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Unit 6
Exponents and Polynomials
Introduction
Symbols are very important in Algebra. There was a time when mathematicians
did not have a standard set of symbols. Without the use of variables, symbols and all
kinds of notation, mathematics looks much different and is much harder. For instance,
a problem written without any symbols or variables would similar to the example
below.
Find the sum of the square of the number plus the number multiplied by a
factor of two increased by one all divided by the difference of the square
of the number and one.
For most people this problem, when written without the variables and symbols, is very
hard to interpret or solve. Below the same problem is written with our variables and
symbols.
Simplify
x 2  2x  1
x2  1
This representation with variables and symbols can be understood and simplified by
most Algebra I students when they finish the next two units. Symbols, variables, and
notation make it possible for us to use the properties of real numbers, the Properties
of Equality, solve equations, and do many other things.
In this unit we will learn to manipulate expressions, as well as perform operations
with polynomials. Additionally, we will learn to represent and use exponents in new ways.
These abilities are essential to Algebra and mathematics.
Unit 6
Vocabulary and Concepts
Base
The number that is going to be raised to a power.
Binomial
A binomial is an expression with exactly two terms.
Coefficient
A numerical or constant quantity placed before and multiplying the
variable in an algebraic expression.
Constant
A constant is a number on its own, or sometimes a letter such as ‘a’,
‘b’, or ‘c’ to stand for a fixed number.
Degree
The degree of a term is the sum of the exponents in the term.
Exponent
An exponent is a raised value or variable to the right of the base that
tells us how many times to use a factor.
Factors
Factors are numbers we multiply together.
Monomial
A monomial is an expression with exactly one term.
Order
The order of a polynomial is the highest degree in the polynomial.
Polynomial
A polynomial is an expression with one or more terms.
Trinomial
A trinomial is an expression with exactly three terms.
Term
A term is part of an expression separated by a plus or minus sign.
Key Concepts for Exponents and Polynomials
Adding and Subtracting Monomials To add or subtract monomials we combine like
terms.
Multiplying Monomials When we multiply monomials we should multiply coefficients
and add exponents.
Dividing Monomials When we divide monomials we should reduce coefficients and
subtract exponents in the denominator from the exponents in the
numerator.
Negative Exponents A negative exponent means the factor belongs in the
denominator of a fraction.
Power of Zero
Any value or variable raised to the zero power is 1.
Common Factoring Common factoring is reversing the Distributive Property. This is
done by putting the GCF on the outside of parentheses and
finding the terms that belong inside the parentheses.
Unit 6 Section 1
Objective

The student will add and subtract monomials.
The most basic type of expression is called a monomial. To understand what a
monomial is we need to break the word down into its parts. The root word in
monomial is “nomial”. This word means “term” in Algebra. There are several prefixes
that can be attached to the root. The prefixes are: poly, tri, bi, and mono. The
meanings of the prefixes are listed below.
“poly”
“tri”
“bi”
“mono”
means many
means three
means two
means one
Putting the root word and the prefixes together we get the following vocabulary.
A monomial is an expression with exactly one term.
A binomial is an expression with exactly two terms.
A trinomial is an expression with exactly three terms.
A polynomial is an expression with one or more terms.
Polynomial is really the name for the overarching category of expressions. Monomials
binomials and trinomials are all subcategories of expressions. The Venn diagram below
shows these relations ships.
Polynomials
monomials
binomials
trinomials
Examples of each of the categories are given below.
Monomials
Binomials
Trinomials
2xy
-11a2bc
4
2y + 3z
4x2 – 25y2
6x + 1
4a – 2b + 7
3x2 - 16x + 5
5az3 – 9az2 – 2az
Manipulating expressions in Algebra begins with adding and subtracting monomials.
This skill is actually something we already know how to do. Another name for adding
and subtracting monomials is combining like terms. To add or subtract monomials we
must remember the definition of “like” terms. Terms are “like” when they have the
same variables with the same exponents. The way to combine like terms is to add
or subtract the coefficients of the like terms. Several examples follow.
Example A
5z – y + z + 4y
3y + 6z
The like terms in this example are 4y with -1y and
5z paired with 1z. So we add their coefficients.
We should usually put our variable terms in
alphabetical order.
Example B
4x – x2 + 7x + 6x2 The like terms here are 4x with 7x and -1x2 paired
5x2 + 11x
with 6x2. We have two pairs of like terms because
the x2 and x are not like due to the different
exponents. We should usually put the higher order
exponents first.
Example C 6 + 2a – 7a + a – 1 The like terms in this expression are those with ‘a’
-4a + 5
and the constants or plain numbers. We usually
list the constant at the end of the expression.
Example D
x+5+y–y+3
x+8
When we combine the like terms here 1y and -1y
add up to 0 and so we have 0y or 0 but we don’t
need to write the 0y. There is nothing to
combine with the ‘x’ so it simply stays in the
expression.
When we are working with monomial expressions we will be asked to determine
the degree of a term. The degree of a term is the sum of the exponents in the term.
The examples that follow show a sample of monomials and the degree of the monomial.
Example A
4x2
The monomial is of degree 2.
Example B -2x3yz The monomial is of degree 5. The ‘y’ and ‘z’ have understood
exponents of 1, so 3 + 1 + 1 is 5.
Example C
a2bc4 The monomial is of degree 7.
The order of a polynomial is the highest degree within the polynomial. The examples
show us how to determine the order of a polynomial.
Example A 2x2 – y3 – 1
The polynomial is of order 3 since the 2nd term is of degree
3 and that is the highest in the polynomial.
Example B 5x4y + x3y3z The polynomial is of order 7 since the sum of the exponents
in the 2nd term is 7 and this is the highest degree term.
Whenever possible polynomials should be written in standard form. The general rule
for putting a polynomial into standard form is that the highest degree terms come
first followed by the lower degree terms in descending order. For instance:
9x + 4x4 – 5x2 + 7
4x4 - 5x2 + 9x + 7
Should be written as:
VIDEO LINKS:
Khan Academy Adding and Subtracting Polynomials 1
Khan Academy Adding and Subtracting Polynomials 2
Khan Academy Terms, Coefficients, and Exponents in a Polynomial
Exercises Unit 6 Section 1
Define the following
1. Term
2. Exponent
3. Coefficient
4. Constant
5. Base
6. Monomial
Given the expression
2v + w4 + 6x3 + yz + 9 answer the following questions
7. How many terms are in the expression?
8. What is the coefficient of the third term?
9. What is the coefficient of w?
10. What is the constant?
11. What is the base of the second term?
12. What is the exponent of the third term?
13. What is the exponent of the first term?
Complete the following
14. Two terms are "like" when they have _____________________________________
15. If we want to add monomials then we should ______________________________
___________________________________________________________________
Simplify add or subtract as indicated and write your answers in standard form.
16.
4z + 3z - z
18. -7x + 7 + x + 5 + 11x
2
2
2
20. 8wx + 3x - 2wx + 6x - 10wx + 4x
2
2
2
17.
-5a + 9a - a
19.
3x + 8 - 16x + 9 - x
3
3
21. -6xy + 5a - 6 + 7a + 6 - 3xy + 9xy
22.
15x + 8a + 3a - 15x
24. -7x + 7 + x - 7 + 14x
2
2
2
23.
7z - 5z - 2z
25.
5x + 8x2 - (-4x) + 9 - x2
3
3
26. 3x + 4z - 2x + 6z - x + (-10z )
27. 6y + 4a - (-4) + 9a + 6 + -3y - 10
28. x + x + x + x + a + a - x + a - x + a
29. 2(x + 3y) + 5x + 6y
30. 3(a2 - 5) + a2 – 4
31. -2(3x + 4y - 6z) + x - 2y + z
32. 4(-2x + 4z) + 5(6 - x)
33. 3x + 8y + 7z + 11
34. 1.5x + 7.2x - 3.4x
35. 2a - 4a2 + a + 6a
36.
1
3
a  2a  a
2
2
List the degree of each monomial.
37. 4x
38. -2ab
39. x2y3
40. 11ab4c2
41.
42. 9xayb
5
List the order of each polynomial.
43. 3x + x2 – 6 + 8x3
44. x2yz4 + x5y - 10xyz2
Unit 6 Section 2
Objective

The student will multiply monomials.
Multiplying monomials requires that we understand how to use exponents in products.
We will review how exponents and factors can be used before we look at the rule for
multiplying monomials.
Definition
Factors are values or variables that we multiply together.
The monomial 2abc when broken down into its factors can be written as (2)(a)(b)(c) or
2 . a . b . c. When factors are repeated we can use exponents. A repeated factor is called
a base.
Definition
An exponent is a raised value or variable to the right of the base that tells
us how many times to use a factor.
Below are some examples of products written as factors and in exponential form.
x3 = x . x . x
w5 = w . w . w . w . w
a2b4 = a . a . b . b . b . b
xy2z = x . y . y . z
The understood exponent is 1 (one). The expression xy2z could have been written
as x1y2 z1 because there is only one ‘x’ in the product, additionally there is only
one ‘z’ in the product so both variables can have the understood exponent of 1.
The examples below show us how multiplying monomials can affect exponents.
Example A
x3 . x4 = (x . x . x) . (x . x . x . x) = x . x . x . x . x . x . x = x7
In this example we can use the Associative Property of Multiplication to remove
the parentheses and then count the x’s to find the resulting exponent.
Example B
a2 . a6 = (a . a) . (a . a . a . a . a . a) = a . a . a . a . a . a . a . a = a8
In this example we can use the Associative Property of Multiplication to remove
the parentheses and then count the ‘a’ s to find the resulting exponent.
Example C
a2b . a4b3 = (a . a . b) . (a . a . a . a) . (b . b . b) = a . a . b . a . a . a . a . b . b . b =
a6b4
We could use this technique to do all of our multiplying; however, we can find a better
way. If we summarize the examples we may see the better method.
Example A
Example B
Example C
x 3 . x 4 = x7
a2 . a6 = a8
a2b . a4b3 = a6b4
The questions we need to ask ourselves are:
How
How
How
How
can
can
can
can
we
we
we
we
use
use
use
use
the
the
the
the
values
values
values
values
3
2
2
1
and 4
and 6
and 4
and 3
to get
to get
to get
to get
7?
8?
6?
4?
The arithmetic we need to do is addition! This should make sense because
x3 . x4 has a total of 7 x’s multiplied together so the new exponent should be 7.
Our rule then for multiplying monomials is:
When we multiply monomials we should multiply coefficients and add exponents.
Below are some examples of applying this rule.
Example A
Example B
3ab . 5a2b
(-2xy3z)(4y2z5)
15a3b2
-8xy5z6
In this example we
could have written
the problem as:
3a1b1 . 5a2b1
After putting in the
understood exponent,
we add the exponents.
VIDEO LINK:
In this example we
have a variable ‘x’ that
is only in one of the
monomials. Even
though we don’t add
any exponents for ‘x’, it
must still be in our
answer.
Khan Academy Multiplying Monomials
Example C
(-wx)(3w2)(-7wx6)(2wxy)
42w5x8y
In example C we have
four monomials. We
must still follow our
rule and multiply the
coefficients and add
exponents. The first
monomial could have a
coefficient of -1.
Exercises Unit 6 Section 2
Fill in the blanks
1. An expression with one term is called a _________________________________
2. Part of an expression separated by plus or minus is called a ___________________
3. The number to the right of a variable/value that tells us how often to use a factor
is called the _____________________
4. A plain number is called a _________________________
5. Two terms are "like" when they have _____________________________________
6. If we want to add monomials then we should ______________________________
Simplify and add or subtract as indicated.
7.
3z + 7z - 8z
8.
9. -8x + 9 + x - 5 + 12x
2
10.
2
2
2
2
2
5a - 9a - 2a
13x + 8 - 16x + (-8) + 3x
3
3
11. 5wz + 3z - 8wz - (-5x ) - wz + z
12. -a + 5a - 2 + a + 8 + -3a + 9
13.
5x - 8a + 3a - 15x - 5a
14.
15.
5x + 8x2 - (-4x) + 9 - x2
16. x + 4z - 5x + 11z - (-2z) + z
3
7a - 5b - 2c
2
2
3
17. -6y + 4a - (-4) + 11a + z + (-3y) - 10 + 9y
18. x + x + x + a + a - x + a - x + a + a + a
19. 5(x + 3y) + 3x - 6y
20. -4(a2 - 5) + a2 - 1
21. 7(3x - 2y - 5z) + x - 2y + 2z
22. -(-3x + 4z) + 5(-2 + x)
1
3
24. - a  4a  a
2
2
23. 3.5x + 4.2x - 6.2x
Complete the sentence
25. If we want to multiply monomials then we should ________________________________
________________________________________________________________________
Multiply
26. x
29.
5 .
4
x
2 .
a
3
27. x
7.
.
.
5
b b b a
4
31. (-2a b2c)(-5a bd)
5 .
2.
x
3
.
x
30. (-3x)(-2x)(-2x)(x)(x)
4
3
6.
28. a a
3 2
3 2
32. (-x )(-5y z)(-x z ) (-x z )
a
33. y3
.
y7
34. 3x2
.
36. (1.5w3)(2.4w2)(w)
37.
1 2
z
2
39. (-.2a2b)(.4a)(-.5ab)
40.
3 3
a
2
-x
.
.
.
-4x5
1 5
z
2
1
- az
5
35. z . a5 . z
3
38. - z5
4
1
z
2
.
.
-
.
4 5
z
3
4 5 2 .
az z
3
Solve the following equations
41. 6x + 5 = 37
42.
43. 8 = -4(a - 5)
44.
45. 3(x - 7) = 3x - 21
46.
47. |x – 3| = 12
48.
3a + 7 = a - 25
2(a + 7) = 3(2a + 10) + 24
|x| = 7
2|x + 4| + 3 = 25
49. Which of the values below is the solution to the equation 6(x + 4) – 1 = 47
a) 4
b)
-4
c) 8.3
d) -8.3
50. Which of the values below is a solution to the equation
a) {4, -20}
b) {-4, 20}
-2|x + 8| = -24
c) {8, -12}
d) {12, -12}
Multiply and simplify the following.
51.
(xa)(xb)
54.
(xa)(x)
52. (xa)(xa)
55.
(x2)(xb)
53. (xa)(x-a)
56. (x2a+1)(xa-4)
Unit 6 Section 3
Objective

The student will divide monomials with natural number exponents.
We know how to add, subtract, and multiply monomials. In this section we will learn
how to divide monomials. To understand the dividing of monomials we need to review
how to factor and reduce fractions. Factoring to reduce a fractional expression is a
technique that will be very useful in the future. The method consists of factoring the
numerator and denominator, and dividing out the common factors. Some examples are
given below.
Example A
Example B
.
.
2 5
10
5
=
=
In this example we turned the 10 into 2 times 5
14
7
2 7
and the 14 into 2 times 7; we divided out the two’s and get 5 over 7.
One of the important features of this method is that the numbers
never get larger, only smaller.
. .
. .
1 2 2 3
 12
2
=
=
18
2 3 3
3
We factored both numbers to primes, and the negative one for the sign.
When we divided out the common factors we were left with -1, 2, and 3.
.
The same method for reducing fractions can be used to reduce monomial expressions
in fractions. The examples below show us how to divide monomials.
Example A
Example B
Example C
x5
x2
a4
a3
6a 3 b 2
2ab 2
. . . .
.
a a a a
a a a
x x x x x
x x
x3
. . .
. .
a1 or a
. . . . . .b
. . .
2 3 a a a b
2 a b b
3a 2
This method of factoring and dividing out the common factors is easy but the larger
the monomials, the more time it will take. We need to develop a rule that will make the
process quicker. To find our rule we need to see how we can use arithmetic to find
the new exponents. We have to combine the 5 and 2 to get 3 in example A. In
example B, we combine the 4 and 3 to get 1. And in the last example, the 3 and the
understood exponent of 1 gives us 2. In each case, we can subtract the exponent in
the denominator from the exponent in the numerator.
Our rule then for dividing monomials is:
When we divide monomials we should reduce coefficients and subtract exponents
in the denominator from the exponents in the numerator.
The examples below show us how to use the rule.
Example A
Example B
x7
x3
16a 5
2a 3
x4
8a 2
In this example
7–3=4
so our answer
has an exponent
of 4.
Example C
 10 x 5 y 6 z
 2y2z
5x 5 y 4
In this example
5–3=2
so our answer
has an exponent
of 2.
In this example
6–2=4
1 -1 = 0
so our answer
has an exponent
of 4 and no ‘z’.
Example C also illustrates a fact that will be very important in the future. The variable
divides out in example C. When we say “divides out”, this is not the same thing as
“cancels out”. When terms cancel out during addition they add up to zero and are
gone. When common factors “divide out”, they gives us a “1” and we do not have to
write the “1” because one times any number does not change the number.
The example below illustrates this.
x4
x4
=1
x4
= x0
x4
This must be true since any number divided by itself is 1.
This must be true if we use our rule and subtract exponents.
In conclusion then we know
x0 = 1. IN words this means:
Any number raised to the zero power is 1.
The examples below illustrate this rule.
ao = 1
50 = 1
-2.50 = 1
(2x)0 = 1
We should also be able to divide monomials when the problem is written horizontally
as well as vertically. The problems below are in horizontal form. To do these problems
the divisor’s (2nd monomial) exponents are subtracted from the dividend’s (1st monomial)
exponents.
Example A
Example B
Example C
x8 ÷ x6 = x2
a4b5c ÷ a3b2c = ab3
-15xy7z2 ÷ 5yz2 = -3xy6
VIDEO LINK: Youttube: Dividing Monomials
Exercises Unit 6 Section 3
Complete the following
1. To Divide monomials we should ___________________________________
Divide the following
2.
x7
x3
6.
ax 5
ax 2
7.
9.
12wx 7
3x 3
10.
 21a 2 x 4 y
 3ax 4
3.
a6
a
w8
w7
4.
a 4 x5 y
a4x
5.
8.
y3
y3
w3 x 4 z
 x4 z
11.
16w 2 x 3 z
 4w 2 x 3 z
12.
x7 ÷ x5
13.
a4 ÷ a4
14. x2y4 ÷ xy
15.
24x6 ÷ 3x2
16.
-5a3b ÷ 5a3b
17. -18wx3y9 ÷ -6x2y8
Complete the following
18. To multiply monomials we should ___________________________________
Multiply the following
19. x
3 .
4
x
.
20. x x
5.
x
21. x . x . x . x . x . x
22. x . x . y . x . y . x . x
23. (-5ab)(3b)(a)(-a)(-b3)
25. (.3w2)(.2w5)(w)
26.
1 3
a
3
.
2 5
a
3
.
1
a
2
24. (-4a2bc)(-9a4b3d)
3
27. - w
5
.
3 4
w
2
Complete the following
28. To add monomials then we should ______________________________
Simplify and add or subtract as indicated.
2z + 5z - 9z
30.
6w3 – 4w3 – 2w3
31. 5x + 3 + x - 8 - 4x
32.
14a + 2 – 17a + (-7) + 3x
29.
2
2
2
33. 6ab + 5a – 12ab - (-5a ) - ab + a
34. -11 + 14a4 - 2 + a4 + 4 + -7a2 + 9 + a2
Match each term with its definition.
35.
36.
37.
38.
_____
_____
_____
_____
relation
range
function
domain
A.
B.
C.
D.
the set of x-values we can use
the set of y-values that can be generated as answers
a set of ordered pairs or points
a relation in which any 'x' value is paired with at most
one 'y' value.
39. The important or special feature of a function is that
a)
b)
c)
d)
if we substitute one number for 'y' we get more than one number for 'x'.
if we substitute one number for 'y' we get one number for 'x'.
if we substitute one number for 'x' we get more than one number for 'y'.
if we substitute one number for 'x' we get one number for 'y'.
40. In the equation "g(x) = 2x + 4" the "g(x)" could be replaced by ________.
41. Decide which of the following rosters represent a function and which are nonfunctions.
( Write yes for a function and no for a non-function. )
a.
b.
c.
d.
{(3, 7), (7, 3), (-4, ½) (½, -4)}
{(1.5, 7), (-4,2), (1.5, -½) (3, 8)}
{(4, 2), (-1.5,-2), (-9, 7), (  , 5)}
{(-2, -1), (3,-1), (4.5, -1) (0, -1)}
Divide and simplify
42.
xa
xb
43.
xa
x
44.
xb
xb
45.
w 4a
wa
46.
w 2 a 3
w a 7
Unit 6 Section 4
Objective

The student will divide monomials with integer exponents.
In the previous section the numerator’s exponents were greater than or equal to the
denominator’s exponents. We need to be able to divide when the denominator has
the larger exponents. We will follow the same rule for division that we found in
Section 3.
When we divide monomials we should reduce coefficients and subtract exponents
in the denominator from the exponents in the numerator.
When we subtract a larger number from a smaller number we get a negative value. So
our answers for this type of problem will be negative. The examples below illustrate
this.
Example A
Example B
Example C
x2
x7
15a
3a 4
 14 x 2 yz 3
4y6 z
x-5
5a-3
-
We must subtract
2–7
So we get -5
We must subtract
1–4
So we get -3
7 2 -5 2
xy z
2
The coefficients were
reduced. The ‘x’ was
only in the numerator
and did not change.
For the ‘y’ variable the
exponent is 1 – 6 = -5.
For the ‘z’ variable the
exponent is 3 – 1 = 2.
We need to understand exactly what a negative exponent means. To investigate
this we will go back to our process of dividing out common factors. We will go back
to Example A.
When factors divide out we
x x
1
x2
Example A
=
= 5
are left with 1’s. So we have
7
x x x x x x x
x
x
a 1 in the numerator.
.
. . . . . .
When we use our rule for division we get
this means that
1
= x-5.
x5
x2
= x-5
7
x
This reworking of example A on the previous page tells us that a negative exponent
means the factor belongs in the denominator of a fraction. We should be able to
convert between negative exponents and fractional representations as illustrated
below.
A negative exponent means the factor should be in the denominator of a fraction.
x-n =
Example A
a-2 =
Example B
1
a2
x-4y-3 =
1
xn
Example C
1
4 3
x y
5b-5c =
5c
b5
Example D

 4y3
4 1 3 6
x y z =
3
3 xz 6
We should also be able to start with a fractional representation and convert it to
an expression with negative exponents.
Example A
3
w5
=
Example B
2a 4
= 2a4b-3c-1
b 3c
3w-5
Example C
3
 3xy3
=  w-1xy3z -2
2
5
5wz
Finally, we should be able to divide monomials and list our answer in either or both
the fractional form, or with negative exponents.
Example A
5y
 30 xy3
= 5x-3y = 3
4 2
x
 6x y
Example B
20a 2 bc 4
5a 2 c 3
5 2 -4 3 -1
=  ab cd =
3
 3b 4 d
 12b 5 cd
Example C
w2 y
w 2 yz
= z -1 =
1
z
VIDEO LINKS: Youtube: Simplifying to Negative Exponents
Youtube: Brightstorm: Zero and Negative Exponents
Exercises Unit 6 Section 4
Complete the Following
1. To divide monomials we should __________________________________.
Simplify the following expressions
2.
x5
3.
x3
6.
x6
4.
x
 20a 3x 5y
7.
 4ax 5
x8
5.
x7
6ax 4
2x 2
5w 2 x 3z
 x 3z
Simplify the following expressions and list your answer in fractional form ( do not use
negative exponents )
8.
x5
x7
9.
x
10.
x5
x7
x8
Simplify the following expressions and all answers should be listed with negative exponents
where appropriate.
11.
x5
x7
12.
x
x5
13.
x7
x8
14. A negative exponent tells us that the variable or factor should be placed in the
___________________ of a fraction. ( hint: compare problems 8-10 with 11-13 )
Write each fraction with negative exponents where appropriate.
15.
18.
21.
1
x7
1
a
4b
a2
16.
19.
22.
1
a5
1
a 4b 3
c
a 5b
24. Any base raised to a zero power is ___________
17.
20.
23.
2
z8
3
xy 5 z 8
 ax
yz 3
Simplify
25.
b0
26.
50
27.
x3
x3
Write each monomial as a fraction.
28. a-4
29. x-1
30. x-2y-5
31. a-4b
32. w6x-1y-3
33. 2x-2
34. -5a-4b3c-2
35. -x-2
36. 5-3
Divide and list your answer as both a fraction and with negative exponents.
24a 3 x 2 y
6ax 5 y
37.
 6ab 2 c 4
39.
10a 3 bc 4
4ab 2 c 4
20b 5 cd
38.
Perform each operation and list your answer with negative exponents where needed.
40. (x3)(x-6)
41. 7-3 . 72
43. (ax4)(ax-2)
44.
46.
x 2
x
47.
1
49. When we evaluate the expression
a. -27
b. 9
(a 3 )( a 2 )
a7
50. The expression
a. a 2
51. The expression
b.
1
a 2
(a 4 )( a 2 )
a 4
42. x-2 . x2
9 2  96
45.
92
x 5
x 3
48. (x-2)(x-8) __________
5
x
x4
(a )( a 5 )
with a = -3 the result is
a2
c. -243
d. 81
in its completely simplified for is
c.
1
a2
d.
1
a
in its completely simplified form is
52. Which of the following expression is equivalent to
a.
x2
x2
b.
2
x
c.
x5
3x 2
d.
53. Which of the following expression is equivalent to
a. x3
b. x-3
c. 5x
3
x 1 1 . 2
+
for x ≠ 0
2
x x
x
x2
x5
x6
(5 x 2 ) 0
x 3
for x ≠ 0
d. 0
54. Given the expression 3-2x4y-1z
a. Write the expression in a simplified form without negative exponents.
b. Explain how you found the coefficient of the simplified form.
Unit 6 Section 5
Objective

The student will add or subtract polynomials.
We have learned to perform operations with monomials in section 4. Now we must
move
on to operations with polynomials. The first two operations we will deal with are
addition
and subtraction of polynomials. The process of adding polynomials is simplified by the
Associative Property of Addition. The problem below demonstrates how we add
polynomials.
(6x3 + x2 + 11x – 12) + (-2x3 + 7x2 – 4x + 7)
The Associative Property of Addition means that we can switch or rearrange the
parentheses any way we need to. In this operation we can remove the parentheses
and combine like terms.
6x3 + x2 + 11x – 12 + -2x3 + 7x2 – 4x + 7
So our answer is:
4x3 + 8x2 + 7x – 5
To add polynomials, we remove the parentheses and combine like terms.
Some examples are given below.
Example A
(5a3 + a – 2) + (a2 – a + 11)
5a3 + a – 2 + a2 – a + 11
5a3 + a2 + 9
Example B
(-6x2 + 3xy – 9y2) + (-x2 - 2xy + 14y2)
-6x2 + 3xy – 9y2 + -x2 - 2xy + 14y2
-7x2 + xy + 5y2
Subtraction can be done with several approaches. One of the easiest methods is to
use the Distributive Property. The example below shows how this is done.
Subtraction is the same thing as
(10y3 + 4y2 + 3y – 11) − (-4y3 + 6y2 – 9y + 3) addition of the opposite. This
(10y3 + 4y2 + 3y – 11) + -(-4y3 + 6y2 – 9y + 3) means we can change the problem
to addition.
We can insert the understood
(10y3 + 4y2 + 3y – 11) + -1(-4y3 + 6y2 – 9y + 3) coefficient of 1 and the use the
10y3 + 4y2 + 3y – 11 + 4y3 – 6y2 + 9y – 3
Distributive Property. Then remove
the parentheses.
Our final answer is 14y3 – 2y2 + 12y – 14
We combined like terms.
The effect of multiplying by negative one is to change all the signs in the second
parentheses. Below are some examples of subtracting the polynomials.
Example A
(-3a3 + 9a2 + a – 13) − (4a3 + 7a2 – 8a + 4)
-3a3 + 9a2 + a – 13 + -4a3 – 7a2 + 8a – 4
-7a3 + 2a2 + 9a – 17
Example B
(11m2 – 5mn + n2) – (-5m2 – 3n2)
11m2 – 5mn + n2 + 5m2 + 3n2
16m2 – 5mn + 4n2
Example C
(-3y + 7x + 14) – (-3y + 7x – 14)
-3y + 7x + 14 + 3y – 7x + 14
28
VIDEO LINK: Youtube: Simplifying with Negative Exponents
Exercises Unit 6 Section 5
Simplify the following
1. (4x – 3) + (7x + 8)
2. (3b + 4) + (-2b – 6)
3. (5m + 8) + (4m + 3)
4. (-2t – 7) + (6t – 3)
5. (5k – 6n + 4) + (-5k + 8n + 2)
6. (6x2 - 2xy + 3y2) + (4x2 – xy – y2)
7. (4x – 4) – (3x + 8)
8. (5x – 2) – (4x + 1)
9. (2m2 - 3mn – 5n) – (-8m2 – n2)
10. (5a2 – 6ab) – (-2a2 + 9ab – b2)
11. (5x – 3y + z) – (2x + 5y – z)
12. (4x + 5y – z) – (4x – 3y – z)
13. Given the rectangle below find the expression for its perimeter.
3x - 5
2x + 1
14.
Given the diagram to the right which of the
expressions below represents the perimeter
of the triangle.
x2 + 13x
3x2 + 11
-4x - 4
4x2
a)
+ 17x + 15
2
b) 3x + 9x + 7
c) 4x2 - 17x + 7
d) 4x2 + 9x + 7
15. The building to the right has a parking garage
underneath it ( shown in black ). The distance
from the bottom of the parking garage to the
top of the building is 3x2 + 5x – 1. And the
distance from the parking garage floor to the
floor on the first level of the building is
x2 + x + 1. The expression for the
height of the building is
a)
b)
c)
d)
2x2 + 4x – 2
4x2 + 6x
2x2 + 4x
4x2 + 4x + 2
Write each fraction with negative exponents where appropriate.
16.
1
x6
17.
3
a4
18.
5
z
19.
5x
y3
20.
w
xy 7
21.
2ab 3
9 yz 5
Write each monomial as a fraction.
22. a-7
23. x-8
24. x-1
25. a-1b
26. -9wx-4y-2
27. 2-1x-2
Divide and list your answer as both a fraction and with negative exponents.
28.
28ax 3 y 4
4ax 5 y
29.
 2b 4 cd
 20ab 5 c
30.
16ab 2 c 4
10a 9 b 2 c
31. Given the triangle to the right with the sides as labeled
a. Find the expression for the perimeter of the
triangle and explain how you found it.
2x + 3
x – x2
x2 + x + 5
b. If the perimeter is 52 cm find the value of x
Show your work algebraically and explain how you found your answer.
c. Find the lengths of the three sides and explain how you found your answer.
32. Given the triangle to the right with the sides as labeled
a. Find the expression for the area and explain
how you found it.
b. If the area of the triangle is 144 in2 find the
value of x and explain how you found the answer.
c. Find the base and height of the triangle.
x
2x
Unit 6 Section 6
Objective
 The student will multiply a monomial times a polynomial.
Polynomials can have one or more terms. So this means we will multiply a monomial
times a binomial, a trinomial, or higher number of terms. The method for doing the
multiplication is to use the Distributive Property. The Distributive Property tells us we
can multiply through a parentheses term by term. The examples below show us
how to use the Distributive Property.
Example A
2x(x3 – 9x)
2x4 – 18x2
Example B
Example C
-3a2b(2a – 4b + 1)
-6a3b + 12a2b2 – 3a2b
-(7ab – 11b3)
-1(7ab – 11b3)
-7ab + 11ab3
In example C we have
insert the understood
exponent of 1 and
then multiply. This
has the effect of
changing all the signs.
VIDEO LINKS: Khan Academy Multiplying Monomials by Polynomials
Exercises Unit 6 Section 6
Use the Distributive Property to simplify the following:
1. x(2x - 4)
2. 5a(7a2 – 9a – 1)
3. -3y(5y - 11)
4. 2ab(a2 + 3ab - 4b2)
5. -6a2(4a3 + 8ax - 2x2)
6. –(-4x2 + 6x – 2)
7. 8a2x3(3a2x + ax - 3x4)
8. y3(x2y – xy + y2)
9. a(8a + 5ab)
10.
a-2(4a2 - 3a + 6a-1)
Simplify and add or subtract as indicated.
11.
2
2
2
5z + 11z - 12z
12.
-3a - 8a - 2a
13. -8x + 9 + x - 5 + 12x
14.
15. x(x + 3y) + 4x2 - 6xy
16. -4(a2 - 5) + 4a2 - 20
2x - 5x2 - (-7x) + 3 - x2
Complete the following
17. If we want to multiply monomials then we should ____________________
Multiply
18. x
3 .
4
x
3
19. x
7
3 .
5.
x
.
x
3
21. (-5a b2c)(-3a bd)
.
20. a5 a3 a
4
3 2
2
22. (x )(-5y z)(-3x z ) (-xz )
Simplify the following expressions
23.
x7
x3
24.
x8
x
25.
16a 4 x 6 y
 4ax 5 y
Simplify the following expressions and list your answer in fractional form ( do not use
negative exponents )
26.
x5
x9
27.
x
x6
x5
x6
28.
Write each fraction with negative exponents where appropriate.
29.
1
x8
30.
2
3
x yz 4
31.
x-2y-1
34. 4a-5b3c-2d
5b 3
c4
Write each monomial as a fraction.
32. a-5
33.
Simplify
35. a0
36. -70
37.
a4
a4
Perform each operation and list your answer with negative exponents where needed.
38. (x-2)(x-6)
39. x-2 . x2
40.
42. Given the rectangle to the right with the
sides as labeled find the expression for the
area and explain how you found it.
x 4
x 1
41.
x 3
x 3
4x2 - 5x + 2
7x2
43. What is the product of 6x5 – 2x4 + x3 – 9x2 and -5x-2
a. -30x3 + 10x2 – 5x + 45
b. -30x3 + 10x2 – 5x
c. -30x7 + 10x6 – 5x5 + 45x4
d. 30x3 – 10x2 + 5x – 45
Unit 6 Section 7
Objective

The student will find the Greatest Common Factor of monomials.
To find the greatest common factor (GCF) of monomials we need to review the method
for finding the GCF of any two numeric values. While there are several ways to do this
the best method involves factoring the numbers to primes. A prime number is divisible
only by one and itself. The factoring is best shown by factor trees. Below are examples
of factor trees.
Example A
Example B
12
2
2
.
Example C
42
.
6
.
2
2
3
2
.
.
3
-45
21
.
7
-1 . 5
.
9
.
3
.
-1
5
.
3
We can use these factor trees to find the greatest common factor for sets of numbers.
Once we have the factor trees for our values then we look to use the pairs of numbers
that the two trees have in common and multiply these factors together. Below are
examples of finding the CGF for pairs of values.
Example A
12
Example B
18
-30
-20
4
.
2
.
2
.
3
2
.
3
.
3
-1
. 2. 3 .
5
2
.
2
.
3
2
.
3
.
3
-1
.
5
.
3 =6
3
2
GCF = 2
.
9
-1
-1 . 2
. 2 . 15
2 .3
.
-1 . 2
-1
.
.
2
.
10
2
.
.
2
5
.
5
GCF = -1 . 2 . 5 = -10
In these examples once we factored the numbers, we circle and joined the pairs of
common factors. We also listed them in the GCF and then we multiplied.
We must also find the greatest common factors of variable terms as well as numbers.
The process can be similar to working with plain numbers. We factor the terms
completely and then find what the two terms have in common. The examples below
show us how to find the GCF.
Example A
Example B
x5
x2
y3
y4
x ∙ x ∙ x ∙ x∙ x
x ∙ x
y ∙ y ∙ y
y ∙ y ∙ y ∙ y
x ∙ x ∙ x ∙ x∙ x
x ∙ x
y ∙ y ∙ y
y ∙ y ∙ y ∙ y
GCF = x ∙ x = x2
GCF = y ∙ y ∙ y = y3
The conclusion we can draw from these examples is that the greatest common factor
of variable terms can be found by using the lowest exponent from the variable.
To find the greatest common factor of variable terms, use the lowest exponent
for each variable the two terms have in common.
Below are some examples of applying this conclusion.
Example A
a7
a4
GCF = a4
Example B
b
Example C
b5
x 4 y2
GCF = b
Example D
xy5
a2bc6
GCF = xy2
bc3d
GCF = bc3
We must use the prime factorization for the coefficients, and then find the GCF for
the variables using the lowest exponents for each variable when we work with
monomials. The examples below show use how to use this process.
Example A
14ab3
21a2b3
GCF = 7ab3
VIDEO LINK:
Example B
-15x2y3
-12y2z5
GCF = -3y2
Example C
8ax3
12ax
-16a4x2
GCF = 2ax
Khan Academy Monomial Greatest Common Factor
Exercises Unit 6 Section 7
1. What does the abbreviation GCF stand for?
Use factor trees to find the GCF for each set of numbers. ( Show the work )
2.
20
12
3.
28
42
4. 45
30
5.
-24
-18
6. 98
-42
7.
55
44
7
9.
11
8.
21
13
Find the GCF for each set of variable terms
10.
abc
bcd
11.
a2b4c
12.
x5y
x5y3
13.
ab
a3bc5
xyz2
Find the GCF for each set of monomials
14.
16.
12a2
-6x3y4z5
18. 100mn4
20.
32ab
14a2
15.
18ab3
-15x2y5z
17.
8ab
16a5b
30n5
19.
-9x3y6
-21x2yz
21.
5xy
8xy
12a3b5
7wx
Put the missing factor in the blank.
22. 4
25.
.
____ = 12
23.
3
.
____ = 18
24. 5
.
____ = 25
.
____ = -40
2
.
____ = 24
26. -2
.
____ = -16
27. 4
28. -7
.
____ = 28
29.
9
.
____ = -9
30. 5a
33. x
31.
4a2
.
____ = 24a5
32. ax
.
____ = a5x3
34.
3x2
.
____ = 18x4
35. -2b3
.
____ = 2b6c
.
.
____ = 15a3
____ = x2y
Unit 6 Section 8
Objective

The student will perform common factoring on polynomials.
Common factoring is a way to reverse the Distributive Property. To reverse the
Distributive Property, we first have to find the greatest common factor between the
terms of the polynomial. This GCF will be placed on the outside of a parenthetical
expression. We then need to figure out the terms that belong in the parentheses.
This process is illustrated below.
Example A
4x2 + 3x
x
x(
)
x(4x
)
x(4x + 3)
Example B
The given binomial
This is the GCF.
Which we use as the factor outside the parentheses.
Since x times 4x is 4x2 the 1st term will be 4x.
Because x times 3 is 3x the 2nd term will be 3.
10ax2 + 14ax The given binomial
2ax
This is the GCF.
2ax(
)
Which we use as the factor outside the parentheses.
2ax(5x
)
Since 2ax times 5x is 5ax2 the 1st term will be 5x.
2ax(5x + 7)
Because 2ax times 7 is14ax the 2nd term will be 7.
Example C
-9x3 – 6xy
-3x
-3x(
)
2
-3x(3x
)
2
-3x(3x + 2y)
The given binomial
This is the GCF.
Which we use as the factor outside the parentheses.
Since -3x times 3x2 is -9x3 the 1st term will be 3x2.
Because -3x times 2y is -6xy the 2nd term will be 2y.
Example D
7a3 – 14a2b – 21ab
7a
7a(
)
2
7a(a
)
7a(a2 – 2ab
)
2
7a(a – 2ab – 3b)
The given trinomial
This is the GCF.
Which we use as the factor outside the parentheses.
Since 7a times a2 is 7a3 the 1st term is a2.
Because 7a times -2ab is -14a2b the 2nd term is -2ab.
Finally 7a times -3b is -21ab the 3rd term is -3b.
We can check to see if our answer is correct by using the Distributive Property and
multiplying through the parentheses. We will check Example B below.
2ax(5x + 7)
10ax2 + 14ax
The binomial is the same as what we started with so the factoring is correct.
VIDEO LINK: Kahn Academy: Factoring and the Distributive Property 2
Exercises Unit 6 Section 8
Perform common factoring on the following polynomials (reverse the Distributive
Property). If the polynomial cannot be factored write “prime”.
2
1. 12a - 15b
2. -4x – 10
3. x - x z
4. 28ab + 7a
5. 4aby - 12by2
6. 8y2z + 2yz
8. 2a2c4 – 10a3c3
9. 2x2 - 9xyz
11. 3ax2 – 5ax + 2a
12. -40a c - 10ac4
14. 5cd -10c2d + 15cd2
15. 11x + 13y
3
7. 18x + 12x2
2
10. 4x2 + 20x z
2
13. 6x - 10x - 18
3
2
2
3
3 2
16. -12x - 2x – 4x
17. 6x y - 22x y + 2xy3
18. x2y + x3y2 – x4y
19. 16v2 – 24vw
20. 24x4 – 18x3 + 36x2 + 12x
Write each fraction with negative exponents where appropriate.
21.
1
z4
22.
4
y8
23.
1
a
24.
 3x 5
y2
25.
a
b 3c
26.
 7 ax 2
3 yz
Write each monomial as a fraction.
27. v-9
28. bx-8
29. u-1
30. a-1b-2
31. -4wx-3y
32. (5x)-1
Divide and list your answer as both a fraction and with negative exponents.
33.
27 axy
 9a 2 x 5 y
34.
5bc 5 d
20b 5 c
36. The best way to factor 18x6y2 – 12x3y8
a. 6x3y2(3x2y – 4xy4)
b. 3(6x6y2 – 4x3y8)
c. 6x3(3x3y2 – 2y8)
d. 6x3y2(3x3 – 4y6)
35.
16a 3 b 7 c 4
14a 3 b 2 c
Unit 6 Section 9
Objective

The student will multiply binomials of the form (x + a)(x + b)
where a and b are constants.
Multiplying binomials is really applying the Distributive Property twice. To see how this
works we will use constants and try the multiplying.
Example A
43
x 25
215
86
1075
In example A we use
the traditional algorithm
for multiplying two digit
numbers.
Example B
40 + 3
20 + 5
200 + 15
800 + 60
1000 + 75
1075
In example B we use a
place value representation
for multiplying. We do
5 x 3 to get 15.
5 x 40 to get 200
20 x 3 to get 60
20 x 400 to get 800
When we add them up we
get 1075.
Example C
(40 + 3)(20 + 5)
800 + 200 + 60 + 15
1075
In example C we still use
place value representation
for multiplying. We are
doing the multiplication
with the Distributive
Property. The 40 must be
multiplied times both the 20
and 5. The 3 must be
multiplied times the 20 and 5
as well.
We get the same answer from all three representations. The difference is that if we
use the method in Example C we can do the multiplication with variables as well as
constants. Example C has arrows that show us how to use the Distributive Property.
The first product was from (40)(20), the second was (40)(5). In this way we
multiplied the first term times both terms in the second binomial. Then we used
the 3 and multiplied it by the 20, and then the 3 times 5. Now that we have used the
Distributive Property with both the terms in the first binomial we can add to find our
answer. The order in which we do the multiplying is critical to much of what we will
do in the future. Below is an illustration of this order.
Given the binomials (x + a)(x + b)
We start with
(x + a)(x + b)
We multiply the
Then we multiply
(x + a)(x + b)
We multiply the OUTER terms in each binomial.
Next comes
(x + a)(x + b)
We multiply the INNER terms in each binomial.
Last we multiply
(x + a)(x + b)
We multiply the LAST terms in each binomial.
This gives us x2 + bx + ax + ab as an answer.
FIRST terms in each binomial.
This way of multiplying binomials is actually the exact same method we saw in
example C above. When we have variables, we can’t add all the terms up. Instead
we will get a polynomial as an answer. There is an abbreviation we can use to help
us remember how to multiply the binomials.
This order for multiplying binomials is F.O.I.L. for First, Outer, Inner, and Last.
Below are some examples of using the FOIL method to multiply binomials.
Example A
Given (x + 5)(x + 7) find the product.
(x + 5)(x + 7)
First terms give us
x2
Outer terms give us 7x
Inner terms give us 5x
Last terms give us 35
The answers written horizontally are x2 + 7x + 5x + 35
We can combine like terms and get our final answer as x2 + 12x + 35
Example B
Given (x – 4)(x – 3) find the product.
(x – 4)(x – 3)
The answer from FOIL written horizontally is x2 – 3x – 4x + 12
We can combine like terms and get our final answer as x2 – 7x + 12
Example C
Given (a – 7)(a + 2) find the product.
(a – 7)(a + 2)
The answer from FOIL written horizontally is a2 + 2a – 7a – 14
We can combine like terms and get our final answer as a2 – 5a – 14
Example D
Given (a – 6b)(a + 9b) find the product.
(a – 6b)(a + 9b)
The answer from FOIL written horizontally is a2 + 9ab – 6ab – 54b2
We can combine like terms and get our final answer as a2 + 3ab – 54b2
VIDEO LINK: Youtube: Using FOIL
Exercises Unit 6 Section 9
Complete the following.
1. If we are multiplying binomials then we must apply the ____________________
Property twice.
2. The abbreviation we can use to tell us how to multiply binomials is _______________
3. The abbreviation FOIL stands for _________________________________________
Multiply the following binomials:
4. (x + 7)(x + 3)
5. (x + 10)(x - 6)
6. (x + 9)(x + 2)
7. (a - 8)(a - 5)
8. (y + 7)(y + 4)
9. (a + 10)(a - 3)
10. (x - 5z)(x - 5z)
11. (x - 6)(x + 6)
12. (a + 1)(a - 9)
13. (a + 8b)(a - 8b)
14. (y + 7)(y + 7)
15. (x - 4y)(x – 4y)
16. (x – 11)(x + 4)
17. (z – y)(z – 6y)
18. (b + 12)(b + 1)
19. (w – 2z)(w + 9z)
20. (a + 1)(a + 2)
21. (y – 1)(y – 1)
22. (t + 4s)(t + 5s)
23. (y + 10)(y – 12)
24. (v – 3w)(v – 3w)
25. (a – 8)(a + 7)
26. Given the rectangle to the right as labeled
Find the expression for the area and
explain how you found it.
x + 15
x-2
Unit 6 Section 10
Objective

The student will multiply binomials of the form (ax + b)(cx + d)
where a,b,c and d are constants.
This section still involves binomial multiplication. The first terms in each binomial
variable can have coefficients, which does not change the process, although it will
change our product. Below are examples of binomial multiplication with these
coefficients.
Example A
Example B
(2x + 5y)(3x – y)
(4x2 – 5z)(-7x2 + 3z)
6x2 – 2xy + 15xy – 5y2
-28x4 + 12x2z + 35x2z – 15z2
6x2 +13xy – 5y2
-28x4 + 47x2z – 15z2
Example C
Example D
(2a3 – 9)(5a3 – 4)
(3y2 + 2z3)(4y2 + z3)
10a6 – 8a3 – 45a3 + 36
12y4 + 3y2z3 + 8y2z3 + 2z6
10a6 – 53a3 + 36
12y4 + 11y2z3 + 2z6
In these examples we see that the resulting trinomial has a leading coefficient. We
can also see that the variables in the binomials have exponents other than the
understood one. We must be careful how we apply the rule for multiplication and
add the exponents when appropriate.
VIDEO LINK: Youtube: Using FOIL
Exercises Unit 6 Section 10
Multiply the following binomials:
1. (2x + 5)(x + 5)
2. (6z – 1)(7z – 2)
3. (x + 7)(4x – 3)
4. (-2x + 1)(3x – 8)
5. (3x + 2y)(5x – 2y)
6. (a – 8b)(9a - b)
7. (5y2 + 7)(y2 + 4)
8. (9a2 + b2)(3a2 – 4b2)
9. (x - 2)(x - 11)
10. (x - 8)(x + 2)
11. (a + 4)(a - 5)
12. (a + 12)(a - 12)
13. (y + z)(y + z)
14. (x - 6y)(x – 10y)
15. (x – 9)(x + 5)
16. (z + 15y)(z – y)
17. (b + 7)(b + 4)
18. (ab – 3z)(ab + z)
19. (a – 1.5)(a + 1.5)
20. (a –
1
2
)(a + )
3
3
Multiply
21. x
24.
5 .
3 .
a
6
x
22. x
2.
.
.
4
b b b a
4 .
2.
x
.
.
23. a2 a2 a2
x
4
25. (-2x5y2z)(7w xy)
26. (-3x)(-4x)(2x)(x)(x)(x)
Use the Distributive Property to simplify the following:
27. 2(3a - 11)
28. 4z(3z2 – z – 1)
29. -8y(2y2 – 10y)
30. 2ab(a2b+ 5ab2 - ab)
31. a2(a3 + 8a2 – 2a)
32. –(x2 – 12x + 5)
Perform common factoring on the following polynomials (reverse the Distributive Property). If
the polynomial cannot be factored write “prime”.
33. 14z - 16
2
36. 5x2 + 25x y
34. -6x2 – 18x
35. x3y – x2y2 + xy3
37. 3ax2 – 5ax3 + 2ax4
38. -4a3c - 18a
39. Given the rectangle to the right as labeled
a. Find the expression for the area and
explain how you found it.
5x - 2
3x - 4
b. If x = 5 cm use the expression you found in part ‘a’ to find the area.
Show your work algebraically.
c. Calculate the lengths of the sides in the rectangle with x = 5 cm and then
find the area.
d. Compare your two answers and explain the result.
Unit 6 Section 11
Objective

The student will multiply special binomial products.
There are two special binomial products. The first one we will study is a binomial
square. To understand how we should multiply out a binomial square we need to
review how exponents work. An exponent of two, means to write down the factor
twice and multiply as shown below.
x2 = x ∙ x
or
(3x)2 = 3x ∙ 3x
When we have a parentheses raised to a second power we write down the expression
in the parentheses twice and multiply. The binomial will work the same way.
(x + 4)2 = (x + 4)(x + 4)
One way to multiply out the binomial square then is to write the binomial twice and use
the FOIL process. Examples of this process are given below.
Example A
Example B
(a – 5)2
(a – 5)(a – 5)
2
a – 5a – 5a + 25
a2 – 10a + 25
(3x + 7y)2
(3x + 7y)(3x + 7y)
9x2 + 21xy + 21xy + 49y2
9x2 + 42xy + 49y2
To multiply binomial squares we write the binomials twice and use the FOIL process.
The second special binomial product is called “sum and difference binomials. The
corresponding terms in the binomials are the same except that one of the binomials is a
sum and the other binomial is a difference. The examples below are sum and difference
binomials.
(x + 7)(x – 7)
(2w – 11y)(2w + 11y)
(5z2 + 3)( 5z2 – 3)
In each example above the binomials are the same except for the operation, addition
versus subtraction.
We can use the FOIL process to multiply out the sum and difference binomials. The
examples below show us how to do the multiplication.
Example A
Example B
(x + 9)(x – 9)
(2w – 5)(2w + 5)
x2 – 9x + 9x - 81
4w2 + 10w – 10w – 25
x2 – 81
4w2 – 25
In example A we can
see that the middle
terms in the product
were opposite in sign
and added up to zero.
We don’t have to
write the 0 so we get
a binomial answer.
In example B we can
see that the middle
terms in the product
were opposite in sign
and added up to zero.
We don’t have to
write the 0 so we get a
binomial answer.
Example C
(3a2 + 4b)(3a2 – 4b)
9a4 – 12a2b+ 12a2b – 16b2
9a4 – 16b2
In example C we can
see that the middle
terms in the product
were opposite in sign
and added up to zero.
We don’t have to
write the 0 so we get a
binomial answer.
In each of these examples the middle terms added up to zero and we did not need
to write anything because they cancelled out. This means that there is a short cut
for finding the product of sum and difference binomials. Because the middle terms
will always cancel out, the FOIL process can become FL or just First and Last. The
terms from multiplying the Outer and Inner parts of the binomials will always add
up to zero. Some examples of sum and difference binomial multiplication are given
below.
Example A
Example B
Example C
(a – 10)(a + 10)
(2x + 11)(2x – 11)
(3y2 +5z)(3y2 – 5z)
a2 – 100
4x2 – 121
9y4 – 25z2
VIDEO LINK: Khan Academy Special Products of Binomials
Exercises Unit 6 Section 11
Examine each problem below and decide if it is a “sum and difference” problem,
a “binomial square” or a problem which can be worked just with “FOIL”. Do
not work the problem, just list the problem type. ( Use the quoted words in
the instructions to identify the problems. )
1. (x + 4)(x – 4)
2. (y + 5)(y – 7)
3. (3x – y)(2x + y)
4. (z + 8)2
5. (9a – 2b)(9a + 2b)
6. (4w2 – 3v)2
7. (q + u)(q – u)
8. (a – 10)(a – 9)
9. (10 – 6z)2
Multiply the following sum and difference binomials.
10. (x + 3)(x – 3)
11. (2w – 5z)( 2w + 5z)
12. (7a + 9)(7a – 9)
13. (a + b)(a – b)
14. (x3 – 4)(x3 + 4)
15. (8 – a)(8 + a)
16. (11m2 + 3n2)(11m2 – 3n2)
18. (w + 12)(w – 12)
17. (6ab – 10c)(6ab + 10c)
19. (1.5x + .7)(1.5x – .7)
20. (
1
1
x + 2)( x – 2)
2
2
Multiply out the following binomial squares.
21. (x + 11)2
22. (a – 6)2
23. (a + 3b)2
24. (2b + 9)2
25. (-3y + 5)2
26. (a + b)2
27. (x + 6)(x – 11)
28. (a + 2)(a + 15)
29. (b – 7)(b – 8)
30. (y – 9z)(2y – 3)
31. (5n + 8)(n + 1)
32. (z + 2)(7z – 4)
33. (2w + 8)(3w – 1)
34. (6x + 1)(3x – 1)
35. (d + 1)(d – 1)
36. (w + 8y)2
37. (3y – 10)2
38. (2a – .5)(2a + .5)
Multiply the following
39. Given the rectangle to the right with the dimensions
as labeled. Find the area of the rectangle using the
dimensions.
a.
b.
c.
d.
3x2 – x – 4
3x2 + x + 4
3x2 + x – 4
4x – 3
3x - 4
x+1
40. Given the triangle to the right with the dimensions
as labeled. Find the area of the triangle using the
dimensions.
a.
b.
c.
d.
4x2 – 11x – 3
4x2 + 11x + 3
4x2 + 11x – 3
8x2 – 22x - 6
2x + 6
4x - 1
41. The expression for the side of a square is given as 2x + 7. Find the expression
for the area of the square and explain how you found the answer.
Unit 6 Section 12
Objective

The student will multiply polynomials.
We will at times need to multiply a binomial times a trinomial or perhaps two trinomials.
We should be able to multiply any polynomial times any polynomial. To multiply any
polynomials we simply need to use the Distributive Property. Each term from the first
polynomial must be multiplied times all the terms in the second polynomial. The example
below illustrates this process.
Example A
Multiply (x + 4)(x2 – 5x + 7)
(x + 4)(x2 – 5x + 7)
We begin by multiplying the ‘x’ times the second polynomial
we get x3 – 5x2 + 7x.
(x + 4)(x2 – 5x + 7)
We continue by multiplying ‘4’ times the second polynomial
we get 4x2 – 20x + 28
When we combine all the products we get x3 – 5x2 + 7x + 4x2 – 20x + 28
Now we can combine the like terms. The final answer is
x3 – x2 – 13x + 28
Example B
Expand the following (x – 5)3 ( expand means to multiply out )
This means (x – 5)(x – 5)(x – 5)
So we multiply one set of binomials and get x2 – 10x + 25
The problem then becomes (x – 5)(x2 – 10x + 25)
(x – 5)(x2 – 10x + 25) We begin by multiplying the ‘x’ times the second polynomial
we get x3 – 10x2 + 25x.
(x – 5)(x2 – 10x + 25) We continue by multiplying ‘-5’ times the second polynomial
we get -5x2 + 50x – 125
When we combine all the products we get x3 – 10x2 + 25x – 5x2 + 50x – 125
Now we can combine the like terms. The final answer is
x3 – 15x2 + 75x – 125
VIDEO LINK: Khan Academy Multiplying Polynomials Example
Exercises Unit 6 Section 12
Multiply the following polynomials
1. (x+ 2)(x2 – 7x + 4)
2. (2x + 1)(x2 + 6x – 3)
3. (a – 4)(a2 + 5a + 9)
4. (a + b)(a2 + 8ab + 12b2)
5. (3y – 4)(y3 + 2y2 + y)
6. (a +2b + 1)(2a – b + 3)
7. (x + 3)3
8. (2a – 1)3
Multiply the following binomials
9. (x + 7)(x + 10)
10. (a – 2)(a + 9)
11. (z + 3)(z – 11)
12. (m – 6)(m – 8)
13. (2x + 7)(x + 12)
14. (a + 4b)(a – 4b)
15. (n + 6)2
16. (x + y)(3x -8y)
17. (2z – 5)(2z + 5)
18. (3n – 1)2
19. (a2 + 5)(a2 – 8)
20. (5 – 9a)(5 + 9a)
Perform common factoring in the problems below. If the polynomial cannot be
factored write “prime”.
21. 16x – 12y
22. -9x2 + 3x
23. x3y – x2z
24. 15a2c4 – 10a3c3
25. 2w2 - 9xyz
26.
27. 14ax3 – 21ax2 + 28ax
28. 6x - 21x - 15
2
3x2 + 27x2z
3
Divide and list your answer as both a fraction and with negative exponents.
30.
32ax 5 y
4a 4 x 5 y
31.
bc 5
b 5 c 1
 18ab 2 c 4
32.
9ab 7 c
33. Any number or variable raised to the zero power is _______
x5
34. What does 5
x
simplify to?
2
29. -16x - 8x – 6x
35. The expression for the side of a cube is given as 3x – 1. Find the expression for
the volume of the cube and explain how you found it.
36. Given the box to the right as labeled
Find the expression for the volume of
the cube and explain how you found it.
3x + 5
2x + 1
2x - 1
37. A fish tank has a width of x inches, a length of 3x inches and a height
of x + 10 inches. Find the expression for the volume of the tank and
explain how you found it.