6.1 Greatest Common Factors and Factoring by
R.1 Fractions
Grouping
Objectives
Factoring
6
1. Factor out the greatest common factor.
2. Factor by grouping.
Section16.1,
1
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1-1
Factor Out the Greatest Common Factor
Writing a polynomial as a product of two or more simpler
polynomials is called factoring the polynomial.
3x(5x – 2) = 15x2 – 6x
Multiplying
15x2 – 6x = 3x(5x – 2)
Factoring
Section26.1,
2
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1-2
Factoring Out the Greatest Common Factor
The first step in factoring is to find the greatest common
factor (GCF) – the largest term that divides each term of
the polynomial. The GCD is a factor of all the terms of
the polynomial.
Factor out the greatest common factor.
7m – 21
Notice that both multiplying and factoring use the
distributive property, but in opposite directions.
Factoring “undoes,” or reverses, multiplying.
7m 21 7 m 7 3
7 is the GCF.
7 m 3
7 can by factored from each term.
Section36.1,
3
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1-3
Section46.1,
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Factoring Out the Greatest Common Factor
Example
Example 3 Factoring Out a Binomial Factor
Factor out the greatest common factor.
Factor out the greatest common factor.
a. 22c – 66d
22c 66d
11 2c 11 6d
11 is the GCF.
x a b
11 2c 6d
b. 9q + 17r
8a
8a
2
a b x
a b y
a b x
y
There is no other common factor than 1.
c. 8a3 – 8a2 + 48a
3
y a b
48a
8a is the GCF
8a a
8a a
2
2
8a
a 6
a
8a 6
8a can be
factored from
each term.
Section56.1,
5
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1-5
3a b c
2
2a b c
3
a b c
a b c
a b c
2
2
2
3
a b c
2
2 b c
3 2 b c
3 2b 2c
Section66.1,
6
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1-6
1
Example 4 Factoring Out a Negative Common Factor
Factoring by Grouping
When the coefficient of the term of greatest degree is
negative, it is sometimes preferable to factor out the –1
that is understood along with the GCF.
Factor out the greatest common factor.
3w 4
Factor only
the 3 out.
Or factor
the –3 out.
9w 3 15
3
3
3w
4
9w
3
15
w
3 w
Either is
correct.
w4
3 3w 3
4
3
3w
4
3
4
3
3 w
Many polynomials have no greatest common factor
other than the number 1. Some of these can be
factored using the distributive property if those terms
with a common factor are grouped together. Consider
the polynomial:
3w
3
5
5
1. The first two terms have a 5 in common, whereas,
2. The last two terms have an x in common.
3w
3
3 5
Applying the distributive property, we have
5
Section76.1,
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Section86.1,
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1-8
Factoring by Grouping
Factoring by Grouping
This last expression can be thought of as having two
terms,
and
.
.
Applying the distributive property again to factor (x + y)
from each term:
Section96.1,
9
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1-9
Example 5
Example 9
Factoring by Grouping
Factor ax – by + ay – bx.
ax by
ay
bx
ax ay
ax by
ay
bx
by
bx
y
b y
x
a x
y
b x
y
y a b
ax bx
x a b
a b x
Factoring by Grouping
Factor by grouping. y 7
a x
x
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1-10
ay
by
y a b
y7
y 6z y 5
y 4z
y 6z y 5
y4 y3
y4
Often there is
more than one
way to group. It
does not matter
which one we
use.
y
Section
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1-11
y
4
y
4
y 4z
y 2z y
y3
y
y
2
z
y 2z
y
z
z y
y
y
2
z
z
1
Do not stop at the step 4 2
y y y z
y z .
This expression is not in factored form because it is a
sum of two terms, y2(y – z) and (y – z), not a product.
Section
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2
6.2 Factoring Trinomials
R.1 Fractions
Objectives
1. Factor trinomials when the coefficient of the
second-degree term is 1.
2. Factor trinomials when the coefficient of the
second-degree term is not 1.
3. Use an alternative method for factoring
trinomials.
4. Factor by substitution.
The product of two binomials sometimes gives a trinomial.
For example:
x 5 x 7
x2
x
2
7 x 5 x 35
2 x 35
So, we have two processes that “undo” each other.
Multiplying
Product
Factored
form
Factoring
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1-13
Factoring Trinomials When the Coefficient of the Squared Term is 1
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1-14
Factoring Trinomials When the Coefficient
of the Squared Term is 1
Multiplying binomials uses the FOIL method, and factoring
involves using the FOIL method backwards.
Product of x and x
is
x2.
F
L Product of 5 and –7
is
–35.
Sum of the product of outer and inner terms
O
I
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1-15
Factoring Trinomials in x2 + bx + c Form
Example 1
Step 1
–32 + 1 = –31
32 + (–1) = 31
–8(4)
8(–4)
You should always check your answer by multiplying the
factors to see if you get the original polynomial.
Write sums of those
pairs of integers
32(–1)
–16(2)
Continued.
The required numbers are –8 and 4, so
Step 2
–32(1)
16(–2)
Factoring Trinomials in x2 + bx + c Form
Factor the polynomial x2 – 4x – 32.
Find pairs of numbers
whose product is –32.
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1-16
Guidelines for Factoring Trinomials
1. If the last term is positive, the factors will have the form
–16 + 2 = –14
( ___ + ___ ) ( ___ + ___ ) or ( ___ – ___ ) ( ___ – ___ )
16 + (–2) = 14
–8 + 4 = –4
8 + (–4) = 4
The + or – sign is determined by the coefficient of the middle term.
Coefficient
of middle term
2. If the last term is negative, the factors will have the form
( ___ + ___ ) ( ___ – ___ ) or ( ___ – ___ ) ( ___ + ___ )
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1-17
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3
Factoring a Trinomial With A Common Factor
Example 4
Factor 3q3 + 12q2m – 36qm2.
Factoring Trinomials When the Coefficient
of the Squared Term is Not 1
Example 5
Start by factoring out the greatest common factor, 3q.
Factor 3x2 – x – 2.
To factor q2 + 4qm – 12m2 , look for two expressions
whose product is –12m2 and whose sum is 4m.
The quantities 6m and –2m have the necessary product
and sum, so 3q3 + 12q2m – 36qm2 = 3q(q + 6m)(q – 2m).
Section
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1-19
Factoring Trinomials When the Coefficient
of the Squared Term is Not 1
Continued.
Factor
3x2 –
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1-20
Factoring Trinomials When the Coefficient
of the Squared Term is Not 1
Continued.
x – 2.
Since the first two terms have the product
3x2
, we have
Factor 3x2 – x – 2.
Here are the possibilities, each of which produces the
correct first and last term, 3x2 and –2, respectively.
Since the product of the last two terms must be –2, we
must place the1 and 3 in the remaining blanks.
Section
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1-21
Factoring Trinomials When the Coefficient
of the Squared Term is Not 1
Example 6
Factor 3x2 – 10x – 8.
Factoring Trinomials When the Coefficient
of the Squared Term is Not 1
Continued.
The trinomial 3x2 – 10x – 8 has the form ax2 + bx + c,
where a = 3, b = –10, and c = –8.
Step 1 The product ac is 3(–8) = –24.
Step 2
Section
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1-22
We must find two numbers whose product is –24
and whose sum is –10.
Section
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1-23
Listing all the pairs
of numbers whose
product is –24 to find
a pair whose sum is
–10, only 2 and –12
have a sum of –10.
Section
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1-24
4
Factoring Trinomials When the Coefficient
of the Squared Term is Not 1
Factoring Trinomials When the Coefficient
of the Squared Term is Not 1
Continued.
Section
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1-25
Example 8
Section
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1-26
Example 10 Factoring a Polynomial Using Substitution
Factor –5x2 + 34x + 7.
Sometimes we can factor more complicated problems by
substituting a variable for an expression.
While we could try to factor this trinomial directly, if is
helpful to first factor our –1.
5x 2
34 x 7
1 5x 2
Factor 2(x – 2)2 – 5(x – 2) – 12.
34 x 7
We can proceed as in earlier examples.
5 x 2 34 x 7
1 5 x 2 34 x 7
1 5x 1 x 7
2 x 2
2
5 x 2
2y 2
12
The answer can also be expressed as
(–5x – 1)(x – 7) or (5x+ 1)(–x + 7).
2y
5y 12
3 y
4
Section
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1-27
Factoring a Polynomial Using Substitution
Section
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1-28
Example 11 Factoring a Trinomial in ax4 + bx2 + c Form
Continued.
Factor 10m4 – 21m2 – 10.
Now, we replace y with x – 2 to get
2 x 2
2
5 x 2
12
2x
2 x 2
3
x 2
4
4 3 x 6
The variable m appears in powers in which the larger
exponent is twice the smaller exponent.
We can gain substitution, letting the variable y stand for m2.
2x 1 x 6
10m4
21m2 10 10 m2
10y 2
CAUTION
Remember to make the final substitution of (x – 2) for y.
2y
21m2 10
21y 10
5 5y
2m 2
Section
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1-29
2
Substitute.
2
5 5m 2
Factor.
2
y = m2
Section
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5
6.3 Special Factoring
R.1 Fractions
The Difference of Squares
Objectives
1.
2.
3.
4.
Factor a difference of squares.
Factor a perfect square trinomial.
Factor a difference of cubes.
Factor a sum of cubes.
Difference of Squares
x2 – y2 = (x + y)(x – y)
Section
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Example 1
EXAMPLE
Factoring Differences of Squares
Section
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1-32
Continued. 1
EXAMPLE
Factoring Differences of Squares
Factor each polynomial.
Factor each polynomial.
(b) 9g2 – 16
(a) 2n2 – 50
9g2 – 16 = (3g)2 – (4)2 = (3g + 4)(3g – 4)
There is a common factor of 2.
A2
2n2 – 50 = 2(n2 – 25)
Factor out the common factor.
= 2(n + 5)(n – 5) Factor the difference of squares.
(c)
4h2
– (w +
– B2
(A
+ B)(A – B)
5)2
2
2
4h2 – (w + 5)2 = (2h) – (w + 5) = (2h + w + 5)(2h – [w + 5])
= (2h + w + 5)(2h – w – 5)
A2
–
B2
(A +
B)
Slide 33 of
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1-33
Caution
(A –
B)
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1-34
Section 6.1, Slide 33
Section 6.1, Slide 34
Perfect Square Trinomial
CAUTION
Perfect Square Trinomial
Assuming no greatest common factor except 1, it is not possible to
factor (with real numbers) a sum of squares, such as x2 + 16.
x2 + 2xy + y2 = (x + y)2
x2 – 2xy + y2 = (x – y)2
Section
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Section
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6
Example 2
EXAMPLE
Factoring Perfect Square Trinomials
Factor each polynomial.
(a) 9g2 – 42g + 49
Factoring Perfect Square Trinomials
Continued.
Factor each polynomial.
(b) 25x2 + 60xy + 64y2
Here 9g2 = (3g)2 and 49 = 72. The sign of the middle term is
negative, so if 9g2 – 42g + 49 is a perfect square trinomial,
the factored form will have to be
(3g – 7)2.
Take twice the product of the two terms to see if this is correct.
2(3g)(–7) = –42g
This is the middle term of the given trinomial, so
9g2 – 42g + 49 = (3g –7)2.
If this is a perfect square trinomial, it will equal (5x + 8y)2.
By the pattern described earlier, if multiplied out, this squared
binomial has a middle term of 2(5x)(8y), which does not
equal 60xy. Verify that this trinomial cannot be factored by
the methods of the previous section either. It is prime.
Section
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1-37
Continued.
Factoring Perfect Square Trinomials
Factor each polynomial.
(c) (n –
4)2
+ 18(n – 4) + 81 = [ (n – 4) + 9
Section
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38
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1-38
Factoring Perfect Square Trinomials
EXAMPLE
Continued.2
Factor each polynomial.
(d) c2 – 6c + 9 – h2
]2
Since there are four terms, we will use factoring by grouping.
The first three terms here form a perfect square trinomial.
Group them together, and factor as follows.
= (n + 5)2,
since 2(n – 4)9 = 18(n – 4), the middle term.
(c2 – 6c + 9) – h2 = (c – 3)2 – h2
The result is the difference of squares. Factor again to get
= (c – 3 + h)(c – 3 – h).
Section
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1-39
Section
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Example 3
Difference of Cubes
Factoring Difference of Cubes
Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).
Difference of Cubes
(a) a3 – 125
x3 – y3 = (x – y)(x2 + xy + y2)
a3
53
a 5 a2
5a 52
a 5 a2
5a 25
a3
Check:
= (a –
–5a
Section
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1-41
5)(a2
–125
+ 5a + 25)
Opposite of the product of the cube
roots gives the middle term.
Section
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42
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1-42
7
Continued.
Factoring Difference of Cubes
Factoring Difference of Cubes
Continued.
Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).
Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).
(c) 64m3 – 27n3
(b) 8g3 – h3
2g
2g
2g
3
h3
h
4m
2
2g
h 4g
2
h2
2g h
2gh h
3
4m 3n
3n
3
4m
2
4m 3n
3n
2
2
4m 3n 16m 2 12mn 9n 2
Section
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1-43
Section
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1-44
Sum of Cubes
Note on Signs
NOTE
Sum of Cubes
The sign of the second term in the binomial factor of a sum or difference
of cubes is always the same as the sign in the original polynomial.
In the trinomial factor, the first and last terms are always positive;
the sign of the middle term is the opposite of the sign of the second term
in the binomial factor.
x3 + y3 = (x + y)(x2 – xy + y2)
Difference of Cubes
x3 – y3 = (x – y)(x2 + xy + y2)
Sum of Cubes
x3 + y3 = (x + y)(x2 – xy + y2)
Section
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1-45
Factoring Sums of Cubes
Continued.
Example 4
Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2).
(a) n3 + 8
n3
Factoring Sums of Cubes
Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2).
23
n 2 n2
2n 22
n 2 n2
2n 4
(b) 64v3 + 27g3
Section
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1-46
4v
3
3g
(c) 2k3 + 250
3
4v
2
4v
3g
4v 3g
4v
3g 16v 2 12gv
3g
2 k 3 125
2 k3
53
2 k
5 k2
5k
25
2
9g 2
Section
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1-47
Section
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1-48
8
6.4 A General Approach
to Factoring
R.1 Fractions
Factoring Summary
Objectives
Special Types of Factoring
1.
2.
3.
4.
Difference of Squares
x2 – y2 = (x + y)(x – y)
Perfect Square Trinomial
x2 + 2xy + y2 = (x + y)2
x2 – 2xy + y2 = (x – y)2
Difference of Cubes
x3 – y3 = (x – y)(x2 + xy + y2)
Sum of Cubes
x3 + y3 = (x + y)(x2 – xy + y2)
Factor out any common factor.
Factor binomials.
Factor trinomials.
Factor polynomials of more than three terms.
Section
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1-49
Section
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1-50
Factoring Out a Common Factor
Example 1
Factoring a Polynomial
A polynomial is completely factored when it is written as a
product of prime polynomials with integers coefficients.
This step is always the same, regardless of the number of
terms in the polynomial.
Factor each polynomial.
(a)
9p
(b)
2
45
9( p 5)
8m p
2
GCF
9
4mp
4mp(2mp 1)
(c)
5 x (a
b ) y (a
(a
b)
b )(5 x
y)
Factor out a b.
Slide 51 of
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Section
6.1,of
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52
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1-52
Section 6.1, Slide 51
Example 2
Factoring Binomials
Factoring Binomials
Use one of the rules to factor each binomial if possible.
(a)
64m2 9n 2
(8m)2 (3n )2
Difference of squares.
(8m 3n )(8m 3n )
(b)
8 p3
8 p3
27
(2p 3)[(2p)2
(2p 3)(4 p
Section
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53
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1-53
2
(2p)3 ; 27
33
(2p)(3) 32 ]
6 p 9)
Section
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54
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1-54
9
Factoring Binomials
Continued.
Factoring Trinomials
Use one of the rules to factor each binomial if possible.
(c) 1000m3
1
(10m)3 13 Sum of cubes
(10m 1)[(10m)2 (10m)(1) 12 ]
(10m 1)(100m2 10m 1)
(d) 25m2
121 is prime.
It is the sum of squares.
The binomial 25m2 + 625 is the sum of squares. It can be factored,
however as 25(m2 + 25) because it has a common factor, 25.
Section
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55
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1-55
Example 3
Factoring Trinomials
(c)
y2
Factor each trinomial.
(d)
10 p 25
( p 5)2
(b) 49z 2
Factoring Trinomials
Continued.
Factor each trinomial.
(a) p2
Section
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56
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1-56
2( p )(5) 10 p
42z 9
(7z 3)2
2(7z )(3)
5y 6
( y 6)( y 1)
42z
(e)
The numbers –6 and 1 have a
product of –6 and a sum of –5.
2k 2 k 6
(2k 3)(k
28z 2
2)
6z 10
2(14z2 3z 5)
2(7z 5)(2z 1)
Section
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1-57
Section
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1-58
Factoring Polynomials with More than Three Terms
Factoring Polynomials with More than Three Terms
Example 4
Factor each polynomial.
Factor each polynomial.
(a)
20k 3 4k 45k 9
(20k 3 4k 2 ) (45k
9)
4k 2 (5k 1) 9(5k 1)
(5k 1)(4k 2
(5k 1)(2k
(b)
4a
2
4a
(4a2
(2a
(2a
9)
3)(2k
(c) 8m3
Factor each group.
5k + 1 is a common factor.
3)
4m2
n3
n2
Group the terms.
(8m3
n3 ) (4m2
n2 )
(2m n )(4m2
2mn n 2 )(2m n )(2m n )
(2m n )(4m2
2mn n 2
Difference of squares
2
1 b
Group the first three terms.
4a 1) b2
Perfect square trinomial
1)2 b2
1 b )(2a 1 b ) Difference of squares
Section
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59
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1-59
Rearrange and group
the terms.
Factor each group.
2m n )
Factor out 2m – n.
Section
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1-60
10
6.5 Solving Equations
by Factoring
R.1 Fractions
Learn to use the zero-factor property.
Objectives
1. Learn and use the zero-factor property.
2. Solve applied problems that require the zerofactor property.
3. Solve a formula for a specified variable, where
factoring is necessary.
The two tools used to solve linear equations, adding the
same number to both sides and multiplying both sides by
the same nonzero number, are not enough to solve higherdegree equations. However, our number system has the
following property which provides another tool necessary
to solve higher-degree equations.
It is important to note that the Zero-Factor Property only
works for a product equal to zero.
Section
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61
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1-61
Section
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62
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1-62
Using the Zero-Factor Property
Using the Zero-Factor Property
Example 1
Continued.
Solve (x – 3)(x – 2) = 0.
The Zero-Factor Property says that this statement will be
true if either (x – 3) or (x – 2) is zero.
Check the solutions 2 and 3 by substitution in the original
equation.
If x = 2 then
x 3
x
0
3
or
x
2
x
0
2
x 3 x 2
0
2 3 2 2
0
10
If x = 3 then
?
0
x 3 x 2
0
3 3 3 2
0
0 1
0
True
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?
True
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Quadratic Equations and Solving Quadratic Equations
Solving a Quadratic Equation by Factoring
Example 2
Solve 2x2 + 5x = 3.
Step 1
2x 2
Step 2
2x 1 x 3
2x 1
Step 3
Step 4
2x
x
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2x 2 5 x 3
5x 3 0
1
0 or
x
3
Standard form
0
Factor
x 3
0
Zero-factor property
Solve each equation
1
2
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Solving Quadratic Equations by Factoring
Solving a Quadratic Equation by Factoring
Continued.
Example 3
Solve 2x2 + 5x = 3.
Step 5
2
Solve 9x2 = 24x – 16.
Check each solution in the original equation.
1
If x
, then
If x
3, then
2
2
2x 5 x 3
2x 2 5 x 3
1
2
Note there are
two solutions.
2
1
3 ?
2
1 5
+
3 ?
2 2
6
3 True
2
2
5
3
2
5
3
3
18 15
3
9x 2
24 x 16
2
3x 4
?
3
3
0
3x 4
0
3x
4
4
3
?
x
True
There is only one distinct
solution, which we call a
double solution because
the trinomial is a perfect
square.
0
The solution is
4
.
3
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Solving an Equation of Degree 3
Solving a Quadratic Equation with a Missing Constant Term
Example 7
Example 4
Solve x3 + 2x2 – 9x – 18 = 0.
Solve 7x2 + 42x = 0.
Comparing this quadratic equation to the standard form
ax2 + bx + c = 0, we see that c = 0. We can still apply the
zero-factor property.
7 x 2 42 x 0
7x x
7x
x
6
0
0 or x
6
The solutions are {–6,0}.
2
9 x
2
0
2 x2
9
0
2 x 3 x 3
0
x
x
0
0 or x 6
x3 + 2x2 – 9x – 18 = 0
x2 x
Factor by grouping
The difference of two squares
Using the zero-factor property to extended to include the
three variable factors.
CAUTION
Remember to include 0 as a solution.
The solution set is {–3, –2, 3}.
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Using a Quadratic Function in an Application
Example 9
The manufacturer of a weather radio has determined
that the revenue R from the sale of radios can be
approximated by using R = –100p2 + 1300p where p
is the selling price of each radio.
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Using a Quadratic Function in an Application
Continued.
100 p2 1300 p
100 p
2
p 5 p 8
What is the selling price of the radios if the revenue is
$4000?
p 5
0 or
p
We know that R = 4000, so we have
–100p2 + 1300p = 4000.
4000
1300 p 4000 0
p2 13 p 40 0
p 8
5 or
p
0
0
8
Standard form
Divide each term by –100
Factor
Zero-factor property
Solve each equation
If the weather radios are sold for either $5 or $8, the
revenue will be $4000.
We now solve this quadratic equation for p.
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