Transcript Section 1.5

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Hawkes Learning Systems:
College Algebra
Section 1.5: Polynomials and Factoring
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Objectives:
o The terminology of polynomial expressions
o The algebra of polynomials
o Common factoring methods
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The Terminology of Polynomial Expressions
o Coefficient: A number multiplied by a variable in any
of the terms of a polynomial.
o Degree of the term: The sum of the exponents of the
variables in that term.
o Constant term: Any non-zero number that is not
multiplied by a variable.
Note: Constant terms have a degree of zero.
o Degree of a polynomial: The largest degree of all the
individual terms.
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The Terminology of Polynomial Expressions
o Monomials: Polynomials consisting of a single term
2
17
x
, 5xy, 3
Ex:
o Binomials: Polynomials consisting of two terms
Ex: 25 x 2  2, 16 x  1
o Trinomials: Polynomials consisting of three terms
Ex: 37 xy 2  4 x  2, 16 y 2  11xy  13
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Example 1: The Terminology of Polynomial
Expressions
Expression
2 3 5
x y  y10  3
3
Terms
3
Type
Trinomial
Degree
Explanation
10
The degree of the first
term is 8, the degree of
the second term is 10,
and the degree of the
third term is 0.
3x 4 y 2  5.4 x3 y 4
2
Binomial
7
The degree of the first
term is 6 and the
degree of the second
term is 7.
5
1
Monomial
0
The degree of a
constant is always 0.
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The Terminology of a Single Variable
Polynomials of a Single Variable
A polynomial in the variable of a degree n can be
written in the form
an xn  an1xn1  ...  a1x  a0
where an, an1,..., a1, a0 are real numbers, an  0 and n
is a positive integer. This form is called
descending order because the powers descend
from left to right. The leading coefficient of this
polynomial is an .
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The Algebra of Polynomials
Like or similar terms: The terms among all the
polynomials being added that have the same
variables raised to the same powers.
Ex: What are the like terms in the polynomial below?
x  2 z  y  z  y  3x
3
2
2
3
x3  2 z  y 2  z  y 2  3x3
x3 and 3x3 , 2 z and z, y 2 and 
3
3
Notice that x and 3x both
include the variable x
raised to the third power.
These are like or similar
terms. Can you find any
y 2 others?
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Example 1: The Algebra of Polynomials
Add or subtract the polynomials, as indicated.
 2x y  3 y  z x   3 y  z
3
2
2
 3xz  4 
2
= 2 x y  (3 y  3 y)  ( xz  3xz )  z  4
3
2
2
2
3
2
2
2
x
y

y

3

3

xz
(1

3)

z
4


=
3
2
2
2
x
y

2
xz

z
4
=
The first step is to identify
like terms and group these
together.
2
Note that the terms z x
and 3xz 2 are similar, as
multiplication is
commutative.
2
Pull out the y and xz
variables and simplify.
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Example 2: The Algebra of Polynomials
Add or subtract the polynomials, as indicated.
3
3
3
4
ab

b
c

4

b

  c
3
3
3
4
ab

b
c

4

b
c
=
3
3
3
4
ab

4

(

b
c

b
c)
=
3
4
ab
4
=
Again, the like terms are identified and grouped together after distributing
the minus sign over all the terms in the second polynomial.
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Example 3: The Algebra of Polynomials
Multiply the polynomials, as indicated.
(2 x2 y  z 3 )(4  3z  3xy)
= 2 x2 y(4  3z  3xy)  z 3 (4  3z  3xy)
= 8x y  6 x yz  6 x y  4 z  3z  3xyz
2
2
3
2
3
4
Use the distributive property
first to multiply each term of
the first polynomial by each
term of the second.
3
None of the resulting terms
are similar, so the final
answer is a polynomial of 6
terms.
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Example 4: The Algebra of Polynomials
Multiply the polynomials, as indicated.
(3ab  a 2 )(ab  a 2 )
= 3ab(ab  a 2 )  a 2 (ab  a 2 )
= 3a 2b 2  3a 3b  a 3b  a 4
2 2
3
4
= 3a b  2a b  a
The four terms that result from
the initial multiplication contain
two similar terms.
We combine these to obtain the
final trinomial.
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The Algebra of Polynomials
When a binomial is multiplied by a binomial, the
acronym FOIL is commonly used as a reminder of the
four necessary products. Consider the product:
First
Outer
(3ab  a 2 )(ab  a 2 )
Inner
Last
The solution to the product above would be
First + Outer + Inner + Last
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The Algebra of Polynomials
Consider the product:
(3ab  a 2 )(ab  a 2 )
The product of the First terms is (3ab  a 2 )(ab  a 2 )  3a 2b 2
3
2
2
3a
b
(
3
ab

a
)(
ab

a
)

The product of the Outer terms is
3
2
2
The product of the Inner Terms is (3ab  a )(ab  a )   a b
4
2
2

a
(3
ab

a
)(
ab

a
)

The product of the Last terms is
So, First + Outer + Inner + Last
2 2
3
3
4
3a
b
3a
b

a
b

a
=
+
+
+
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Common Factoring Methods: Terminology
o Factoring: Reversing the process of multiplication in
order to find two or more expressions whose
product is the original expression.
o Factorable: A polynomial with integer coefficients is
factorable if it can be written as a product of two or
more polynomials, all of which also have integer
coefficients.
o Irreducible (over the integers) or prime: A
polynomial that is not factorable.
o Completely Factor: To write a polynomial as a
product of prime polynomials.
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Common Factoring Methods
o Method 1: Greatest common factor.
o Method 2: Factoring by grouping.
o Method 3: Factoring special binomials.
o Method 4: Factoring trinomials.
o
Case 1: Leading coefficient is 1.
o
Case 2: Leading coefficient is not 1.
o Method 5: Factoring Expressions Containing
Fractional Exponents
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Common Factoring Methods: Method 1
Method 1: Greatest Common Factor
The Greatest Common Factor (GCF) among all the terms is
simply the product of all the factors common to each. The
Greatest Common Factor method is a matter of applying the
distributive property to “un-distribute” the greatest common
factor.
Find the greatest common factor of 12 x 4 y 2  8 x3 y 2  4 x 3 y
3
GCF: 4x y
What do the three terms in the polynomial have in
3
common? 4, x , and y . The product of these terms
is the GCF of the polynomial.
So,12 x 4 y 2  8x3 y 2  4 x3 y  4 x3 y  3xy  2 y  1
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Example 5: Common Factoring Methods
Use the Greatest Common Factor method to factor the
following polynomial.
12 x 5  4 x 2  8 x 3 z 3
2
3
2
2
3
(
4
x
)(3
x
)

(
4
x
)(

1)

(
4
x
)(2
xz
)
=
2
3
3
(4
x
)(3
x

1

2
xz
)
=
Applying the distributive
property in reverse
leads to the factored
form of this degree 6
trinomial.
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Example 6: Common Factoring Methods
Use the Greatest Common Factor method to factor the
following polynomials.
24ax 2  60a
2
(

12
a
)(2
x
)  (12a)(5)
=
= 12a(2 x2  5)
An alternative form of the final
2
answer is12a(2 x  5) .
We would have obtained this
answer if we had factored out 12a
initially. These two answers are
equivalent.
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Example 7: Common Factoring Methods
Use the Greatest Common Factor method to factor the
following polynomials.
(a 2  b)  3(a 2  b)
2
2
(
a

b
)(1)

(
a
 b)(3)
=
2
= (a  b)(1  3)
= 2(a 2  b)
In factoring out the greatest
common factor (a 2  b)
remember that it is being multiplied
first by 1 and then by –3. One
common error in factoring is to
forget factors of 1.
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Common Factoring Methods: Method 2
Method 2: Factoring by Grouping.
Factoring by Grouping: A trial and error process
applied when the first factoring method is not
directly applicable. If the terms of the polynomial
are grouped in a suitable way, the GCF method may
apply to each group, and a common factor might
subsequently be found among the groups.
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Common Factoring Methods
Ex: Use Method 2, factoring by grouping, to factor
the following polynomial:
5xy  3  15x  y
= (15x  5xy)  (3  y)
= 5x(3  y)  1(3  y)
= (5x  1)(3  y)
Note: the GCF of the four terms is 1, so you
cannot use the GCF method. Try rearranging
the way they are grouped.
Once you group (15 x  5 xy ) and(3  y) notice
that the two groups each have a GCF: 5x and
1. Don’t forget to factor out 1 from (3  y) as
it is a common mistake to not factor out a 1.
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Example 8: Common Factoring Methods
Use the Factor by Grouping method to factor the
following polynomials.
6 x 2  y  2 x  3xy
2
(6
x
 2 x)  ( y  3xy)
=
= 2 x(3x  1)  y (1  3 x)
= 2 x(3x  1)  y (3x  1)
= (3 x  1)(2 x  y )
The GCF of the four terms in the
polynomial is 1, so the Greatest
Common Factor Method does
not directly apply. So we must
use Method 2, Factoring by
Grouping.
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Example 9: Common Factoring Methods
Use the Factor by Grouping method to factor the
following polynomial.
ax  ay  bx  by
=  ax  bx    ay  by 
= (ax  ay )  (bx  by )
= a ( x  y )  b( x  y )
= ( x  y )(a  b)
OR
= x a  b  y a  b
= ( x  y )(a  b)
We could also group the first and third terms and the second and fourth
terms to obtain the same result.
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Common Factoring Methods
Method 3: Factoring Special Binomials
Three types of binomials can always be factored following
certain patterns. In the following, A and B represent algebraic
expressions.
o Difference of Two Squares:
A2  B2  ( A  B)( A  B)
o Difference of Two Cubes:
A3  B3  ( A  B)( A2  AB  B2 )
o Sum of Two Cubes:
A3  B3  ( A  B)( A2  AB  B 2 )
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Example 10: Common Factoring Methods
Factor the following binomials.
49 x 2  9 y 6
2
3 2
(7
x
)

(3
y
)
=
3
3
(7
x

3
y
)(7
x

3
y
)
=
The first step is to realize that the
binomial is a difference of two squares
and to identify the two expressions that
are being squared. Then follow the
pattern to factor the binomial.
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Example 11: Common Factoring Methods
Factor the following binomials.
27a 6b12  c 3
= (3a 2b4 )3  (c)3

= (3a b  c) (3a b )  (3a b )(c)  (c)
2 4
2 4 2
2 4
2 4
4 8
2 4
2
= (3a b  c)(9a b  3a b c  c )
2

First recognize the
binomial as a sum
of two cubes.
Then follow the
pattern to factor
the binomial.
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Example 12: Common Factoring Methods
Factor the following binomials.
64  ( x  y)3
In this difference of two cubes the second cube is
itself a binomial. But the factoring pattern still
applies, leading to the final factored form of the
original binomial.
= 43  ( x  y)3

= 4  ( x  y)  4  4( x  y)  ( x  y)
2
2

2
2
= (4  x  y)(16  4 x  4 y  x  2 xy  y )
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Common Factoring Methods
Method 4, Case 1: Leading Coefficient is 1.
In this case, p and r will both be 1 so we only need q
2
2
and s such that x  bx  c  x  (q  s) x  qs .
That is, we need two integers whose sum is b, the
coefficient of x, and whose product is c, the constant
term.
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Common Factoring Methods
Use Method 4, Case 1 to factor the following polynomial.
Ex:
o Factor: x 2  3 x  2
o Begin by writing x 2  3x  2   x  ?  x  ?  .
o We need to find two integers to replace the question
marks. The two integers we seek must have a product of 2.
Because the product is positive, both integers must be
either positive or negative. Therefore, the only possibilities
are 2,1or2, 1 .
o Additionally, the sum of these two integers must be 3.
Therefore, they must be 2,1 .
2
x
o Thus,  3x  2  ( x  2)( x  1).
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Example 13: Common Factoring Methods
o Factor: x  x  12
2
x
 x  12  ( x  ?)( x  ?)
o Begin by writing
o We need to find two integers to replace the question
marks. The two integers we seek must have a product of
–12. Because the product is negative, one integer must
be positive and one must be negative. Therefore, the
only possibilities are
1, 12 , 1,12 , 2, 6 , 2,6 , 3, 4 , 3, 4
o Additionally, the sum of these two integers must be 1.
Therefore, they must be{3,4} .
2
o Thus, x  x  12  ( x  3)( x 4).
2
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Common Factoring Methods
Method 4, Case 2: Leading Coefficient is not 1.
Factoring Trinomials by Grouping
For the trinomial ax 2  bx  c :
Step 1: Multiply a and c.
Step 2: Factor ac into two integers whose sum is b. If
no such factors exist, the trinomial is irreducible over
the integers.
Step 3: Rewrite b in the trinomial with the two
integers found in step 2. The resulting polynomial of
four terms may now be factored by grouping.
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Example 14: Common Factoring Methods
Factor the following trinomial by grouping:
6 x 2  x  12
1. (6)(12)  72
Multiply a and c.
2. 9 and 8
Factor ac into two integers whose sum is
b.
2
6
x
 9 x  8 x  12
3.
Rewrite b in the trinomial with the
two integers found in step 2 and
distribute.
Group.
4. 3 x(2 x  3)  4(2 x  3)
5. (2 x  3)(3x  4)
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Common Factoring Methods
Perfect Square Trinomials: trinomial expressions whose
factored form is the square of a binomial expression.
There are two forms of Perfect Square Trinomials:
A2  2 AB  B 2  ( A  B ) 2
A2  2 AB  B 2  ( A  B ) 2
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Common Factoring Methods: Method 5
Method 5: Factoring Expressions Containing
Fractional Exponents
To factor an algebraic expression that has fractional
exponents, identify the least exponent among the
various terms and factor the variable raised to that
least exponent from each of the terms.
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Example 15: Common Factoring Methods
Factor the following algebraic expressions:
3x
2
3
1
3
 6 x  3x
2
3
4
3
=3x (1  2 x  x )
2
3
2
=3x ( x  1)( x  1)
2
3
=3 x ( x  1) 2
2
3
First we factor out 3x .
Next, notice that the second factor is
a second-degree trinomial and is
factorable. In fact, it is a perfect
square trinomial.
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Example 16: Common Factoring Methods
Factor the following algebraic expression:
1
2
( x  1)  ( x  1)
=( x  1)
1
2
1
2
1
2
 ( x  1)  1
= ( x  1) ( x  2)
1
2
Factor out ( x  1) using the properties
of exponents to obtain the terms in
the second factor.
Simplify the second term.