College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson P Prerequisites.
Download ReportTranscript College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson P Prerequisites.
College Algebra
Fifth Edition
James Stewart
Lothar Redlin
Saleem Watson
P
Prerequisites
P.7
Factoring
Factoring
We use the Distributive Property to expand algebraic expressions. We sometimes need to reverse this process (again using the Distributive Property) by: • Factoring an expression as a product of simpler ones.
Factoring
For example, we can write:
x
2 (
x
2)(
x
2)
Expanding
• We say that
x
– 2 and
x
+ 2 are factors of
x
2 – 4.
Common Factors
Factoring
The easiest type of factoring occurs when:
• The terms have a common factor.
E.g. 1 —Factoring Out Common Factors
Factor each expression.
(a) 3
x
2 – 6
x
(b) 8
x
4
y
2 + 6
x
3
y
3 – 2
xy
4
E.g. 1 —Common Factors
Example (a) The greatest common factor of the terms 3
x
2 and –6
x
is 3
x.
So, we have: 3
x
2 – 6
x
= 3
x
(
x
– 2)
E.g. 1 —Common Factors
We note that:
Example (b) • 8, 6, and –2 have the greatest common factor 2. •
x
4 ,
x
3 , and
x
have the greatest common factor
x.
•
y
2 ,
y
3 , and
y
4 factor
y
2 .
have the greatest common
E.g. 1 —Common Factors
Example (b) So, the greatest common factor of the three terms in the polynomial is 2
xy
2 .
Thus, we have: 8
x
4
y
2 + 6
x
3
y
3 – 2
xy
4 = (2
xy
2 )(4
x
3 ) + (2
xy
2 )(3
x
2
y
) + (2
xy
2 )( –
y
2 ) = 2
xy
2 (4
x
3 + 3
x
2
y
–
y
2 )
E.g. 2 —Factoring Out a Common Factor
Factor : (2
x
+ 4)(
x
– 3) – 5(
x
– 3)
E.g. 2 —Factoring Out a Common Factor
The two terms have the common factor
x
– 3.
(2
x
+ 4)(
x
– 3) – 5(
x
– 3) = [(2
x
+ 4) – 5](
x
– 3) = (2
x
– 1)(
x
– 3) (Distributive Property) (Simplify)
Factoring Trinomials
Factoring x 2 + bx + c
In order to factor a trinomial of the form
x
2 +
bx
+
c
, we note that: (
x
+
r
)(
x
+
s
) =
x
2 + (
r
+
s
)
x
+
rs
• So, we need to choose numbers
r
and
s
so that
r
+
s
=
b
and
rs
=
c
.
E.g. 3 —Factoring x 2 + bx + c by Trial and Error
Factor:
x
2
+ 7
x
+ 12
• We need to find two integers whose product is 12 and whose sum is 7. • By trial and error, we find that they are 3 and 4. • Thus, the factorization is:
x
2 + 7
x
+ 12 = (
x
+ 3)(
x
+ 4)
Factoring ax 2 + bx + c
To factor a trinomial of the form
ax
2 +
bx
+
c
with
a
≠ 1, we look for factors of the form
px
+
r
and
qx
+
s
:
ax
2 +
bx
+
c
= (
px
+
r
)(
qx
+
s
) =
pqx
2 + (
ps
+
qr
)
x
+
rs
Factoring ax 2 + bx + c
Thus, we try to find numbers
p
,
q
,
r
, and
s
such that:
pq
=
a rs
=
c ps
+
qr
=
b
• If these numbers are all integers, then we will have a limited number of possibilities to try for
p
,
q
,
r
, and
s
.
E.g. 4 —Factoring ax 2 + bx + c by Trial and Error
Factor: 6
x
2
+ 7
x
– 5
• We can factor: 6 as 6 ∙ 1 or 3 ∙ 2 –5 as –25 ∙ 1 or –5 ∙ (–1) • By trying these possibilities, we arrive at the factorization 6
x
2 + 7
x
– 5 = (3
x
+ 5)(2
x
– 1)
E.g. 5 —Recognizing the Form of an Expression
Factor each expression.
(a)
x
2 – 2
x
– 3 (b) (5
a
+ 1) 2 – 2(5
a
+ 1) – 3
E.g. 5 —Recognizing the Form
x
2
– 2
x
– 3
Example (a)
= (
x
– 3)(
x
+ 1)
(Trial and error)
E.g. 5 —Recognizing the Form
Example (b) This expression is of the form __ 2 – 2__ – 3 where __ represents 5
a
+ 1.
E.g. 5 —Recognizing the Form
Example (b) This is the same form as the expression in part (a).
So, it will factor as ( __ – 3)( __ + 1).
• (5
a
= [(5 + 1) 2
a
– 2(5
a
+ 1) – 3 + 1) – 3][(5
a
+ 1) + 1] = (5
a
– 2)(5
a
+ 2)
Special Factoring Formulas
Special Factoring Formulas
Some special algebraic expressions can be factored using the following formulas. • The first three are simply Special Product Formulas written backward.
E.g. 6 —Factoring Differences of Squares
Factor each polynomial.
(a) 4
x
2 – 25 (b) (
x
+
y
) 2 –
z
2
E.g. 6 —Differences of Squares
Example (a) Using the Difference of Squares Formula with
A
= 2
x
and
B
= 5, we have: 4
x
2 – 25 = (2
x
) 2 – 5 2 = (2
x
– 5)(2
x
+ 5)
E.g. 6 —Differences of Squares
Example (b) We use the Difference of Squares Formula with
A
=
x
+
y
and
B
= z. (
x
+
y
) 2 –
z
2 = (
x
+
y
–
z
)(
x
+
y
+
z
)
E.g. 7 —Factoring Differences and Sums of Cubes
Factor each polynomial.
(a) 27
x
3 – 1 (b)
x
6 + 8
E.g. 7 —Difference of Cubes
Example (a) Using the Difference of Cubes Formula with
A
= 3
x
and
B
= 1, we get: 27
x
3 – 1 = (3
x
) 3 – 1 3 = (3
x
– 1)[ (3
x
) 2 + (3
x
)(1) + 1 2 ] = (3
x
– 1)(9
x
2 + 3
x
+ 1)
E.g. 7 —Sum of Cubes
Example (b) Using the Sum of Cubes Formula with
A
=
x
2 and
B
= 2, we have:
x
6 + 8 = (
x
2 ) 3 + 2 3 = (
x
2 + 2)(
x
4 – 2
x
2 + 4)
Perfect Square
A trinomial is a perfect square if it is of the form
A
2 + 2
AB
+
B
2 or
A
2 – 2
AB
+
B
2 • So, we recognize a perfect square if the middle term (2
AB
or –2
AB
) is plus or minus twice the product of the square roots of the outer two terms.
E.g. 8 —Recognizing Perfect Squares
Factor each trinomial.
(a)
x
2 + 6
x
+ 9 (b) 4
x
2 – 4
xy
+
y
2
E.g. 8 —Perfect Squares
Here,
A
=
x
and
B
= 3.
So, 2
AB
= 2
.
x
.
3 = 6
x
. Example (a) • Since the middle term is 6
x
, the trinomial is a perfect square. • By the Perfect Square Formula, we have:
x
2 + 6
x
+ 9 = (
x
+ 3) 2
E.g. 8 —Perfect Squares
Here,
A
= 2
x
and
B
=
y.
So, 2
AB
= 2
.
2
x
.
y
= 4
xy
. Example (b) • Since the middle term is –4
xy
, the trinomial is a perfect square. • By the Perfect Square Formula, we have: 4
x
2 – 4
xy
+
y
2 = (2
x
–
y
) 2
Factoring an Expression Completely
Factoring an Expression Completely
When we factor an expression, the result can sometimes be factored further. In general, • We first factor out common factors.
• Then, we inspect the result to see if it can be factored by any of the other methods of this section.
• We repeat this process until we have factored the expression completely.
E.g. 9 —Factoring an Expression Completely
Factor each expression completely.
(a) 2
x
4 – 8
x
2 (b)
x
5
y
2 –
xy
6
E.g. 9 —Factoring Completely
Example (a) We first factor out the power of
x
with the smallest exponent.
2
x
4 – 8
x
2 = 2
x
2 (
x
2 – 4) = 2
x
2 (
x
– 2)(
x
+ 2) (Common factor is 2
x
2 ) (Factor
x
2 – 4 as a difference of squares)
E.g. 9 —Factoring Completely
Example (b) We first factor out the powers of
x
and
y
with the smallest exponents.
x
5
y
2 –
xy
6 =
xy
2 (
x
4 –
y
4 ) =
xy
2 (
x
2 +
y
2 )(
x
2 –
y
2 ) (Common factor is
xy
2 ) (Factor
x
4 –
y
4 as a difference of squares) =
xy
2 (
x
2 +
y
2 )(
x
+
y
)(
x
–
y
) (Factor
x
2 –
y
2 as a difference of squares)
Factoring Expressions with Fractional Exponents
In the next example, we factor out variables with fractional exponents.
• This type of factoring occurs in calculus.
E.g. 10 —Factoring with Fractional Exponents
Factor each expression.
(a) 3
x
3/2 – 9
x
1/2 + 6
x
–1/2 (b) (2 +
x
) –2/3
x
+ (2 +
x
) 1/3
E.g. 10 —Fractional Exponents
Example (a) Factor out the power of
x
with the smallest exponent —that is,
x
–1/2 .
3
x
3/2 – 9
x
1/2 + 6
x
–1/2 = 3
x
–1/2 (
x
2 – 3
x
+ 2) = 3
x
–1/2 (
x
– 1)(
x
– 2) (Factor out 3
x
–1/2 ) (Factor the quadratic
x
2 – 3
x
+ 2)
E.g. 10 —Fractional Exponents
Example (b) Factor out the power of 2 +
x
with the smallest exponent —that is, (2 +
x
) –2/3 .
(2 +
x
) –2/3
x +
(2 +
x
) 1/3
=
(2 +
x
) –2/3 [
x
+ (2 +
x
)]
=
(2 +
x
) –2/3 (2 + 2
x
) = 2(2 +
x
) –2/3 (1 +
x
) (Factor out (2 +
x
) –2/3 ) (Simplify) (Factor out 2)
Factoring by Grouping Terms
Factoring by Grouping
Polynomials with at least four terms can sometimes be factored by grouping terms. • The following example illustrates the idea.
E.g. 11 —Factoring by Grouping
Factor each polynomial.
(a)
x
3 +
x
2 +4
x
+ 4 (b)
x
3 – 2
x
2 – 3
x
+ 6
E.g. 11 —Factoring by Grouping
x
3 +
x
2 +4
x
+ 4 Example (a) = (
x
3 +
x
2 ) + (4
x
+ 4) (Group terms) =
x
2 (
x
+ 1) + 4(
x
+ 1) = (
x
2 + 4)(
x
+ 1) (Factor out common factors) (Factor out
x
+ 1 from each term)
E.g. 11 —Factoring by Grouping
x
3 – 2
x
2 – 3
x
+ 6 Example (b) = (
x
3 – 2
x
2 ) – (3
x
– 6) (Group terms) =
x
2 (
x
– 2) – 3(
x
– 2) = (
x
2 – 3)(
x
– 2) (Factor out common factors) (Factor out
x
– 2 from each term)