example 1e
Factor. 150x3y – 120x2y2 + 24xy3
Solution
6xy(25x2 – 20xy + 4y2) Factor out the monomial GCF, 6xy.
6xy(5x – 2y)2
Factor the perfect square trinomial.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 97
Example 1f
Factor. x5 – 2x3 – 27x2 + 54
Solution
No common monomial, factor by grouping.
(x5 – 2x3) + (– 27x2 + 54)
x3(x2 – 2) – 27(x2 – 2)
(x2 – 2)(x3 – 27)
Difference of cubes
(x2 – 2)(x – 3)(x2 + 3x + 9)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 98
Factor. 6x2 + 17x + 5
a) (6x + 1)(x + 5)
b) (3x + 1)(2x + 5)
c) (6x + 1)(x – 5)
d) (3x – 1)(2x – 5)
6.5
Copyright © 2011 Pearson Education, Inc.
Slide 6- 99
Factor. 6x2 + 17x + 5
a) (6x + 1)(x + 5)
b) (3x + 1)(2x + 5)
c) (6x + 1)(x – 5)
d) (3x – 1)(2x – 5)
6.5
Copyright © 2011 Pearson Education, Inc.
Slide 6- 100
Factor. 7y4 + 49y2
a) 7y(y3 + 7y)
b) 7y2(y2 + 49)
c) y2(7y2 + 49)
d) 7y2(y2 + 7)
6.5
Copyright © 2011 Pearson Education, Inc.
Slide 6- 101
Factor. 7y4 + 49y2
a) 7y(y3 + 7y)
b) 7y2(y2 + 49)
c) y2(7y2 + 49)
d) 7y2(y2 + 7)
6.5
Copyright © 2011 Pearson Education, Inc.
Slide 6- 102
6.6
Solving Quadratic Equations by Factoring
1. Use the zero-factor theorem to solve equations
containing expressions in factored form.
2. Solve quadratic equations by factoring.
3. Solve problems involving quadratic equations.
4. Use the Pythagorean theorem to solve problems.
Copyright © 2011 Pearson Education, Inc.
Zero-Factor Theorem
If a and b are real numbers and ab = 0, then a = 0
or b = 0.
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Slide 6- 104
Example 1
Solve. (x + 4)(x + 5) = 0
Solution
According to the zero-factor theorem, one of the two
factors, or both factors, must equal 0.
x+4=0
or
x + 5 = 0 Solve each equation.
x = 4
x = 5
Check
For x = 4:
For x = 5:
(x + 4)(x + 5) = 0
(x + 4)(x + 5) = 0
(4 + 4)(4 + 5) = 0
(5 + 4)(5 + 5) = 0
0(1) = 0
(1)(0) = 0
Copyright © 2011 Pearson Education, Inc.
Slide 6- 105
Solving Equations with Two or More Factors
Equal to 0
To solve an equation in which two or more factors
are equal to 0, use the zero-factor theorem.
1. Set each factor equal to zero.
2. Solve each of those equations.
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Slide 6- 106
Example 2
Solve.
a. y(5y + 2) = 0
b. x(x + 2)(5x – 4) = 0
Solution
a. y(5y + 2) = 0
b. x( x  2)(5 x  4)  0
y = 0 or 5y + 2 = 0
x  0 x  2  0 5x  4  0
5y = 2
x  2
5x  4
This equation
is already
solved.
2
y
5
4
x
5
To check, we verify that the solutions satisfy the
original equations.
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Slide 6- 107
Quadratic equation in one variable: An equation that
can be written in the form ax2 + bx + c = 0, where a,
b, and c are all real numbers and a  0.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 108
Solving Quadratic Equations Using Factoring
To solve a quadratic equation using factoring,
1. Write the equation in standard form
(ax2 + bx + c = 0).
2. Write the variable expression in factored form.
3. Use the zero-factor theorem to solve.
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Slide 6- 109
Example 3
Solve. 2x2 – 5x – 3 = 0
Solution
The equation is in standard form, so we can factor.
2x2 – 5x – 3 = 0
(2x + 1)(x – 3) = 0 Use the zero-factor theorem to solve.
2x + 1 = 0 or x – 3 = 0
2x  1
1
x
2
x3
To check, we verify that the
solutions satisfy the original
equations.
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Slide 6- 110
Example 4
Solve. x3 + 4x2 – 21x = 0
Solution
x3 + 4x2 – 21x = 0 Factor out the monomial GCF, x.
x(x2 + 4x – 21) = 0 Use the zero-factor theorem to solve.
x(x – 3)(x + 7) = 0
x=0
or
x–3=0
or
x+7=0
x3
x  7
To check, we verify that the
solutions satisfy the original
equations.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 111
Example 5a
Solve. 6y2 + 11y = 10 + 4y
Solution
Write the equation in standard form.
6y2 + 11y = 10 + 4y
6y2 + 7y = 10
Subtract 4y from both sides.
6y2 + 7y – 10 = 0
Subtract 10 from both sides.
(6y – 5)(y + 2) = 0
Factor.
6y – 5 = 0 or y + 2 = 0 Use the zero-factor theorem.
6y  5
5
y
6
y  2
Copyright © 2011 Pearson Education, Inc.
Slide 6- 112
Example 5b
Solve. y(y – 7) = –12
Solution
Write the equation in standard form.
y ( y  7)  12  0
y 2  7 y  12  0
( y  3)( y  4)  0
y 3  0
y3
or
y4 0
y4
Copyright © 2011 Pearson Education, Inc.
Slide 6- 113
Example 6
The product of two consecutive odd natural numbers is
323. Find the numbers.
Understand Odd numbers are 1, 3, 5,…
Let x = the first odd number
Let x + 2 = consecutive odd number
The word product means that two numbers are
multiplied to equal 323.
Plan Translate to an equation, then solve.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 114
continued
Execute
x(x + 2) = 323
x(x + 2) – 323 = 0
x2 + 2x – 323 = 0
(x + 19)(x – 17) = 0
x + 19 = 0
x – 17 = 0
x = –19
x = 17
Answer Because –19 is not a natural number and 17
is, the first number is 17. This means that the
consecutive odd natural number is 19.
Check 17 and 19 are consecutive odd natural numbers
and their product is 323.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 115
The Pythagorean Theorem
Given a right triangle, where a and b represent the
lengths of the legs and c represents the length of
the hypotenuse, then a2 + b2 = c2.
c (hypotenuse)
a (leg)
b (leg)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 116
Example 9
Find the length of the missing side.
Solution
Use the Pythagorean theorem, 15
a2 + b2 = c2
152 + 362 = c2
Substitute.
36
2
225 + 1296 = c
Simplify exponential forms.
1521 = c2
Add.
c2 – 1521 = 0
Standard form.
(c – 39)(c + 39) = 0 Factor.
c – 39 = 0
or c + 39 = 0
c = 39
or
c = –39
Only the positive solution is sensible.
Copyright © 2011 Pearson Education, Inc.
?
Slide 6- 117
Solve. x2 = 6x – 8
a) 2 and 4
b) 2 and 4
c) 2 and 4
d) 1 and 8
6.6
Copyright © 2011 Pearson Education, Inc.
Slide 6- 118
Solve. x2 = 6x – 8
a) 2 and 4
b) 2 and 4
c) 2 and 4
d) 1 and 8
6.6
Copyright © 2011 Pearson Education, Inc.
Slide 6- 119
Solve. One natural number is four times
another. The product of the two numbers is
900. Find the larger number.
a) 15
b) 30
c) 35
d) 60
6.6
Copyright © 2011 Pearson Education, Inc.
Slide 6- 120
Solve. One natural number is four times
another. The product of the two numbers is
900. Find the larger number.
a) 15
b) 30
c) 35
d) 60
6.6
Copyright © 2011 Pearson Education, Inc.
Slide 6- 121
Find the length of the hypotenuse.
?
14
a) 15
48
b) 46
c) 50
d) 62
6.6
Copyright © 2011 Pearson Education, Inc.
Slide 6- 122
Find the length of the hypotenuse.
?
14
a) 15
48
b) 46
c) 50
d) 62
6.6
Copyright © 2011 Pearson Education, Inc.
Slide 6- 123
6.7
Graphs of Quadratic Equations and
Functions
1. Graph quadratic equations in the form y = ax2 + bx + c.
2. Graph quadratic functions.
Copyright © 2011 Pearson Education, Inc.
Quadratic equation in two variables: An equation that
can be written in the form y = ax2 + bx + c, where a, b,
and c are real numbers and a  0.
Axis of symmetry: A line that divides a graph
into two symmetrical halves.
Vertex: The lowest point on a
parabola that opens up or the
highest point on a parabola that
opens down.
axis of symmetry
x = 0 (y-axis)
vertex
(0, 0)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 125
Graphing Quadratic Equations
To graph a quadratic equation,
1. Find ordered pair solutions and plot them in
the coordinate plane. Continue finding and
plotting solutions until the shape of the
parabola can be clearly seen.
2. Connect the points to form a parabola.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 126
Example 1a
Graph. y = 2x2 + 1
Solution
Complete a table of solutions.
x
y
Plot the points.
2
9
Connect the points.
1
3
0
1
1
3
2
9
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Slide 6- 127
Example 1b
Graph. y = x2 – 2x – 3
Solution
Complete a table of solutions.
x
y
Plot the points.
2
5
Connect the points.
1
0
0
3
1
4
2
3
3
0
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Slide 6- 128
Example 1c
Graph. y = 3x2 + 4
Solution
Complete a table of solutions.
x
y
2
8
Plot the points.
Connect the points.
1
1
0
4
1
1
2
8
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Slide 6- 129
Opening of a Parabola
Given an equation in the form y = ax2 + bx + c, if
a > 0, then the parabola opens upward; if a < 0,
then the parabola opens downward.
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Slide 6- 130
Graphing Quadratic Functions
To graph a quadratic function,
1. Find enough ordered pairs by evaluating the
function for various values of x so that when
those ordered pairs are plotted, the shape of the
parabola can be clearly seen.
2. Connect the points to form the parabola.
Copyright © 2011 Pearson Education, Inc.
Slide 6- 131
Example 3a
Graph. f(x) = 2x2 – 1
Solution
Complete a table of solutions.
x
y
Plot the points.
2
7
1
1
0
1
1
1
2
7
Connect the points.
This parabola
opens upward since
a > 0.
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Slide 6- 132
Example 3b
Graph. f(x) = 2x2 + 8x – 1
Solution
Complete a table of solutions.
x
y
1
11
0
1
1
5
2
7
3
5
4
1
Plot the points.
Connect the points.
This parabola
opens downward
since a < 0.
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Slide 6- 133
Example 4a
Determine whether the graph is the graph of a
function. Give the domain and range.
Solution
This is a function because any
vertical line intersects the
graph in at most one point.
Domain: all real numbers or (, )
Range: {y|y ≥ 3} or [3, )
Copyright © 2011 Pearson Education, Inc.
Slide 6- 134
Example 4b
Determine whether the graph is the graph of a
function. Give the domain and range.
Solution
This is NOT a function
because a vertical line can be
drawn that intersects the graph
in more than one point.
Domain: {x|x ≥ 1} or [1, )
Range: all real numbers or (, )
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Slide 6- 135
Graph. y = x2 – 2
6.7
a)
b)
c)
d)
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Slide 6- 136
Graph. y = x2 – 2
6.7
a)
b)
c)
d)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 137
Graph. f(x) = x2 + 2x – 2
6.7
a)
b)
c)
d)
Copyright © 2011 Pearson Education, Inc.
Slide 6- 138
Graph. f(x) = x2 + 2x – 2
6.7
a)
b)
c)
d)
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Slide 6- 139