Chapter 1 — Expressions
Section 1.5
Simplifying Algebraic Expressions
TERMINOLOGY
1.5
Previously Used: This is a review of terms with which you should already be familiar.
Associative Properties
Distributive Property
Commutative Properties
Exponent
Prerequisite Terms:
This is a review of terms with which you should already be familiar. In order to be sure that you have an
understanding of their meanings before proceeding, provide a short definition of each term.
Term
Your definition
New Terms to Learn:
For each of the following terms, provide 1) a definition in your own words, 2) the formal definition (as
provided by your text or instructor), and 3) an example of the term using a drawing or problem. A sample
completed form is available in the Introduction.
Unary Operator
Your definition
Formal definition
Example
Binary Operator
Your definition
Formal definition
Example
42
Equivalent Expression
Your definition
Formal definition
Example
READING ASSIGNMENT
READING AND SELF-DISCOVERY QUESTIONS
1.5
Sections 1.1 and 1.8
1.5
1. What are like terms?
2. What is a fully simplified algebraic expression?
3. How do you combine like terms?
KEY CONCEPTS
1.5
A unary operator operates on one number. Taking the negative of a number (–) is a unary operator. Taking
the negative of 3 is –3.
A binary operator requires two numbers. Subtraction (–) is a binary operator. For example, 5 subtract 3
is (5 − 3).
Equivalent expressions exist if substituting the same domain value from any of the expressions into each
expression gives the same output value for all of the expressions.
43
Chapter 1 — Expressions
Convention
A Simplified Algebraic Expression: the sequence of terms begins with the highest power term on the left,
no parentheses, and no double negatives.
TECHNIQUE
REMOVING PARENTHESES
1.5
Removing the parentheses is a technique for rewriting an expression or equation.
Limitation: If there are more than two factors involved with parentheses, errors can occur.
Use the Distributive Property to substitute for parts of an expression that have parentheses and validate the
signs when removing the parentheses.
Example 1: Remove the parentheses:
3(a – 2b) + 4(2a – b) – 5
3a – 6b + 8a – 4b – 5
Here 3(a – 2b) is equivalent to 3a – 6b and 4(2a – b) is equivalent to 8a – 4b by the Distributive
Property. We replace 3(a – 2b) and 4(2a – b) with their equivalent expressions.
Example 2: Remove the parentheses
2 − 3 x(2 − 5 x − 4 y ) + 7
2 − (3 x(2 − 5 x − 4 y )) + 7
2 − (6 x −15 x 2 −12 xy ) + 7
2 − 6 x +15 x 2 +12 xy + 7
The expression 3x(2 – 5x – 4xy) is replaced by the equivalent expression (6x – 15x2 – 12xy) and
we validate the signs when removing the parentheses. In this process, –3x was distributed over
(2 – 5x – 4y). Note that this expression has not yet been fully simplified.
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METHODOLOGY
SIMPLIFYING ALGEBRAIC EXPRESSIONS
1.5
The methodology for solving the problem in Example 1 is shown in the following table. Students should
complete Example 2 in the same table, following each step in the methodology.
Note: Algebraic expressions come in many forms, some with a few terms, and some with many terms. In
working with algebraic terms, experts like to simplify the expression into its easiest interpretable
form. The specification for such forms is to have only one of each type of term.
Example 2
Simplify: 2( x − 4) − 3 x 2 + x( x − 4) − 5 + 3 x 2 x
Simplify: − 3 x + 2(5 − x) + 4 x 3 − x(5 + x) − 7 − x 2
1
Distribute through the
expression so as to remove
all parentheses in the
expression
2
3
Remove
parentheses
Check signs
Identify the
number of terms
for simplified
answer
Review all coefficients for
accuracy in signs, carefully
checking for negatives and
double negatives
Look through every term
to inventory how many like
terms you will have and
produce that number of
groupings
Example 1
Discussion
− 3 x + 2(5 − x) + 4 x3 − x(5 + x) − 7 − x 2
2 x −8 − 3x 2 + x 2 − 4 x − 5 + 3x 2 x
Checked.
2 x −8 − 3x 2 + x 2 − 4 x − 5 + 3x 2 x
2 x, − 4 x
− 3x 2 , x 2
3x 2 x
− 8, − 5
Example 2
Steps
Example 2 Example 1 Example 2 Example 1
Example 1
45
Discussion
4
Set up buckets
Set up structure for the
final simplified expression,
matching the identified like
terms, and starting with the
highest power
5
Rearrange the
expression
Rearrange the expression by
grouping coefficients within
the appropriate parentheses
6
Validate all terms
used
Make sure that all terms in
the previous expression
are accounted for by lightly
crossing out, term by term,
each term that has been
incorporated
Example 1
Simplify each
grouping
Convention says that
each variable has a single
coefficient and that the
computations needed to
remove parentheses should
be carried out
(
) x2 + (
)x+(
) x2x + ( )
constant
(–3x2 + x2) + (2x – 4x) + (3x2x) + (–8 – 5)
Example 2
Example 1
Example 2
Example 1
Steps
Example 1
Chapter 1 — Expressions
2 x −8 − 3x 2 + x 2 − 4 x − 5 + 3x 2 x
Example 2
(− 3 x 2 + x 2 ) + (2 x − 4 x) + (3 x 2 x ) + ( − 8 − 5)
Example 2
7
46
− 2 x 2 + (− 2 x) + 3 x 2 x + (− 13)
− 2 x 2 − 2 x + 3 x 2 x −13
Let x = 2.
2( x − 4) − 3 x 2 + x( x − 4) − 5 + 3 x 2 x
Steps
8
Validate by
substitution
2(2 − 4) − 3  22 + 2(2 − 4) − 5 + 3  22  2
Discussion
4
− 2)
− 3variable
− 2)
− 5 +the
3 original
2
 4 + 2(in
Pick a number other than 0,1, and -1 to substitute2(for
the
both
and simplified expressions. If both values are equivalent,
you
have
validated
your
− 4 −12 − 4 − 5 + 3  16
work.
Example 1
Let x = 2.
2
2( x − 4) − 3 x + x( x − 4) − 5 + 3 x
2x
2(2 − 4) − 3  22 + 2(2 − 4) − 5 + 3  22  2
2(− 2) − 3  4 + 2(− 2) − 5 + 3  24
− 4 −12 − 4 − 5 + 3  16
− 13 + 48
23
And
− 2 x − 2 x + 3 x −13
2
− 2(2 ) − 2  2 + 3  2
And
− 2 x 2 − 2 x + 3 x 2 x −13
− 2(22 ) − 2  2 + 3  22  2 −13
− 2  4 − 4 + 3  24 −13
− 8 − 4 + 3  16 −13
− 8 − 4 + 48 −1 3
− 25 + 48
23
2x
2
− 13 + 48
23
2  2
−13
Example 2
− 2  4 − 4 + 3  24 −13
− 8 − 4 + 3  16 −13
− 8 − 4 + 48 −1 3
− 25 + 48
23
CRITICAL THINKING QUESTIONS
1.5
1. What are equivalent expressions?
2. What changes can you make to an expression to yield an equivalent expression?
47
Chapter 1 — Expressions
3. Why do you remove parentheses?
4. How do you ensure that the signs are correct?
5. How do you ensure that all the terms are used?
6. Why shouldn’t you use zero in validating that terms are equivalent?
7. How might you use graphing with a calculator to validate an expression in one variable?
48
DEMONSTRATE YOUR UNDERSTANDING
1.5
Simplify each of the following expressions.
1. –3x(5 – x) + 3x – 3(x + 2)
2. 4x2 – 3x + 2x2a – 8 + 7x(x – 2) + 5
3. –(x – 2) + 3(2 – x) + 7x3 – x + 2x(x + 3)
IDENTIFY AND CORRECT THE ERRORS
1.5
In the second column, identify the error you find in each of the following worked solutions and describe
the error made. Solve the problem correctly in the third column.
Problem
Describe Error
Correct Process
1. Simplify:
− 3 x 2 − 2 x(3 − x) + 7
Worked Solution
(What is wrong here?)
− 3 x 2 − 2 x(3 − x) + 7
− 3x 2 − 6 x + 6 x 2 + 7
(− 3 x 2 + 6 x 2 ) + (− 6 x) + (7)
3x 2 − 6 x + 7
49
Chapter 1 — Expressions
Problem
2. Simplify and validate:
2 x − 3x + 5 x 2 + 4 x − 5 x + 7
Worked Solution
(What is wrong here?)
2 x − 3x + 5 x 2 + 4 x − 5 x + 7
(5 x 2 ) + (2 x − 3 x − 5 x) + (7)
5x2 − 6 x + 7
Validate :
Let x = 0.
2 x − 3x + 5 x 2 + 4 x − 5 x + 7
2  0 − 3  0 + 5  02 + 4  0 + 7
0−0+0+0+7
7
And
5x2 − 6 x + 7
5  02 − 6  0 + 7
0−0+7
7
50
Describe Error
Correct Process