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3 POLYNOMIALS 3.1 Definitions and Examples A polynomial in one variable, say x, is an algebraic expression whose terms
are of the form nna x , where na (read as “a sub n”) is a real number and n is a
nonnegative integer. We can use other variables in place of x. Examples The following are polynomials:
1. 2x
2. 2 2 3y y+ −
3. 26 n n− −
4. 10 140.1 0.2w w−
5. 2 5a a−
6. 3 28 4 5 6t t t− − +
7. 81 9
4 25m −
The following are not polynomials:
8. 123 2x x− + 9.
+
−
1
5
x
x 10.
2
5 2
xx+
Classification of Polynomials Based on the Number of Terms
A polynomial can be classified based on the number of terms. A polynomial with only one term is called a monomial, a polynomial with two terms is called a binomial, a polynomial with three terms is called a trinomial. A polynomial with four or more terms is simply called a polynomial or a multinomial. Examples The examples above are classified as follows:
1. 2x is a monomial since it has only one term 2x .
2. 2 2 3y y+ − is a trinomial since it has three terms 2, 2 , and 3.y y −
3. 26 n n− − is a trinomial since it has three terms 26, , and .n n− −
4. 10 140.1 0.2w w− is a binomial since it has two terms 10 140.1 and 0.2 .w w
5. 2 5a a− is a binomial since it has two terms 2 and 5 .a a−
6. 3 28 4 5 6t t t− − + is a multinomial or simply, a polynomial, since it has four
terms 3 28 , 4 , 5 , and 6.t t t− −
7. 81 9
4 25m − is a binomial since it has two terms 81 9
and .4 25
m −
46
The Degree of Each Term of a Polynomial in One Variable Each term of a polynomial has a degree. The degree of a term is the exponent of the variable. A constant term has no variable part and its degree
is zero (the variable part is 0x ). Examples
1. There is only one term 2x and the degree of this term is 1.
2. There are three terms: 2y of degree 2; 2y of degree 1; −3 of degree 0.
3. There are three terms: 6 of degree 0; −n of degree 1; 2n− of degree 2.
4. There are two terms: 100.1w of degree 10; 140.2w− of degree 14.
5. There are two terms: 2a of degree 2; −5a of degree 1.
6. There are four terms: 38t of degree 3; 24t− of degree 2; −5t of degree 1;
6 of degree 0.
7. There are two terms: 81
4m of degree 8; −
9
25of degree 0.
The Degree of a Polynomial in One Variable
The degree of a polynomial in one variable is defined to be the highest degree of a term. Examples
1. 2x is of degree 1.
2. 2 2 3y y+ − is of degree 2.
3. 26 n n− − is of degree 2.
4. 10 140.1 0.2w w− is of degree 14.
5. 2 5a a− is of degree 2.
6. 3 28 4 5 6t t t− − + is of degree 3.
7. 81 9
4 25m − is of degree 8.
Classification Based on the Arrangement of the Terms We can also talk of polynomials as being arranged in ascending or descending order, depending on whether the terms are arranged from lowest degree to highest degree or from highest degree to lowest degree, respectively. Examples
1. 2x has only one term and we don’t talk about the arrangement of the terms.
2. 2 2 3y y+ − has three terms arranged in descending order.
3. 26 n n− − has three terms arranged in ascending order.
4. 10 140.1 0.2w w− has two terms arranged in ascending order.
47
5. 2 5a a− has two terms arranged in descending order.
6. 3 28 4 5 6t t t− − + has four terms arranged in descending order.
7. 81 9
4 25m − has two terms arranged in descending order.
Polynomials in More Than One Variable A polynomial in 2 variables, say x and y, is an algebraic expression whose terms are of the form
m nka x y ,
where ka is any real number and m and n are nonnegative integers. Other
variables can be used in place of x and y. Examples
8. 2 5 3 43 7x y xy xy x y+ − − −
9. 2 2a b−
10. 3 3
8 27
m n+
11. 2 23 10y yz z− −
12. 6 5 4 34 24 9x x y x y+ +
A polynomial in three or more variables is defined similarly.
The Degree of a Multivariable Polynomial The degree of a term of a polynomial in several variables is defined as the sum of the exponents of the variables in a term. The degree of the polynomial is the largest degree of a term. Let us discuss these for the examples above. Examples
8. There are five terms: 2x y of degree 2 1 3,+ =
53xy of degree 1 5 6,+ =
−7xy of degree 1 1 2,+ =
3x− of degree 3,
4y− of degree 4.
The degree of the polynomial is 6.
9. There are two terms: 2a of degree 2,
2b− of degree 2.
The degree of the polynomial is 2.
48
10. There are two terms: 3
8
mof degree 3,
3
27
nof degree 3.
The degree of the polynomial is 3.
11. There are three terms: 2y of degree 2,
−3yz of degree 1 1 2,+ =
210z− of degree 2.
The degree of the polynomial is 2.
12. There are three terms: 64x of degree 6,
524x y of degree 5 1 6,+ =
4 39x y of degree 4 3 7.+ =
The degree of the polynomial is 7.
3.2 Adding or Subtracting Polynomials As explained in Chapter 2, the basic principle used in adding or subtracting algebraic expressions is combining like terms. We use the same principle for adding or subtracting polynomials. Adding Polynomials We can add two or more polynomials using either the horizontal format or the vertical format. Examples
1. Add 2 23 5 1 and 4 6 9x x x x+ − − − + .
A. Horizontal Format
( ) ( )
( ) ( ) ( )
2 2
2 2
2 2
2
3 5 1 + 4 6 9
3 5 1 4 6 9
3 4 5 6 1 9
8
x x x x
x x x x
x x x x
x x
+ − − − +
= + − − − +
= − + − + − +
= − − +
B. Vertical Format - align the like terms
2
2
2
3 5 1
4 6 9
8
x x
x x
x x
+ −
− − +
− − +
49
2. Add 4 2 62a a b b− − to 4 2 23 5a ab a b+ − .
A. Horizontal Format
( ) ( )
( ) ( )
4 2 6 4 2 2
4 2 6 4 2 2
4 4 2 2 6 2
4 2 6 2
2 + 3 5
2 3 5
3 2 5
4 3 5
a a b b a ab a b
a a b b a ab a b
a a a b a b b ab
a a b b ab
− − + −
= − − + + −
= + + − − − +
= − − +
B. Vertical Format - align the like terms
4 2 6
4 2 2
4 2 6 2
2
3 5
4 3 5
a a b b
a a b ab
a a b b ab
− −
− +
− − +
Subtracting Polynomials We can subtract two polynomials using either the horizontal format or the vertical format. Examples
3. Subtract: ( ) ( )2 23 5 1 4 6 9x x x x+ − − − − +
A. Horizontal Format
( ) ( )
( ) ( ) ( )
2 2
2 2
2 2
2
3 5 1 4 6 9
3 5 1 4 6 9
3 4 5 6 1 9
7 11 10
x x x x
x x x x
x x x x
x x
+ − − − − +
= + − + + −
= + + + + − −
= + −
B. Vertical Format
2 2
2 2
2
3 5 1 3 5 1
( 4 6 9) 4 6 9
7 11 10
x x x x
x x x x
x x
+ − + −
− − − + → + −
+ −
4. Subtract 2 22 8 from 3 2 1y y y y+ − − − − .
A. Horizontal Format
( ) ( )
( ) ( ) ( )
2 2
2 2
2 2
2
3 2 1 2 8
3 2 1 2 8
3 2 2 1 8
5 3 7
y y y y
y y y y
y y y y
y y
− − − − + −
= − − − − − +
= − − + − − + − +
= − − +
50
B. Vertical Format
2 2
2 2
2
3 2 1 3 2 1
(2 8) 2 8
5 3 7
y y y y
y y y y
y y
− − − − − −
− + − → − − +
− − +
We can also combine addition and subtraction.
5. Subtract 2 2 23 4 3 from the sum of 5 6 and 2 8x x x x x+ + − − − .
Solution: First we find the sum of 2 25 6 and 2 8x x x− − − :
2
2
2
5 6
2 8
3 5 14
x x
x
x x
− −
+ −
− −
Then we perform the subtraction:
( )
2 2
2 2
3 5 14 3 5 14
3 4 3 3 4 3
9 17
x x x x
x x x x
x
− − − −
− + + → − − −
− −
Let us practice translating from English to algebra. Translate and perform the operation in the following examples.
6. The sum of 6 3 and 5 2, increased by 4x x x− − −
Solution: Translating, we get
( ) ( )�
6 3 5 2 +
The sum of 6 3 and 5 2, increased by 4
x x
x x x
− + −
− − −��������������� �������
We see that all we need to do is add all three polynomials. Verify that the
answer is 12 9x − .
7. 2 23 subtracted from twice 7x x x x+ − − −
Solution:
( )2
2 2
this means the 2nd 2 7minus the 1st
3 subtracted from twice 7
x x
x x x x
− −
+ − − −��������� ���������
( ) ( )2 2
2 2 2
Thus we get 2 7 3 , and removing parentheses,
we get 2 2 14 3 which simplifies to 3 11.
x x x x
x x x x x x
− − − + −
− − − − + − −
51
3.3 Multiplying Polynomials
Multiplying Monomials Use the following exponent rules:
( )
Product Rule
Power Rule
n m n m
nm mn
a a a
a a
+=
=
Examples
1. Multiply: ( )( )3 72 5x x−
Solution:
( )( ) ( ) ( ) ( )3 7 3 7Apply commutative and associative properties.
10Apply the product rule.
2 5 2 5
10
x x x x
x
− = ⋅ − ⋅
= −
2. Multiply: ( )( )3 4 75 2x y xy
Solution:
( )( ) ( ) ( )( )3 4 7 3 4 7Apply commutative and associative properties.
4 11Apply the product rule.
5 2 5 2
10
x y xy x x y y
x y
= ⋅ ⋅ ⋅
=
3. Multiply: ( ) ( )5 3
3 2 32p q p q−
Solution:
( ) ( ) ( )5 3 53 2 3 15 5 6 9
Apply the distributivity of exp over mult.
15 6 5 9Apply the commutative property.
21 14Apply the product rule.
2 2
32
32
p q p q p q p q
p p q q
p q
− = − ⋅
= − ⋅ ⋅ ⋅
= −
Multiplying a Monomial by a Polynomial
Apply the distributive property + = +( )a b c ab ac and then use the power
rule n m n ma a a += .
Examples
4. Multiply: ( )3 22 3 4 5x x x+ −
Solution:
( )( )
( )( ) ( )( ) ( )( )
3
3 2
3 2 3 3Distribute 2 to each of the 3 terms inside the .
3 2 3 3Apply the comm and asso prop.
5 4 3Apply the power rule.
2 3 4 5
2 3 2 4 2 5
2 3 2 4 2 5
6 8 10
x
x x x
x x x x x
x x x x x
x x x
+ −
= ⋅ + ⋅ − ⋅
= ⋅ ⋅ + ⋅ ⋅ − ⋅
= + −
52
5. Multiply ( )6 84 5 2 3y y y y− − − +
Solution:
( )6 84 5 2 3y y y y− − − +
( ) ( ) ( ) ( ) ( ) ( )6 84 5 4 2 4 4 3y y y y y y y= − ⋅ − − ⋅ − − + −
( ) ( ) ( )( ) ( )( ) ( )( )6 84 5 4 2 4 4 3y y y y y y y= − ⋅ − − ⋅ ⋅ − − ⋅ + − ⋅ ⋅
2 7 920 8 4 12y y y y= − + + −
Of course one may omit showing some of the middle steps in the process.
Multiplying a Polynomial by a Polynomial
Apply the distributive property + = +( )a b c ab ac and then use the power
rule n m n ma a a += . Distribute each term from one polynomial to each term
from the other. Like addition and subtraction of polynomials, multiplication can be also be done either horizontally or vertically. Examples
6. Multiply ( )( )3 2 5x y x y+ − + +
A. Horizontal Format
( ) ( )
( ) ( ) ( )
distribute each termof this polynomial
2 2
distribute distribute 3distribute
2 2
3 2 5
2 5 2 5 3 2 5
2 5 2 5 3 6 15
3 2 2 15
yx
x y x y
x x y y x y x y
x xy x xy y y x y
x xy x y y
−
+ − + +
= + + + + + − + +
= + + + + + − − −
= + + + − −
�����
���������������������
B. Vertical Format
( )( )2
Distribute from 1st.
2Distribute from 1st.
2 2
Distribute 3 from 1st.
3 2 5
2 5
2 5
3 6 15
3 2 2 15
x
y
x y x y
x xy x
xy y y
x y
x xy x y y
−
+ − + +
+ +
+ +
− − −
+ + + − −
Expanding Polynomials This is really doing exponentiation with polynomials. Just use the fact that exponentiation is repeated multiplication, i.e.
factors
n
n
a a a a= ⋅ ⋅ ⋅…�����
.
53
Examples
7. Expand: ( )2
6x +
Solution:
( ) ( )( )2 2
2
2
6 6 6 since
6 Distribute .
6 36 Distribute 6.
12 36 Combine like terms.
x x x a a a
x x x
x
x x
+ = + + = ⋅
= +
+ +
+ +
8. Expand: ( )3
2a b+
Solution:
( ) ( )( )( )3 32 2 2 2 since a b a b a b a b a a a a+ = + + + = ⋅ ⋅
Multiply the first two binomials:
( )( )2
2
2 2
2 2
4 2 Distribute 2 .
2 Distribute .
4 4 Combine like terms.
a b a b
a ab a
ab b b
a ab b
+ +
= +
+ +
+ +
Multiply this answer by +2a b :
( )( )2 2
3 2 2
2 2
2 3 2
3 2 2 3
4 4 2
8 4 Distribute 4 .
8 4 Distribute 4 .
2 Distribute .
8 12 6 Combine like terms.
a ab b a b
a a b a
a b ab ab
ab b b
a a b ab b
+ + +
= +
+ +
+ +
+ + +
We can again practice translating from English to algebra.
9. Translate and perform the operation: The product of 4 3 and 6 1x x+ −
Solution: The word “product” denotes multiplication so we multiply the two
polynomials: ( )( )4 3 6 1x x+ − and use either format. Verify that the answer is
224 14 3x x+ −
.
10. The square of 4x − multiplied by twice 1 2x−
Solution:
( )( )2 2 1 2
4
The square of 4, multiplied by twice 1 2
xx
x x
−⋅−
− −������������������ �������
54
Thus, we need to do three things:
(a) Expand ( )2
4x − .
(b) Multiply ( )2 1 2x− .
(c) Find ( ) ( )2
4 2 1 2x x− ⋅ − using the answers from (a) and (b).
Verify that the answer is 3 24 34 80 32x x x− + − + .
3.4 Multiplying Binomials Using Special Product Formulas Recall that binomials are polynomials with two terms. For these polynomials we have special product formulas that can help us get the product faster.
FOIL ( )( )+ + = + + +
a b c d ac ad bc bd
F O I L (1)
where F stands for First (multiply the first terms a and c) O Outer (multiply the outer terms a and d) I Inner (multiply the inner terms b and c) L Last (multiply the second or last terms b and d) Note: This is really the same method of distributing each term of one binomial to each of the terms of the other binomial so it is not really a special formula. Examples
1. Multiply: + +( 3)( 2)x x Solution:
This step may be omitted.
2
2Combine like terms.
( 3)( 2) 2 3 3 2
F O I L
2 3 6
5 6
x x x x x x
x x x
x x
+ + = ⋅ + ⋅ + ⋅ + ⋅
= + + +
= + +
2. Multiply: + −(2 3 )(5 2 )x y x y Solution:
( ) ( ) This step may be omitted.
2 2
2 2Combine like terms.
(2 3 )(5 2 )
2 5 2 2 3 5 3 2
10 4 15 6
10 11 6
x y x y
x x x y y x y y
x xy xy y
x xy y
+ −
= ⋅ + ⋅ − + ⋅ + ⋅ −
= − + −
= + −
3. Multiply: ( )( )+ −3 3x x
Solution: ( )( ) 2 23 3 3 3 9 9x x x x x x+ − = + − − = −
4. Multiply: ( )( )+ −2 3 2 3x y x y
Solution: ( )( ) 2 2 2 22 3 2 3 4 6 6 9 4 9x y x y x xy xy y x y+ − = + − − = −
55
FOIL Leading to a Difference of Squares In examples 3 and 4 the O and the I terms in FOIL cancelled out each other. Notice that the 2 binomials being multiplied have exactly the same first and last terms. We can generalize this result as follows:
( )( ) 2 2a b a b a b+ − = − (2)
This special product formula says that multiplying the sum of two quantities and the difference of the same two quantities will give the difference of the squares of the quantities. Examples
5. Multiply: ( )( )+ −2 2x x
Solution: ( )( ) 2 2 22 2 2 4x x x x+ − = − = −
6. Multiply: ( )( )− +4 5 4 5a b a b
Solution: ( )( ) ( ) ( )2 2 2 24 5 4 5 4 5 16 25a b a b a b a b− + = − = −
Squaring Binomials
Consider the expression ( )2
a b+ . Carrying out the expansion, we get:
( ) ( )( )2
2 2
2 2
FOIL
2 Combine like terms.
a b a b a b
a ab ba b
a ab b
+ = + +
= + + +
= + +
We can conclude from the above that squaring a binomial gives the square of the first term plus twice the product of the two terms plus the square of the second or last term. Note that the square of a binomial is always a trinomial. Examples
7. Expand: ( )2
5x +
Solution: ( ) � � �
2 2
the 1st term twice the product of the 2nd term squared the 1st and 2nd squared
5 10 25x x x+ = + +
8. Expand: ( )2
7y −
Solution: ( )� � �
2 2
the 2nd term the 1st term twice the product of squaredsquared the 1st and 2nd
7 14 49y y y− = − +
9. Expand: ( )2
3 4x y−
Solution: ( ) � �
2 2 2
the 1st term the 2nd term twice the product of squared squaredthe 1st and 2nd
3 4 9 24 16x y x xy y− = − +���
56
Cubing Binomials
Consider the expression ( )3
a b+ . Carrying out the expansion, we get:
( ) ( ) ( )
( )( ) ( )2
2 2
2 2Square the binomial .
3 2 2 Distribute .
2 2 3 Distribute .
3 2 2 3
2
2
2
3 3
a b
a
b
a b a b a b
a ab b a b
a a b ab
a b ab b
a a b ab b
+
+ = + +
= + + +
= + +
+ + +
= + + + Combine like terms.
We can conclude that cubing a binomial gives a polynomial with four terms. The 1st term is the cube of the first term, the 4th term is the cube of the second term, and the two middle terms have a factor of 3. The terms are arranged in descending powers of the first and ascending powers of the second. Examples
10. Expand: ( )3
1x +
Solution: ( ) � � � �
3 3 2 2 3
1st term 1st term squared 1st term times 2nd term cubed times 2nd term 2nd term squared cubed
1 3 1 3 1 1x x x x+ = + ⋅ ⋅ + ⋅ ⋅ +
3 2 3 3 1x x x= + + +
11. Expand: ( )3
2n −
Solution: ( ) � ( ) ( ) ( )�
3 2 33 2
1st term 2nd term1st term squared 1st term timescubed cubedtimes 2nd term 2nd term squared
2 3 2 3 2 2n n n n− = + ⋅ ⋅ − + ⋅ ⋅ − + −����� �����
3 2 6 12 8n n n= − + −
12. Expand: ( )3
2 5a +
Solution: ( ) ( )�
( ) ( ) �
3 3 2 2 3
2nd term1st term 1st term squared 1st term times cubedcubed times 2nd term 2nd term squared
2 5 2 3 2 5 3 2 5 5a a a a+ = + ⋅ ⋅ + ⋅ ⋅ +����� �����
3 28 60 150 125a a a= + + +
13. Expand: ( )3
3 2x y−
Solution: ( ) ( )�
( ) ( ) ( ) ( ) ( )3 3 2 2 3
1st term 1st term squared 1st term times 2nd term cubed times 2nd term 2nd term squared cubed
3 2 3 3 3 2 3 3 2 2x y x x y x y y− = + ⋅ ⋅ − + ⋅ ⋅ − + −������� ������� �����
3 2 2 327 54 36 8x x y xy y= − + −
57
We can translate and perform the operation. 14. The square of the sum of three times a number and twice another number Solution:
( )2
3 2
3 2
3 2
The square of the sum of three times a number and twice another number
x y
x y
x y
x y
+
+
����� ���������
����������� �����������
�����������������������������
�������������������������������������
Performing the expansion, we have
( ) ( ) ( ) ( ) ( )2 2 2 2 23 2 3 2 3 2 2 9 12 4x y x x y y x xy y+ = + + = + + .
3.5 Dividing Polynomials
Dividing a Monomial by a Monomial When dividing a monomial by another monomial, use the quotient rule for exponents
−= ≠, 0
mm n
n
aa a
a.
For example, 3 6
3 1 6 3 2 3
3
24 246
44
x yx y x y
xy
− −= ⋅ ⋅ = .
Dividing a Polynomial by a Monomial When dividing a polynomial by a monomial, use
±
= ± ≠, 0a b a b
cc c c
,
and perform the individual division of a monomial by another monomial. . In this section whenever we are dividing a polynomial P by a polynomial D, we shall write the answer in the form
+R
QD
,
where P is the dividend, D is the divisor, Q is the quotient, and R is the remainder. Thus, we have
P R
QD D
= + .
58
Examples
1. Divide 2 7x x− − by x.
Solution: 2 2
Write the problem as 3 monomial division problems.
Divide using the quotient rule for exponents.
7 7
71
x x x x
x x x x
xx
− −= − −
= − −
2. Divide: ( )3 4 212 27 3x y xy xy+ ÷
Solution:
3 4 2
3 4 2
Write the problem into 2 monomial division problems.
2 3Divide using the quotient rule for exponents.
12 27
3
12 27
3 3
4 9
x y xy
xy
x y xy
xy xy
x y y
+
= +
= +
Note: There is no remainder in Example 2; in this case, we say that the division is exact.
Dividing a Polynomial by a Binomial If there are 2 terms in the divisor we have to use long division. Let us review how it works for numbers. Let us divide 345 by 2. The steps taken were:
1. Divide 3 by 2 to get 1. 2. Multiply 1 by 2 to get 2. 3. Subtract 2 from 3 to get 1. 4. Bring down 4. 5. Divide 14 by 2 to get 7. And so on.
We could have done Example 1 above as follows: The steps taken were:
1. Divide 2x by x to get x.
2. Multiply x by x to get 2x .
3. Subtract 2x from2x .
4. Bring down −x .
5. Divide −x by x to get −1 .
6. Multiply −1 by x to get −x . 7. Subtract −x from −x .
8. Bring down −7 which cannot be divided by x so it is the remainder.
−
−
−
172
2 345
2
14
14
5
4
1
2
2
1
7
( )
7
x
x x x
x
x
x
−
− −
−
−
− −
−
59
We are now ready to divide with binomial divisors. It is similar to the above except that we ignore the second term in the divisor each time we divide.
Examples
3. Divide: ( ) ( )2 3 5 1x x x+ − ÷ +
Solution:
The steps taken were:
1. Divide 2x by x to get x. Note: Ignore the 2nd term 1 in the divisor.
2. Multiply x by + 1x to get 2x x+ .
3. Subtract 2x x+ from 2 3x x+ to get 2x .
4. Bring down −5 .
5. Divide 2x by x to get 2. Note: Ignore the 2nd term 1 in the divisor.
6. Multiply 2 by + 1x to get +2 2x .
7. Subtract +2 2x from −2 5x to get −7 which becomes the remainder.
Thus, the answer is + −+
72
1x
x. To check the answer, multiply the quotient
by the divisor and add the remainder to get the dividend. Check:
2
2
( 2)( 1) 7
2 2 7
3 5
x x
x x x
x x
+ + −
= + + + −
= + −
4. Divide: 28 3 1
3
x x
x
+ −
+
Solution: Just like when dividing whole numbers we should have the digits arranged according to decreasing place value, for polynomial division we have to make sure that both the dividend and divisor are arranged in descending order. Thus, the division problem should look like
23 8 1
3
x x
x
+ −
+
and we can proceed as follows:
2
2
2
1 3 5
( )
2 5
(2 2)
7
x
x x x
x x
x
x
+
+ + −
− +
−
− +
−
60
Thus, 23 8 1 2
3 13 3
x xx
x x
+ −= − +
+ +.
5. Divide: ( ) ( )3 1 2x x x+ − ÷ +
Solution: Note that the 2x term is missing from the dividend. We supply the missing term by using 0 for the numerical coefficient:
Thus,
( ) ( )3 2 111 2 2 5
2x x x x x
x+ − ÷ + = − + −
+.
6. Divide: ( ) ( )4 32 5 2y y y y− − + ÷ −
Solution:
Thus, we get ( ) ( )4 3 3 32 5 2 1
2y y y y y
y− − + ÷ − = − +
−.
2
2
3 1
3 3 8 1
(3 9 )
1
( 3)
2
x
x x x
x x
x
x
−
+ + −
− +
− −
− − −
( )
( )
( )
2
3
3 2
2
2
2 5
2 1
2
2
2 4
5 1
5 10
11
x x
x x x
x x
x x
x x
x
x
− +
+ + + −
− +
− +
− − −
−
− +
−
20x
( )
( )
3
4 3
4 3
2
1
2 2 5
2
0 0 5
2
3
y
y y y y
y y
y y
y
−
− − + − +
− −
+ − +
− − +
20y
61
Sometimes the divisor is of degree 2 or more.
7. Divide: 2 5 32 1 2 3 1x x x x− − + −
Solution:
Thus, the quotient is 3
2
1
2 1x x
x− −
−.
Polynomials involving two variables as well as divisors that have more than two terms will not be covered in this book.
We can translate and divide:
8. The quantity 1 less than the sum of the fourth power of a number and the number, divided by the sum of the square of the number and 1
Solution: We need to translate the dividend:
4
4
4 1
1 less than the sum of the fourth power of a number and the number
xx
x x
x x
+
+ −
����������������������
�����������������������������
���������������������������������
We also need to translate the divisor:
2
2 1
the sum of the square of the number and 1
x
x +
�������������
���������������������
Thus, we have the division problem
( ) ( )4 21 1x x x+ − ÷ +
and one can verify that the quotient is 2
21
1
xx
x− +
+.
( )
( )
3
2 5 3
5 3
3 2
3
2 1 2 3 1
2
2 0
2
x x
x x x x
x x
x x x
x x
−
− + − + + −
− −
− + +
− − +
4 20x 0x
0 1−
62
3.6 Chapter Review A polynomial in x is an algebraic expression whose terms are of the
form nna x , where na is a real number and n is a nonnegative integer.
A polynomial can be classified based on the number of terms. A polynomial with only one term is called a monomial, a polynomial with two terms is called a binomial, a polynomial with three terms is called a trinomial. Polynomials with four or more terms are simply called polynomials or sometimes multinomials. Each term of a polynomial has a degree. The degree of a term is the exponent of the variable. A constant term has no variable part and its
degree is zero (the variable part is 0x ). The degree of the polynomial is defined to be the highest degree of a term.
Examples 38x is a monomial in x of degree 3 in x.
42 8 5y y− + − is a trinomial in y; the degrees of the terms are 4, 1,
and 0, respectively; the degree of the polynomial is 4. A polynomial in two variables x and y is an algebraic expression whose
terms are of the form m nax y , where a is any real number and m and n are
nonnegative integers. Polynomials in three or more variables are defined similarly. The degree of a term is defined as the sum of the exponents of the variables in a term. The degree of the polynomial is the largest degree of a term.
Example 2 23x y y− is a binomial in x and y; the degree of the terms are 3 and
2, respectively; the degree of the polynomial is 3
Evaluating a Polynomial - substitute the given values of the variable(s)
Example Evaluate 3 1 for 2x x+ = .
Solution: 3 31 2 1 8 1 9x + = + = + =
Adding or Subtracting Polynomials - combine like terms - remember to change the signs of each term of the subtrahend - can be done horizontally or vertically
Example ( ) ( )2 22 5 3 6x x x x− − − + −
Solution: or
( ) ( )2 2
2 2
2
2 5 3 6
2 5 3 6
6 3
x x x x
x x x x
x x
− − − + −
= − − − − +
= − +
2
2
2
2 5 3
6
6 3
x x
x x
x x
− −
− − +
− +
63
Multiplying Polynomials
- use the distributive property
- use the rule m n m na a a +⋅ =
- use the rule factors
n
n
a a a a= ⋅ ⋅ ⋅������
for problems like ( )4
2x +
Example
( ) ( )2 3 2 2 3 24 7 1 4 4 4 7 7 7 4 3 3 7a a a a a a a a a a a− + + = + + − − − = − − −
Special Products for Multiplying Binomials (1) FOIL
Example ( )( ) 2 23 2 4 3 12 2 8 3 10 8y y y y y y y− + = + − − = + −
(2) ( )( ) 2 2a b a b a b+ − = −
Example ( )( ) 25 5 25x x x− + = −
(3) ( )2 2 22a b a ab b± = ± +
Example ( ) ( )2 2 2 23 2 3 3 6 9w w w w w− = − + = − +
Dividing a Polynomial by a Monomial
- use , 0a b a b
cc c c
+= + ≠
- use , 0m
m nn
aa a
a
−= ≠
Example 3 2 2 4 3 2 2 4
22 2 2
10 5 10 52
5 5 5
a b a b a b a ba ab
ab ab ab
+= + = +
Dividing a Polynomial by a Polynomial - make sure both dividend and divisor are arranged in
descending order - use long division (D-M-S-B)
Example
Thus, the answer is4
42
xx
− −−
.
( )
( )
2
2
4
2 6 4
2
4 4
4 8
4
x
x x x
x x
x
x
−
− − +
− −
− +
− − +
−
64
4. FACTORING POLYNOMIALS
Let us review the notion of factoring for integers. When we multiply ⋅3 5 we
get 15; 3 and 5 are called the factors of 15 and the result 15 is called the product. When we are asked to factor an integer we want to write the integer as a product of 2 or more integers. Thus, we can think of factoring as “undoing” the operation of multiplication. Factoring a polynomial means writing a polynomial as a product of two or more polynomials. We usually need to factor completely. This is equivalent to the
notion of prime factorization for integers. Thus, = ⋅24 4 6 is a factorization of
24 but the prime factorization of 24 is ⋅ ⋅ ⋅2 2 2 3 or 32 3⋅ .
We will consider various types of factoring techniques.
4.1 Factoring out the Greatest Common Factor The simplest type of factoring is factoring out the greatest common factor (GCF). This is really just the distributive property applied in reverse:
+ = +( )ab ac a b c
Examples
1. Factor: +2 2x y
Solution: ( )+ = + since 2 is a common factor of both terms2 2 2 x y x y
2. Factor: + +tz pz yz
Solution: + + = + + since is common to the three terms( ) ztz pz yz z t p y
3. Factor: − − +5 5 5 5ay ab ax aq
Solution: 5 5 5 5ay ab ax aq− − +
5 ( ) since 5 is common to the four termsa y b x q a= − − +
4. Factor: 2 2 221 21 21z yh z yw z y− −
Solution: 2 2 221 21 21z yh z yw z y− −
22since 21 is common to the three terms21 ( 1) z yz y h w= − −
Note: To check the answer to a factoring problem, multiply the polynomials in the proposed factored form. Let us now look at GCF problems where the GCF is not obvious. For instance,
consider 3 2 2 4 336 120 48x y x y x y− + .
There are 3 terms with numerical coefficients −36, 120, and 48 and they seem
to have a common factor; likewise the variable parts 3 2 2 4 3, , and x y x y x y
seem to share some common factor(s). Let us look at each part separately.
65
For the numerical coefficients we appeal to prime factorization:
2 2
3
4
2
36 2 2 3 3 2 3
120 2 2 2 3 5 2 3 5
48 2 2 2 2 3 2 3
GCF 2 2 3 2 3 12
= ⋅ ⋅ ⋅ = ⋅
= ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅
= ⋅ ⋅ ⋅ ⋅ = ⋅
= ⋅ ⋅ = ⋅ =
The answer 12 could of course be obtained by inspection but the above method gives us a way to get the GCF for the variable parts. Notice that we align the like factors and the GCF will consist of those factors that are present in all three rows. When the prime factorization is written in exponential form this is equivalent to taking the smallest power of the factor. Applying the same technique for the variable parts, we get
3 2
2
4 3
2GCF
x y x x x y y
x y x x y
x y x x x x y y y
x x y x y
= ⋅ ⋅ ⋅ ⋅
= ⋅ ⋅
= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
= ⋅ ⋅ =
Again, note that we choose the smallest powers of x and y.
The factored form of ( )3 2 2 4 3 236 120 48 is 12 ___ ___ ___x y x y x y x y− + − + .
Note that there are three blanks to fill in inside the quantity and this is because the polynomial we are factoring has three terms. To fill in each blank we perform three division problems:
3 2 2 4 3
2 2
2 2 2
36 120 483 ; 10; 4
12 12 12
x y x y x yxy x y
x y x y x y= = =
Thus, the factored form is ( )2 2 212 3 10 4x y xy x y− + .
Instead of dividing one can also do the following:
3 2 2 4 3
2 2 2
36 120 48
12 _____ 12 _____ 12 _____
x y x y x y
x y x y x y
− +
= ⋅ − ⋅ + ⋅
and one can figure out what the missing factor is by “eyeballing” the problem. One should then get:
( )
3 2 2 4 3
2 2 2
2 2 2
36 120 48
12 12 12
12 3 10 4
x y x y x y
x y x y x y
x y xy x y
− +
= ⋅ − ⋅ + ⋅
= − +
2 23xy 10 4x y
66
We will use the exponential form in getting the GCF in the next examples.
6. Factor: 2 4 3 4 2 420 30 10a b a b c a b+ +
Solution:
2 4 2 2 4
3 4 3 4
2 4 2 4
2 4 2 4
20 2 5
30 3 2 5
10 2 5
GCF 2 5 10
a b a b
a b c a b c
a b a b
a b a b
= ⋅ ⋅ ⋅
= ⋅ ⋅ ⋅ ⋅ ⋅
= ⋅ ⋅ ⋅
= ⋅ ⋅ ⋅ =
Thus,
( )
2 4 3 4 2 4
2 4 2 4 2 4
2 4
20 30 10
10 10 10
10 2 3 1
a b a b c a b
a b a b a b
a b ac
+ +
= ⋅ + ⋅ + ⋅
= + +
2 3ac 1 .
7. Factor: 3 2 2 4 3 2 218 36 45x yz x y z x yz+ −
Solution:
3 2 2 3 2
2 4 3 2 2 2 4 3
2 2 2 2 2
2 2 2 2 2
18 2 3
36 2 3
45 5 3
GCF 3 9
x y z x y z
x y z x y z
x yz x z
x y z x yz
= ⋅ ⋅ ⋅ ⋅
= ⋅ ⋅ ⋅ ⋅
= ⋅ ⋅ ⋅ ⋅
= ⋅ ⋅ ⋅ =
Thus,
( )
3 2 2 4 3 2 2
2 2 2 2 2 2
2 2 3
18 36 45
9 9 9
9 2 4 5
x yz x y z x yz
x yz x yz x yz
x yz x y z
+ −
= ⋅ + ⋅ − ⋅
= + −
32x 4y z 5
So far in the above examples the GCFs have been monomials. We can also have GCFs that are quantities having two or more terms. We can use substitution to help us factor these polynomials.
8. Factor: ( ) ( )+ + +2 a b x a b
Solution: We notice that the binomial +a b is common to the two terms.
We can do a substitution, say, let = +n a b (any letter can be used); then the
problem will look like +2n xn . Clearly, n is common to the two terms, so we
can factor it out to get ( )+2n x and replacing n by +a b , we get the factored
form ( )( )+ +2a b x .
67
9. Factor: − + −(2 ) 4 (2 )x y z p y z
Solution:
( )
( )( )
(2 ) 4 (2 ) 4 Replace 2 by .
4 Factor out .
2 4 Replace by 2 .
x y z p y z xa pa y z a
a x p a
y z x p a y z
− + − = + −
= +
= − + −
10. Factor: − − −2 ( ) 5 ( )a x y b x y
Solution:
One does not have to do the substitution if one can clearly see the binomial GCF. We will omit the substitution process in the next examples. Sometimes the suspected GCF does not look the same in the terms.
11. Factor: + − +5( 7) (7 )m x m
Solution: + 7m and +7 m are the same by the commutative property for addition. Thus,
( ) ( )
( ) ( )
( )( )
+ − +
= + − +
= + −
5 7 7
5 7 7
7 5
m x m
m x m
m x
12. Factor: ( ) ( )− + −5 5w y z y
Solution: This time − −5 and 5y y are not the same but are opposites of each
other since the terms have opposite signs. Note that
( )− = − −5 5y y or ( )− = − −5 5y y .
Using the second, we have
13. Factor: ( ) ( )− − −4 8 11 8a p p
Solution: Again note that − −8 and 8p p are opposites of each other, i.e.
− = − −8 (8 )p p or ( )− = − −8 8p p .
Using the second one (one can also use the first), we get
( ) ( )
( ) � ( )
( )( )
−−
−
− − −
= − + −
= − +
�����the changed
8 changedto +to 8
4 8 11 8
4 8 11 8
8 4 11
pp
a p p
a p p
p a
( ) ( )
( )
( )( )
2 5 2 5 Replace by .
2 5 Factor out .
2 5 Replace by .
a x y b x y ap bp x y p
p a b p
x y a b p x y
− − − = − −
= −
= − − −
( ) ( )
( ) � ( )
( ) ( )
−−
− + −
= − − −
= − −
�����the + sign
5 becamechanged to a -5
5 5
5 5
5
yy
w y z y
w y z y
y w z
68
Sometimes the GCF has to be factored out more than once.
14. Factor: ( ) ( )− + −6 3 9 3x a b y a b
Solution: ( ) ( )6 3 9 3x a b y a b− + −
( ) ( )
( ) ( )
( )( )
3 6 9 Factor out the GCF 3 .
3 3 2 3 Factor out 3 from 6 9 .
3 3 2 3 Apply commutativity of multiplication.
a b x y a b
a b x y x y
a b x y
= − + −
= − ⋅ + +
= − +
15. Factor: ( ) ( )215 2 10 2ax y x y− − −
Solution: ( ) ( )215 2 10 2ax y x y− − −
( ) ( )
( )( )
( )( ) 2
2
2
Note that 2 and 2 are opposites.
Factor out the GCF 2.
Factor out 5 from 15 10 .
15 2 10 2
15 10 2
5 3 2 2
y y
y
x ax x
ax y x y
ax x y
x ax y
− −
−
+
= − + −
= + −
= + −
We can also translate from English to algebra and factor.
16. The sum of twice a number and 8
Solution:
2
2 8
The sum of twice a number and 8
x
x+
���������
�����������������
Factoring out the GCF 2 from 2 8x + , we get ( )2 4x + .
17. The difference of twice a number and its square
Solution:
2
2
2
2
The difference of twice a number and its squarex
x
x x−
��������� �����
�����������������������
Factoring out the GCF x from 22x x− , we get ( )2x x− .
18. 6 added to the sum of twice the square of a number and four times the number Solution:
�
2
2
2
2 2
++ 4
2
2 4
6 2 4 or 2 4 6
6 added to the sum of twice the square of a number and four times the number
xx
x
x x
x x x x
+
+ + + +
����� ����������������������
���������������
���������������������������������
������������� ������� �������������������
Factoring out the GCF 2 from either expression, we get
( ) ( )2 22 3 2 or 2 2 3x x x x+ + + + .
69
4.2 Factoring by Grouping In some cases there is no GCF that can be factored out from all the terms but when certain terms are grouped one finds a GCF. Consider the polynomial
+ + +2 2x y ax ay .
There is no GCF for the four terms but if we consider the first two terms we see that the GCF is 2 and if we consider the last two terms we see that the GCF is a. Thus, if we group the first two terms and the last two terms, we have
( ) ( )
( ) ( )
( )( )
Group the 1st two and the last two.
Factor out 2 from the 1st grp; from the 2nd.
Factor out .
2 2
2 2
2
2
a
x y
x y ax ay
x y ax ay
x y a x y
x y a +
+ + +
= + + +
= + + +
= + +
Examples
1. Factor: − + −3 5 15xt t x
Solution:
( ) ( )
( ) ( )
( )( )
Group the 1st two terms and the last two terms.
Factor out from the 1st group and 5 from the 2nd group.
Factor out 3.
3 5 15
3 5 15
3 5 3
3 5
t
x
xt t x
xt t x
t x x
x t −
− + −
= − + −
= − + −
= − +
2. Factor: + − −2 2a ac b bc
Solution:
( ) ( ) ( )
( ) ( )
( )( )
Group 1st two and last two; change signs inside for the 2nd grp. (Why?)
Factor out from the 1st and from the 2nd.
Factor out 2 .
2 2
2 2
2 2
2
a b
c
a ac b bc
a ac b bc
a c b c
c a b +
+ − −
= + − +
= + − +
= + −
It may be necessary to rearrange the terms first before grouping:
3. Factor: − + −cx dy cy dx
Solution:
( ) ( )
( ) ( )
( )( )
− + −
= + − −
= + − +
= + − +
= + −
Apply commutativity of +.
Group 1st two and last two.
Factor out from 1st, from 2nd.
Factor o
c d
cx dy cy dx
cx cy dy dx
cx cy dy dx
c x y d y x
x y c d +ut .x y
It may be necessary to factor some more after factoring out the binomial GCF:
70
4. Factor: 3 2 22 2 4 4x x y x xy− − +
Solution:
( ) ( )( ) ( )
( )( )
( )( )
2
3 2 2
3 2 2
2
2
Group 1st two and last two.
Factor out 2 from 1st grp, 4 from 2nd.
Factor out .
2 2 4 4
2 2 4 4
2 4
2 4
2 2
x x
x y
x x y x xy
x x y x xy
x x y x x y
x x x y
x x x y
−
− − +
= − − −
= − − −
= − −
= − −2
Factor out 2x from 2 4 . x x−
4.3 Factoring Trinomials of the Form 2x bx c+ +
In Section 3.4 we saw that the product, say, of ( )( )+ +3 5x x using FOIL is
2 5 3 15x x x+ + + which simplifies to 2 8 15x x+ + . Now the problem is to
factor 2 8 15x x+ + and we see, of course, that the answer is ( )( )+ +3 5x x .
We want to develop a general way of factoring any trinomial of the
form 2x bx c+ + .
We note the following:
1. 2 8 15x x+ + is the product of two binomials of the form ( )( )+ +? ?x x .
2. The middle coefficient 8 is equal to +3 5 and the constant term 15 is equal
to ⋅3 5 .
3. The factored form is ( )( )+ +3 5x x .
Thus, to factor 2x bx c+ + , we want to find two numbers whose product is c
and whose sum is b, say m and n and write the factored form as
( )( )+ +x m x n . This process can be represented by the following diagram:
In other words, we want to find factors of the constant term c that add up to the middle coefficient b. Note the following: If c is positive and b is positive, then both m and n must be positive.
If c is positive and b is negative, then both m and n must be negative.
If c is negative, then m and n must have opposite signs and the sign of the “larger” is the same as the sign of b.
+ b
i c
m n
71
Examples
1. Factor: 2 5 6x x+ +
Solution:
( )( )2 5 6 2 3x x x x+ + = + +
2. Factor: 2 3 10t t+ −
Solution:
( )( )2 3 10 5 2t t t t+ − = + −
3. Factor: 2 2y y− − Solution:
( )( )2 2 2 1y y y y− − = − +
4. Factor: 2 3 2n n− + Solution:
( )( )2 3 2 2 1n n n n− + = − −
It may not always be possible to find two numbers that will satisfy the two conditions as in the following example.
5. Factor: 2 5x x+ +
Solution: We want to find two numbers that multiply to 5 and add up to 1 which is impossible. In this case, we say the polynomial is not factorable or that it is prime or irreducible. Sometimes the polynomial involves two variables.
6. Factor: 2 27 12x xy y− + Solution:
( )( )2 27 12 4 3x xy y x y x y− + = − −
7. Factor: 2 26 9a ab b+ + Solution:
( ) ( )2 26 9 3 3a ab b a b a b+ + = + +
Note that the answer in Example 7 can be written as ( )2
3a b+ . This is an
example of a perfect square trinomial. Next, let us see how we can deal with large numbers.
+ 5
i 6
2 3
+ 3
i -10
5 -2
+ -1
i -2
-2 1
+ -3
i 2
-2 -1
+ -7
i 12
-4 -3
+ 6
i 9
3 3
72
8. Factor: 2 3 108x x− −
Solution: We want to find two factors of −108 that add up to −3 which is not
a very easy task. One strategy is to break down 108 into its prime factors and
“play” with the factors and see which pair will give −3 when subtracted (note
that we are really subtracting since we want one positive factor and one negative factor). One can verify that the prime factorization of 108 is given by
⋅ ⋅ ⋅ ⋅2 2 3 3 3
so we can “play” with two 2’s and three 3’s as follows:
Try �⋅ ⋅ ⋅ ⋅���
4 27
2 2 3 3 3 .
We note that there is no way 4 and 27 will give −3 when subtracted.
So we try another combination:
�⋅ ⋅ ⋅ ⋅���
912
2 2 3 3 3
and we see that there is a lot of hope to get a −3 from 12 and 9. Note that if
we add − +12 9 we get 3. Thus, if we go back to our “X” diagram we have
and we have the factored form of 2 3 108x x− − is ( )( )− +12 9x x .
9. Factor: 2 38 360x x+ +
Solution: We want to find the factors of 360 that add up to 38. This time we want to add the two positive factors. So we can break down 360 into its prime factors and “play”. The following shows the correct combination:
What this means is we get the factors 18 and 20 from ⋅ ⋅2 3 3 and ⋅ ⋅2 2 5 ,
respectively. Thus 2 38 360x x+ + factors into ( )( )+ +18 20x x .
We also need to watch out for common factors.
+ −3
i −108 = − ⋅12 9
-12 9
+ 38
i 360 = ⋅ ⋅ ⋅ ⋅ ⋅2 2 2 3 3 5
18 20
73
10. Factor: 25 30 80x y xy y− −
Solution: Note that the three terms have 5y for a GCF so we need to factor
this out first.
( )( )( )
2
2
5 30 80
5 6 16
5 8 2
x y xy y
y x x
y x x
− −
= − −
= − +
We can translate and then factor the resulting expression.
11. The square of a number added to the difference of the number and 12
Solution:
2 + 12
The square of a number added to the difference of the number and 12
xx −
����� ������������������������������
Thus, we get the expression 2 12x x+ − which factors to ( )( )4 3x x+ − .
12. The sum of twice a number and 15, subtracted from the square of the number Solution:
22 this meanssecond quantity - 2 15first quantity
The sum of twice a number and 15, subtracted from the square of the number
x xx+
��������� ��������� �������������
�����������������
Thus the expression is ( )2 2 15x x− + which simplifies to 2 2 15x x− −
and factors to ( )( )5 3x x− + .
4.4 Factoring Trinomials of the Form 2ax bx c+ +
We now look at the case when the numerical coefficient of 2x , called the leading coefficient, is not equal to 1. For instance, when we multiply
( )( )2 1 3 4x x+ − using FOIL we get 26 8 3 4x x x− + − which simplifies to
26 5 4x x− − . It is not trivial to go from 26 5 4x x− − to the factored form
( )( )2 1 3 4x x+ − . One could start either with ( ) ( )2 ? 3 ?x x+ + or
( )( )6 ? ?x x+ + .
We will discuss three methods:
1. trial - and error method
2. grouping method
3. bottoms-up method The first method, as the name suggests, involves guessing while the second and third are more systematic. Each method can assume various forms depending on who is explaining it.
74
The Trial – and – Error Method This method basically considers the leading coefficient a and the constant term
c and all their possible factors. Going back to the example, we have 6a = and
we either have 6 2 3 or 6 6 1= ⋅ = ⋅ ; 4c = − and 4 4 1 or 4 4 ( 1)− = − ⋅ − = ⋅ − or
4 2 2− = − ⋅ . Then we look at the form
( ) ( )? ? ? ?x x+ +
and plug in various combinations of the factors for a and c and hope to get b when we apply O and I of FOIL. For instance we can try,
( )( )6 1 4x x− +
but we get OI as 24 23 5x x x x− = ≠ − so this is an error…so we try say,
( )( )3 1 4x x− +
but we again get another error since OI gives 12 11 5x x x x− = ≠ − …so we try
again until we get
( )( )2 1 3 4x x+ −
where we see that OI is 8 3 5x x x− + = !!! We hit the correct factored form!
Note that we only need to check the OI part of FOIL because the F and L were already taken care of when we did the factoring of a and c. The Grouping Method Like the trial-and-error method, the grouping method also considers a and c but this time it considers the PRODUCT of a and c. Here are the first 2 steps in the grouping method: Step 1 Get the product ac.
Step 2 Find two numbers whose product is ac and whose sum is b. Steps 1 and 2 can be summarized in the following diagram: Note that Step 2 is similar to the procedure done in the preceding section. In the preceding section, a was equal to 1, so the method explained there is a special case of the methods explained in this section.
+ b
i ac
m n
75
Back to the example26 5 4x x− − ,
we see that a = 6 and 4c = − , so 24ac = − . Now we want to find two numbers
that will give 24− when multiplied and 5− when added. These are 8 and 3− .
Step 3 Write the middle term as the sum or difference of the two numbers found in Step 2.
2
middle term 5
6 8 3 4x
x x x−
− + −�����
Step 4 Factor the resulting polynomial with four terms by grouping.
( ) ( )
( ) ( )
( )( )
2
2
2
6 5 4
6 8 3 4
6 8 3 4
2 3 4 3 4
2 1 3 4
x x
x x x
x x x
x x x
x x
− −
= − + −
= − + −
= − + −
= + −
The Bottoms-up Method
Step 1 Get the product ac.
Step 2 Find two numbers whose product is ac and whose sum is b. →
Note: These are the same first steps in the grouping method.
Step 3 Using the two numbers 8− and 3 we can start with the factored
form ( )( )8 3x x− + .
Step 4 Divide 8− and 3 by the leading coefficient 6 to get8 3
6 6x x
− +
.
Step 5 Reduce the resulting fractions to lowest terms giving4 1
3 2x x
− +
.
Step 6 Apply bottoms-up to get ( )( )3 4 2 1x x− + .
+ -5
i 6(-4) =-24
-8 3
+ b
i ac
m n
+ -5
i -24
-8 3
76
The above steps can be abbreviated using the following diagram:
→reduce
Then doing “bottoms-up” yields the factored form ( )( )3 4 2 1x x− + .
Note that everything can be done using one diagram. Why does the bottoms-up method work? Here’s a little “proof” by example:
If we factor out the leading coefficient 6 from 26 5 4x x− − , we will get
2 5 46
6 6x x
− −
.
Note that the quantity inside the parentheses is a trinomial square with leading coefficient of 1 and thus we can factor it as in the preceding section, i.e. we can
look for factors of 4
6− that will add up to
5
6− .
If we look at the “X” diagram we have the following:
We note that 8 3 2
6 6− ⋅ = −
4 3⋅ ⋅
2 3⋅
4
66= −
⋅ and
8 3 5
6 6 6− + = − and thus the factors of
4
6− that will add up to
5
6− are
8 3 and
6 6− . Of course it is not an easy task to
think of these factors! Now if we reduce these to lowest terms we get
4 1 and
3 2− . Thus, the factored form we have so far is
2 5 4 4 16 6
6 6 3 2x x x x
− − = − +
.
+ 5−
i 6(-4) -24
-8 6
3 6
+ 5−
i 6(-4) -24
-4 3
1 2
+ 5
6−
i 4
6−
-8 6
3 6
77
Now, how do we get the answer ( )( )3 4 2 1x x− + ? We need to answer how
the 6 and the denominators 3 and 2 disappeared from the answer. The trick is
to factor 6 into 3 2⋅ and distribute the factors as follows
( )( )
4 1 4 16 3 2
3 2 3 2
4 1 3 3 2 2
3 2
3 4 2 1
x x x x
x x
x x
− + = − ⋅ +
= − ⋅ + ⋅
= − +
Conclusion: The bottoms-up technique is a mathematically sound way of factoring and…in the author’s opinion is a much more efficient way of factoring than the grouping method. ☺ However, one should watch out for the case when there are GCFs that can be factored out! WARNING! Make sure to factor out all GCFs before applying the bottoms-up method.
Example Factor 212 26 10x x+ −
Solution: The CORRECT way to use the bottoms-up technique:
( )2 212 26 10 2 6 13 5x x x x+ − = + −
Thus, ( )( )212 26 10 2 2 5 3 1x x x x+ − = + − .
The INCORRECT way to use the bottoms up technique:
→reduce
+ 13
-2 6
15 6
i 6(-5) -30
+ 26
-4
12
30 12
i 12(-10) -120
+ 26
-1 3
5 2
i 12(-10) -120
reduce→
+ 13
-1 3
5 2
i 6(-5) -30
78
Then, if one concludes that ( )( )212 26 10 2 5 3 1x x x x+ − = + − , this is not
correct since multiplying the binomials will give 26 13 5x x+ − and not the
original problem. The next examples will focus on the grouping and bottoms-up methods.
Examples
1. Factor: 23 8 3x x+ −
Solution:
A. grouping
( ) ( )
( )( )
2
2
3 8 3
3 9 3
3 3 3
3 1 3
x x
x x x
x x x
x x
+ −
= + − −
= + − +
= − +
B. bottoms-up
Using either method, one gets the factored form ( )( )3 3 1x x+ − .
2. Factor: 24 3 10x x− −
Solution:
A. grouping
( ) ( )
( ) ( )
( )( )
2
2
2
4 3 10
4 8 5 10
4 8 5 10
4 2 5 2
4 5 2
x x
x x x
x x x
x x x
x x
− −
= − + −
= − + −
= − + −
= + −
B. bottoms-up
Thus, ( )( )24 3 10 2 4 5x x x x− − = − + .
+ 8
i 3(-3) -9
9 -1
+ 8
i 3(-3) -9
3 1
-1 3
+ -3
i 4(-10)
-40
-8 5
+ -3
i 4(-10)
-40
-2 1
5 4
+ 8
i 3(-3) -9
9 3
-1 3
reduce→
+ -3
i 4(-10) -40
-8 4
5
4 reduce
→
79
Sometimes the product ac is a large number. For such cases it is better to factor a and c and “play” with the factors to see which combination will produce b, i.e. DO NOT multiply a and c and get a really large number!
3. Factor: 214 47 72x x− − Solution:
The prime factorizations of 14 and 72 are 2 7⋅ and 2 2 2 3 3⋅ ⋅ ⋅ ⋅ , respectively.
“Play” with the four 2’s, the two 3’s and the 7 until we get 47− when the factors
are added: 14 and 72 will not give 47− , 56 and 18 will not work as well, but 63 and 16 will work as follows: One can then do either grouping or bottoms-up: grouping:
( ) ( )
( ) ( )
( )( )
2
2
2
14 47 72
14 63 16 72
14 63 16 72
7 2 9 8 2 9
2 9 7 8
x x
x x x
x x x
x x x
x x
− −
= − + −
= − + −
= − + −
= − +
bottoms-up:
→reduce
This method also gives ( )( )214 47 72 2 9 7 8x x x x− − = − + .
+ -47
i 14(-72)
-8 5
+ -47
i 14(-72)
-63
14 16 14
+ -47
i 14(-72)
-9
2 8 7
16
+ -47
i 14(-72)
2 7 2 2 2 3 3= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
-63
80
Again, make sure to factor out any common factors first.
4. Factor: 218 54 40x x− +
Solution: Note that 2 is common to all three terms so factor it out first and then do grouping or bottoms-up.
( ) ( )( )2 218 54 40 2 9 27 20 2 3 5 3 4x x x x x x− + = − + = − − .
5. Factor: 3 230 51 189y y y+ −
Solution: Factor out 3y first and then do either grouping or bottoms-up to get
( ) ( )( )3 2 230 51 189 3 10 17 63 3 2 7 5 9 .y y y y y y y y y+ − = + − = + −
We can translate and factor.
6. Six times the square of a number, added to the difference of 6 and thirteen times the number Solution:
2
26
13
6 13
Six times the square of a number, added to
the difference of 6 and thirteen times the number
x
x
x
x
+
−
����������������
�����������������
�������������
�������������������������
Thus, the expression is 26 6 13x x+ − or 26 13 6x x− + which factors
into ( )( )3 2 2 3 .x x− −
4.5 Factoring the Difference of Squares
In Section 3.4 we saw the special product formula 2 2( )( )a b a b a b+ − = − .
The result is a difference of squares. In order now to factor a difference of squares we reverse the formula to get
2 2 ( )( )a b a b a b− = + − .
Examples
1. Factor: 2 16y −
Solution: ( )( )2 2 216 4 4 4y y y y− = − = + −
2. Factor: 225 t−
Solution: ( )( )2 2 225 5 5 5t t t t− = − = + −
3. Factor: 2 81z −
Solution: ( ) ( )2 2 281 9 9 9z z z z− = − = + −
81
4. Factor: 2 1
4x −
Solution: 2
2 21 1 1 1
4 2 2 2x x x x
− = − = + −
5. Factor: 2 216 9x y−
Solution: ( ) ( ) ( )( )2 22 216 9 4 3 4 3 4 3x y x y x y x y− = − = + −
6. Factor: 2 2
25 4
a b−
Solution:
2 22 2
25 4 5 2 5 2 5 2
a b a b a b a b − = − = + −
Can we factor the sum of two squares? For instance, let us look at 2 9x + .
We can consider 2 9x + as a trinomial of the type discussed in Section 4.3 with
0 for the middle coefficient. Thus we have
2 29 0 9x x x+ = + +
and we want to find two numbers whose product is 9 and whose sum is 0. This
is impossible and so 2 9x + cannot be factored or we say that it is prime or
irreducible. We can again combine this technique with factoring out the GCF first.
7. Factor: 2 232 50a b−
Solution: ( ) ( )( )2 2 2 232 50 2 16 25 2 4 5 4 5a b a b a b a b− = − = − +
8. Factor: 3 2 38 18x y x y−
Solution: ( ) ( )( )3 2 3 2 28 18 2 4 9 2 2 3 2 3 x y x y xy x y xy x y x y− = − = − +
Let us translate and factor the resulting expression. 9. Four times the square of a number, less nine times the square of another number Solution: We translate as follows:
�
2 2
2 24 9
Four times the square of a number, less nine times the square of another number
x y
x y
−����������� ���������������
����������������� ���������������������
Thus, we get the expression 2 24 9x y− which factors into ( )( )2 3 2 3x y x y+ − .
82
4.6 Factoring the Sum or Difference of Cubes
If we perform the multiplication ( )( )2 2a b a ab b+ − + we will get 3 3a b+
(verify!). Likewise, the product ( )( )2 2a b a ab b− + + yields 3 3a b− (verify!).
Thus we have the following factoring formulas for the sum and difference of two cubes, respectively:
( )( )
( )( )
3 3 2 2
3 3 2 2
a b a b a ab b
a b a b a ab b
+ = + − +
− = − + +
Examples
1. Factor: 3 1x +
Solution: ( ) ( ) ( )( )3 3 3 2 2 21 1 1 1 1 1x x x x x x x x+ = + = + − + = + − +
2. Factor: 3 8x −
Solution: ( )( ) ( )( )3 3 3 2 2 28 2 2 2 2 2 2 4x x x x x x x x− = − = − + + = − + +
3. Factor: 3 38 27y m+
Solution: ( ) ( ) ( ) ( ) ( )( ) ( )( )3 3 2 23 38 27 2 3 2 3 2 2 3 3y m y m y m y y m m+ = + = + − +
( )( )2 22 3 4 6 9y m y ym m= + − +
4. Factor: 3 327a b−
Solution: ( ) ( )( )33 3 3 2 227 3 3 9a 3 a b a b a b ab b− = − = − + +
Again, watch out for any common factors:
5. Factor: 4 340 5x xy+
Solution: ( ) ( )( )4 3 3 3 2 240 5 5 8 5 2 4 2x xy x x y x x y x xy y+ = + = + − +
We can translate from English to algebra and factor: 6. Eight times the cube of a number, less 27 Solution: We translate as follows:
3
3
3
8
8 27
Eight times the cube of a number , less 27
x
x
x −
�����������
�����������������
���������������������
Thus, the expression is 38 27x − which we then factor as the difference of two
cubes:
( ) ( )( )33 3 28 27 2 3 2 3 4 6 9x x x x x− = − = − + +
83
4.7 Combined Factoring Techniques In this section we will summarize all the factoring techniques and sometimes combine them in a single problem. The following strategy is useful when dealing with a factoring problem: Step 1: Factor out the GCF, if any. Step 2: Count the number of terms. a. two terms of the form
difference of squares ( )( )2 2a b a b a b− = + −
sum of squares 2 2a b+ is prime or not factorable
difference of cubes ( ) ( )3 3 2 2a b a b a ab b− = − + +
sum of cubes ( )( )3 3 2 2a b a b a ab b+ = + − +
b. three terms of the form 2x bx c+ + will factor to ( )( )x n x m+ +
where and n m b n m c+ = ⋅ =
c. three terms of the form 2ax bx c+ +
Use trial – and – error method or the ac – method:
(i) Find two numbers n and m such that and n m b n m ac+ = ⋅ =
(ii) Do either factoring by grouping or bottoms-up technique d. four or more terms
Try factoring by grouping. Note that when there are four or more terms that it is better to try grouping first before factoring out any GCF.
m
+ b
n m
c
i
+ b
n
ac
i
84
Examples
1. Factor completely: 5 52 8x y xy−
Solution: Following the strategy we first check if there are any GCFs and we see that 2 is the GCF so we get:
( )5 5 4 42 32 2 16x y xy xy x y− = − .
Next, we count the number of terms in the factor 4 416x y− and we see that
there are two terms and we have the form of a difference of squares since
( ) ( )2 2
4 4 2 216 4x y x y− = −
which factors into
( ) ( )2 2 2 24 4x y x y− + .
Now the first factor is again the difference of two squares while the second is prime or irreducible; thus, we can further factor the polynomial into
( )( )( )2 22 2 4x y x y x y− + + .
Thus, the complete factored form for the problem is
( )( )( )2 22 2 2 4xy x y x y x y− + + .
2. Factor completely: 3 2 2 3 43 6 3a b a b ab+ +
Solution: 3 2 2 3 43 6 3a b a b ab+ +
( )( )( )
( )
2 2 2 2
2 2 2
22
3 2 Factor out the GCF 3 .
3 Factor the trinomial 2 .
or 3
ab a ab b ab
ab a b a b a ab b
ab a b
= + +
= + + + +
+
3. Factor completely: 3 2 2 23 3 12 12x x xy y+ − −
Solution: 3 2 2 23 3 12 12x x xy y+ − −
( ) ( )( ) ( )
( ) ( ) ( )
( )( )
3 2 2 2
2 2
2 2
2 2
3 3 12 12 4 terms group terms
3 1 12 1 Factor out the GCF from each group.
3 12 1 Factor out the GCF 1 .
3 4 1
x x xy y
x x y x
x y x x
x x y x
= + + − − ⇒
= + − +
= − + +
= − +
( )( )( )
2 2
2 2
Factor out the GCF 3 from 3 12 .
3 2 2 1 Factor the difference of squares 4 .
x x y
x x y x y x x y
−
= + − + −
85
Sometimes a substitution might help in seeing the form of the problem better.
4. Factor completely: 4 26 7 2x x+ +
Solution: This trinomial is actually in the form 2ax bx c+ + since
( )2
4 2 2 26 7 2 6 7 2x x x x+ + = + + .
.Using the substitution, say, 2x y= , the trinomial becomes
26 7 2y y+ +
and this polynomial can be factored by grouping or bottoms-up to get
( )( )2 1 3 2y y+ + . (VERIFY!)
Replacing y by 2x , we get the factored form
( )( )2 22 1 3 2x x+ + .
5. Factor completely: 5 312 63 15y y y− +
Solution: 5 312 63 15y y y− +
( )
( )( )
( )( )( )
4 2
2 2 4 2
2 2
3 4 21 5 Factor out the GCF 3 .
3 4 1 5 Factor the trinomial square 4 21 5.
3 2 1 2 1 5 Factor the difference of squares 4 1.
y y y y
y y y y y
y y y y y
= − +
= − − − +
= − + − −
6. Factor completely: 5 3 25 45 5 45a a a− + −
Solution: 5 3 25 45 5 45a a a− + −
( ) ( )
( ) ( )
( )( )
( )
5 3 2
3 2 2
2 3 2
2 3
5 45 5 45 Group the first two and last two terms.
5 9 5 9 Factor out the GCF from each group.
9 5 5 Factor out 9.
5 9
a a a
a a a
a a a
a a
= − + −
= − + −
= − + −
= − +( )
( )( )( )( )
3
2 2 3
1 Factor out 5 from the second factor 5 5.
5 3 3 1 1 Factor 9 and 1.
a
a a a a a a a
+
= − + + − + − +
86
4.8 Chapter Review Factoring Techniques (see Section 4.7)
1. Factoring out the greatest common factor or GCF
Examples ( )4 18 2 2 9xy y y x− = −
( )2 212 4 20 4 3 1 5a b ab ab ab a b− + = − +
( ) ( ) ( )( )3 6 5 6 6 3 5a y y y a− + − = − +
( ) ( ) ( ) ( ) ( )( )5 3 6 3 5 3 6 3 3 5 6t m m t m m m t− + − = − − − = − −
2. Factoring by grouping Examples
( ) ( )
( ) ( )
( )( )
3 3
3 3
3
3
x y bx by
x y bx by
x y b x y
x y b
+ + +
= + + +
= + + +
= + +
( ) ( )
( ) ( )
( ) ( )
( )( )
5 10 4 2
5 10 4 2
5 2 2 2
5 2 2 2
2 5 2
m n xn xm
m n xn xm
m n x n m
m n x m n
m n x
− − +
= − − −
= − − −
= − + −
= − +
3. Factoring trinomial squares with leading coefficient of 1 Find the factors of the constant term that add up to the middle coefficient.
Examples ( )( )2 6 8 2 4x x x x+ + = + +
( )( )2 23 10 5 2m mn n m n m n− − = − +
( )( ) ( )22 26 9 3 3 3y xy x y x y x y x− + = − − = −
4. Factoring trinomial squares with leading coefficient not 1
Find the factors of the product of the leading coefficient and the constant term that add up to the middle coefficient. Then do either grouping or bottoms-up.
Examples ( ) ( )210 17 3 2 3 5 1a a a a+ + = + +
( )( )2 26 35 36 3 4 2 9p pq q p q p q− + = − −
5. Factoring the difference of squares
Examples ( )( )2 25 5 5w w w− = + −
( )( )2 216 81 4 9 4 9a b a b a b− = + −
6. Factoring the sum or difference of cubes
Examples ( )( )3 227 3 3 9x x x x+ = + − +
( )( )3 264 4 4 16n n n n− = − + +
