Download M19500 Precalculus Chapter 0: Algebra preliminaries

Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Welcome to Math 19500 Video Lessons
Prof. Stanley Ocken
Department of Mathematics
The City College of New York
Fall 2013
An important feature of the following Beamer slide presentations is that you, the
reader, move step-by-step and at your own pace through these notes. To do so, use
the arrow keys or the mouse to move from slide to slide, forwards or backwards. Also
use the index dots at the top of this slide (or the index at the left, accessible from the
Adobe Acrobat Toolbar) to access the different sections of this document.
To prepare for the Chapter 0 Quiz (September 9th at the start of class), please
Read all the following material carefully, especially the included Examples.
Memorize and understand all included Definitions and Procedures.
Work out the Exercises section, which explains how to check your answers.
Do the Quiz Review and check your answers by referring back to the Examples.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
A basic real-life problem
The area of a square with side length s is s2 .
If we call the area A, we can write a formula for A in terms of s:
A = s2
This simple formula is described in many ways.
• A depends on s.
• s is the independent variable and A is the dependent variable.
• s is the input and s2 is the output.
• A is a function of s.
Obvious basic questions about squares include:
• If the side length of a square is 12, what is its area?
Solution: A = s2 . Since s = 12, then A = s2 = 122 = 144.
Answer: The area is 144.
• If the area of a square is 100, what is its side length?
Solution: A = s2 . Since A = 100, then 100 = s2 and so s = 10.
Answer: The side length is 10.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Quiz review
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Some things to be think about:
• If the question uses words, the answer must use those words.
• In science classes, you must specify units. In math class, we don’t use units
unless the problem includes them.
For example, we are willing to say that a square with side length 12 has area 122 = 144.
If this bothers you, insert the word “unit” and say:
A square with side length 12 units has area 144 square units.
Now let’s think about a square that is growing.
Suppose at a starting time t = 0 the side length of the square is 10 units.
Also suppose that the side length grows at a rate of 3 units per second.
Then the formula for the side is easy to figure out:
s = 10 + 3t
In this formula,
• s depends on t.
• t is the independent variable and s is the dependent variable.
• t is the input and 10 + 3t is the output.
• s is a function of t.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Let’s put everything together. Let t be time, s the side length of a square, and A the
square’s area. Then
A = s2 and s = 10 + 3t
Let’s focus on the algebra: forget the real-life problem. Now A depends on s, which in
turn depends on t. This double dependence generates new questions:
• Find a formula showing how A depends on t.
Solution: Substitute (10 + 3t) for s in A = s2 .
Answer: A = (10 + 3t)2 . Stop here. You weren’t asked to simplify.
• When t = 3, find A.
Solution: since t = 3 and A = (10 + 3t)2 , we get A = (10 + 3 · 3)2 = 192 = 361.
Answer: when t = 3, then A = 361.
• When A = 256, find t.
Solution:(10 √
+ 3t)2 = 256. Therefore
10 + 3t = ± 256 = ±16. This may come as a surprise.
If 10 + 3t = 16, then t = 2. If 10 + 3t = −16, then t = −26/3 .
Answer: t = 2 or t = −26/3. This may come as a bigger surprise.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
The difference between the last two examples:
• The first example asked for the side length of a square. Although s2 = 100 has
two solutions s = 10 and s = −10, we rejected the negative answer (without saying so
at the time) because the side length of a square is a positive number.
• The second example (check the wording of the question) talked about A, s, and
t, not about area, side, and time. Therefore the question must be answered using
algebra, without reference to the underlying physical problem.
Question: Why do these notes consistently use the phrase “side length?” What’s
wrong with just saying ”side?”
This preliminary section of online notes deals with the basics of algebra. An important
goal is to develop machinery for analyzing real-life problems including, but typically
more complicated than, the one discussed above. Please read carefully, since this
material is not always explained in high school classes or in textbooks.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
A possibly surprising ingredient of student success in mathematics
Success in mathematics requires the combined focus of intellect, logic, and memory.
Students in this course will be required not only to do algebra problems, but also to
reproduce on each examination a selection of definitions and procedures in these notes.
The understanding of mathematics requires memorization of definitions. The efficient
application of mathematics requires memorization of procedures.
In recent years, memorization has been denigrated and undervalued. Today’s students
and educators would do well to read Chapter 60 in historian Daniel Boorstin’s
remarkable cultural history entitled The Discoverers (Random House, 1983). Here is
the opening paragraph:
Before the printed book, Memory ruled daily life and the occult learning, and fully deserved the
name later applied to printing, the ”art preservative of all arts.” The Memory of individuals and
of communities carried knowledge through time and space. For millennia personal Memory
reigned over entertainment and information, over conduct of professions. By Memory and in
Memory the fruits of education were garnered, preserved, and stored. Memory was an awesome
faculty everyone had to cultivate, in ways and for reasons we have long since forgotten. In
these last five hundred years we see only pitiful relics of the empire and the power of Memory.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Introduction: Algebra Review
This introduction for Math 19500 contains important review material. The main goal
is to help you understand how and why we use algebra formulas.
Whole numbers have certain obvious properties such as 3 + 4 = 4 + 3 and 3 + 0 = 3.
These basic properties, shared by all numbers, are called the Laws of Algebra.
Here is a familiar example. You know that
3+4=4+3
7+9=9+7
12 + 333 = 333 + 12
To convey the idea that the order of adding numbers doesn’t matter, we use letters to
represent arbitrary numbers, and say that A + B = B + A for all real numbers A and
B. This is the commutative law for addition.
All of algebra is summarized in nine (just 9!) Algebra Laws. In the list below, A,B,
and C stand for any real numbers. We use the · symbol for multiplication.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Four Laws for Addition
A + B = B + A (Addition is commutative)
A + (B + C) = (A + B) + C(Addition is associative)
A + 0 = A = 0 + A (0 is the identity for Addition)
Every number A has a negative, written −A, satisfying A + −A = 0
Four Laws for Multiplication
A · B = B · A (Multiplication is commutative)
A · (B · C) = (A · B) · C(Multiplication is associative)
A · 1 = A = 1 · A (1 is the identity for multiplication)
Every nonzero number A has a reciprocal, written A−1 , satisfying AA−1 = 1
One Distributive Law
A · (B + C) = A · B + A · C
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Quiz review
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
We use some special notation and abbreviations, as follows:
1
is another way of writing A−1 .
The fraction
A
A
1
and A/B, read “A divided by B,” are abbreviations for the product A · .
B
B
We will not use the symbol ÷ for division.
A power such as A5 is just an abbreviation for a product, in this case
A5 = A · A · A · A · A .
1
A−n is an abbreviation for n .
A
√ n
√
n
A (the nth root of A ) satisfies n A = A.
√
When n = 2, we write just A , which we call the square root of A.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Quiz review
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Using substitution to rewrite expressions
The polynomial expression 100 − 16t2 arises in physics as follows. If you drop a rock
from a height of 100 feet, the height of the rock above the ground t seconds later is
100 − 16t2 , feet, provided 0 ≤ t ≤ 2.5.
Since t can vary within the stated range of values, it is common to refer to 100 − 16t2
as a polynomial in the variable t. However, we will often refer to t simply as a letter
(that’s what it is, after all).
Definition
An algebra expression contains letters and numbers, and combines them by using
parentheses and symbols for algebra operations.
Definition
A numerical expression contains numbers, and combines them by using parentheses
and symbols for algebra operations.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
We will use the word “expression” for “algebra expression.” Note that an expression
without any letters is a numerical expression.
Here is a precise statement of how to construct expressions:
Any letter is an expression.
Any whole number is an expression.
If E is an expression, then (E) is an expression.
If E and F are expressions, then so are
E + F, E − F, E · F, E/F, and EˆF .
We use a dot, not the calculator symbol *, for multiplication. In practice, most
multiply signs are omitted to save space. For example, the usual way of writing
3 · A + 2 · B · C · (C + 2) · (C + 3) is 3A + 2BC(C + 2)(C + 3).
Finally, we often write E/F as
E
F
and usually write E F for E ˆF .
Very often, we modify a starting expression by replacing one or more of its letters by
other expressions. This important process is called substitution.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Definition of substitution
To substitute an expression E for a letter in a formula, replace every occurence of
the letter in the formula by (E). The parentheses are crucial!
Example 1: In the expression A2 − B 2 substitute x + 3 for A and y − 2 for B.
Answer : (x + 3)2 − (y − 2)2
Please note the following: both statements are important.
The substituted expression must be enclosed in parentheses. Exception: if the
substituted expression is a letter, the parentheses are not needed.
(x + 3)2 − (y − 2)2 is the only correct answer. The instruction says just
“substitute.” Do not simplify or otherwise modify an expression unless you are
explicitly asked to do so.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Example 2: In the expression A2 − B 2 substitute A + 3 for A and B − 2 for B.
Answer: (A + 3)2 − (B − 2)2
In this example, the statement “Substitute A + 3 for A” cannot sensibly be rephrased
as “Let A = A + 3.” On the other hand, you may say:
In expression (A + 3)2 − (B − 2)2 , let A = 3 and B = x + 2.
But it is best to say
In expression (A + 3)2 − (B − 2)2 , substitute 3 for A and x + 2 for B.
Example 3: In the statement f (x) = 3 − 7x − 2x3 substitute x + h for x.
Answer: f ((x + h)) = 3 − 7(x + h) − 2(x + h)3 .
This looks a little funny. Double parentheses ((...)) should always be replaced by single
parentheses and so we can write this solution more neatly as
f (x + h) = 3 − 7(x + h) − 2(x + h)3 .
Again, don’t multiply out unless you are asked to do so.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Based on your previous math course, you might think that the symbol f (x) in Example
3 means a function of x. That doesn’t have to be the case. Unless you are informed
that a function is involved, f (x) might mean the product of letters f and x. But it
really doesn’t matter: the stated solution is correct in both cases, whether f (x)
indicates a function of x or just a product of variables.
Substitution is a purely mechanical operation that can be done by doing a “find and
replace” operation with a word processor: find every occurrence of “x” and replace it
by “(x + h)”. When you substitute an expression for a variable in a formula, you do
not need to, and perhaps you should not, think about what the algebra operations
mean. Rather, it would be wise to consider the 1911 advice of mathematician and
philosopher Alfred North Whitehead:
It is a profoundly erroneous truism, repeated by all copy-books and by eminent people when
they are making speeches, that we should cultivate the habit of thinking about what we are
doing. The precise opposite is the case. Civilization advances by extending the number of
important operations which we can perform without thinking about them. Operations of
thought are like cavalry charges in battle – they are strictly limited in number, they require
fresh horses, and must only be made at decisive moments.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Correct use of parentheses is crucial for student success
Procedure: three rules for parentheses
1. When you substitute an expression for a letter, replace the letter by the expression
enclosed in parentheses.
2. Insert parentheses whenever you expand an expression. Specifically, each time you
multiply or take a power of a parenthesized expression, put parentheses around the
result.
3. When you substitute an expression for a function value f (x), put parentheses
around the result.
Example 4:
Rule 1: Substitute x + h for x in 3 − x2 .
Answer: 3 − (x + h)2 , not 3 − x2 + h or 3 − x + h2 .
Rule 2: Simplify 3 − 3(x + h).
Answer: 3 − 3(x + h) = 3 − (3x + 3h) = 3 − 3x − 3h , not 3 − 3x + 3h.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Rule 3: ( to be discussed in Section 2.1)
If f (x) = x2 then f (x + h) = (x + h)2 , not x2 + h and not x + h2 .
Definition
An identity, also called an algebra law, is
one of the nine Laws of Algebra listed above , or
a new algebra law obtained by substituting expressions for the letters in a known
algebra law.
Each of the nine Laws of Algebra is a statement that two expressions are equal. New
identities are produced by substitution in previously known identities. Therefore every
identity is a statement that two expressions are equal.
Example 5:
Substituting x + y for A and z for B in (A + B)(A − B) = A2 − B 2
yields the new identity ((x + y) + z)((x + y) − z) = (x + y)2 − z 2
which can be rewritten
(x + y + z)(x + y − z) = (x + y)2 − z 2
In this example, omitting parentheses when you substitute is a mistake that (luckily)
happens not to affect the answer. But it is still a mistake.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Example 6.
Substituting z for A and x + y for B in
yields the new identity
which can be rewritten
Simplifying expressions
Order of operations
Exercises
Quiz review
(A + B)(A − B) = A2 − B 2
(z + (x + y))(z − (x + y)) = z 2 − (x + y)2 ,
(z + x + y)(z − x − y) = z 2 − (x + y)2 .
In this example, omitting parentheses when you substitute is a mistake that yields an
incorrect answer.
Every identity is written as a statement that two expressions are equal. Here is an
example. Assume you already know the FOIL identity
(A + B)(C + D) = AC + AD + BC + BD. Then the familiar identity
(A + B)(A + B) = A2 + 2AB + B 2 can be obtained by substituting A for C and B
for D in the FOIL identity, as follows:
Start with
Substitute A for C and B for D:
Use the commutative law AB = BA:
Rewrite using powers and collect terms:
Stanley Ocken
(A + B)(C + D) = AC + AD + BC + BD
(A + B)(A + B) = AA + AB + BA + BB
(A + B)(A + B) = AA + AB + AB + BB
(A + B)(A + B) = A2 + 2AB + B 2
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
If you think that the above discussion is too detailed, I agree completely.
The identity (A + B)(A + B) = A2 + 2AB + B 2 should be memorized. Similarly,
the identity (A + B)(A − B) = A2 − B 2 should be memorized.
Every identity is written using letters but it is equally true when any expressions are
substituted for those letters . In particular, identities produce true statements when
numbers are substituted for letters.
Example 7: Start with the Distributive Law A(B + C) = AB + AC.
Substitute x for A, x + 1 for B, and x + 1 for C to get
((x))((x + 2) + (x + 1)) = (x)(x + 1) + (x)(x + 2)
Removing doubled parentheses yields
(x)((x + 2) + (x + 1)) = (x)(x + 1) + (x)(x + 2) . This is the answer to the
question: you were not asked to simplify the result of substitution.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Quiz review
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Example 8: Start with the identity: (A + B)(A − B) = A2 − B 2
Substitute 3x for A and 5 for B: ((3x) + (5)) ((3x) − (5)) = (3x)2 − (5)2
Remove unnecessary parentheses: (3x + 5) (3x − 5) = (3x)2 − 52
We will discuss in detail in the next section why (3x) + 5 can be rewritten without
parentheses as 3x + 5.
In contrast, the parentheses in (3x)2 cannot be omitted, since (3x)2 = (3x)(3x) = 9x2
is not the same as 3x2 .
Example 9: Check that the new identities obtained in Examples 7 and 8 yield true
statements when you substitute
−3 for x.
7 for x
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
What does the word ’simplify’ mean?
The identity x2 + 3x + 2 = (x + 1)(x + 2) does not say whether one expression is
simpler than the other.
When you start with (x + 1)(x + 2) and then multiply out and collect terms to get
x2 + 3x + 2, you are expanding (multiplying out) the product and rewriting it as a
simplified sum.
When you start with x2 + 3x + 2 and then factor it to get (x + 1)(x + 2) , you are
rewriting the sum as a simplified product.
Every “simplify” instruction should say which kind of simplified form is desired. Some
books ignore this advice. In this course, we say either
Rewrite as a simplified sum, which means: collect like terms or
Rewrite as a simplified product, which means: the answer should be completely
factored.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Parentheses and PEMDAS: order of operations
When you write a story you organize your narrative by using paragraphs. When you
write mathematics, you organize your computations by using parentheses.
In the last example above, we said that ((3x) + 5) could be rewritten more simply as
(3x + 5) . Why are we allowed to remove the parentheses around the 3x?
The problem is that the expression 3 · 4 + 5 is ambiguous: it doesn’t indicate whether
the result should be (3 · 4) + 5 = 12 + 5 = 17 or 3 · (4 + 5) = 3 · 9 = 27. It seems that
parentheses must be used to make clear what is intended.
The difficulty with this approach is that using a lot of parentheses produces unreadable
expressions. That’s why mathematicians and computer scientists formulated the
PEMDAS rule, which specifies the order of operations. The real purpose of the
PEMDAS rule is to reduce the number of parentheses required needed to write
complicated expressions.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Here’s an easy example. In the expression 3 + 4 + 5 + 6 + 7, there are many ways to
insert parentheses. However, the associative law guarantees that all of these methods
give the same answer. This is obvious from everyday experience.
After all, when you pour containers of 3, 4, 5, 6, and 7 gallons (in the order listed) into
intermediate containers and then into a single barrel, the barrel will contain 25 gallons,
regardless of which intermediate containers you used. Similarly, no matter how you
insert parentheses in the expression 3 + 4 + 5 + 6 + 7, you will get the same answer.
For example, check that (3 + 4) + (5 + (6 + 7)) = 3 + (((4 + (5 + 6)) + 7)).
However, the above example is an exception: the value of many (most) expressions
does change, depending on how parentheses are inserted. Here are a few examples:
8 − 5 + 2 =? Two ways: (8 − 5) + 2 = 3 + 2 = 5
but 8 − (5 + 2) = 8 − 7 = 1
8 · 5 + 2 =? Two ways: (8 · 5) + 2 = 40 + 2 = 42 but 8 · (5 + 2) = 8 · 7 = 56
3 · 42 =? Two ways: 3 · (42 ) = 3 · 16 = 48
but (3 · 4)2 = 122 = 144
To resolve these ambiguities, we need to specify the order of applying arithmetic
operation symbols: ˆ + − · and \.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
The PEMDAS rule explains the order of operations intended when parentheses are
missing. The letter P tells you that the general goal is to get rid of parentheses. The
letters EMDAS, which probably should be written as E(M or D)(A or S), tell you what
is the proper order of operations.
More precisely, EMDAS tells you that at any stage of a simplification process, the first
operation to apply is chosen as follows.
Definition: Order of operations
E: (Exponent): If there is a power symbol ˆ, apply the one furthest to the right.
MD: (Multiply or Divide): Otherwise, if there is a · or / symbol, apply the one
furthest to the left.
AS: (Add or Subtract) Otherwise, if there is a + or − symbol, apply the one
furthest to the left.
After you apply that operation, you get a new expression. Apply these rules to the new
expression, and continue applying the rules until the result has no parentheses.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
In practice, you should remember the following special cases:
Multiply before you Add or Subtract:
2 + 3 · 4 = 2 + (3 · 4) = 2 + 12 = 14,
not (2 + 3) · 4 = 5 · 4 = 20
Exponentiate before you Multiply:
3 · 2ˆ2 = 3 · (2ˆ2) = 3 · 4 = 12,
not (2 · 3)ˆ2 = 6 ˆ 2 = 36. Similarly:
3 · 22 = 3 · (22 ) = 3 · 4 = 12,
not (2 · 3)2 = 62 = 36
Add and/or Subtract from left to right:
100 − 2 + 1 = (100 − 2) + 1 = 98 + 1 = 99,
not 100 − (2 + 1) = 100 − 3 = 97.
Multiply and/or Divide from left to right:
8/2 · 4 = (8/2) · 4 = 4 · 4 = 16,
not 8/(2 · 4) = 8/8 = 1
In practice, this last statement is not well known. It is safer to write (8/2) · 4 or
8/(2 · 4), whichever is intended.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Quiz review
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Special facts about modern math notation
Most multiply signs are omitted: 3 · x · y · (x + 5) · (2 · 3) = 3xy(x + 5)(2 · 3) .
A
Fraction lines are usually written horizontally: A/B = .
B
Substituting x + 2 for A and x + 3 for B gives
x+2
(x + 2)
=
.
(x + 2)/(x + 3) =
(x + 3)
x+3
Using a horizontal fraction line takes more space but is clearer, since you can omit
the parentheses required when you use slash notation.
Use superscript notation: write 34 for 3ˆ4.
If the first symbol in an expression is a minus sign, it means “negate,” i.e.,
multiply by −1. Thus −32 = (−1)(32 ) = (−1)(9) = −9, not (−3)2 (which equals
+9).
We do not follow the practice of some K-12 algebra books that use two different
minus signs, one to indicate subtract, and the other to indicate the sign of a
negative number.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Example 10: Apply PEMDAS to find the value of (3 + 2 · 4)2 + (8/2 · 4 − 3)3 . Here
“find the value” means: rewrite as a single number.
Step 1: 3 + 2 · 4 = 3 + 8 = 11, since we multiply before we add.
Step 2: 8/2 · 4 − 3= 4 · 4 − 3, since we multiply/divide from left to right
= 16 − 3 = 13, since we multiply before we subtract.
Using the results of Steps 1 and 2, we get
(3 + 2 · 4)2 + (8/2 · 4 − 3)3
= (11)2 + (13)3
= 112 + 133 Remove unnecessary parentheses.
= 121 + 2197 Exponentiate before you Add.
= 2318.
All of the above examples involved numerical expressions only. We are also interested
in using PEMDAS as a guide to rewriting algebra expressions with letters.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Example 11: Rewrite (x + 2)(x + 3) − x2 as a simplified sum.
Solution:
(x + 2)(x + 3) − x2 is given. Note that x2 can’t simplify further.
= x2 + 5x + 6 − x2
Multiply before you subtract.
= 5x + 6 .
This example is very misleading, because we got the right answer somewhat by
accident. We didn’t follow
Parenthesis Rule 2:
Whenever you Multiply or Exponentiate, put parentheses around the result.
In the following problem, omitting parentheses will produce a wrong answer and will
earn zero part credit.
Example 12: Rewrite x2 − (x + 2)(x + 3) as a simplified sum.
Solution:
x2 − (x + 2)(x + 3) is given.
= x2 − (x2 + 5x + 6) Multiply before you subtract. Insert parentheses!
= x2 − x2 − 5x − 6
Remove parentheses by distributing the minus sign.
= −5x − 6 .
The answer would have been incorrect had we forgetten to use parentheses.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Example 13: Rewrite x2 (x + 4) − (x + 2)2 as a simplified sum.
In the following solution, we apply PEMDAS by Exponentiating first to expand
(x + 2)2 .
Start:
Exponentiate first: Result in parentheses!
Note that x2 is already simplified, so we ignore it:
Multiply before you subtract:
The first set of parentheses is not needed:
The second set is needed! Distribute the minus sign:
Collect terms:
x2 (x + 4) − (x + 2)2
x2 (x + 4) − (x2 + 4x + 4)
(x3 + 4x2 ) − (x2 + 4x + 4)
x3 + 4x2 − (x2 + 4x + 4)
(x3 + 4x2 ) − x2 − 4x − 4
x3 + 4x2 − x2 − 4x − 4
x3 + 3x2 − 4x − 4
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
You can perform a partial check by substituting a number (say 1) for x. However, you
run the risk that you will make an algebra slip in your check even though your original
work was correct.
x2 (x + 4) − (x + 2)2 =? x3 + 3x2 − 4x − 4
12 (1 + 4) − (1 + 2)2 =? 13 + 3 · 12 − 4 · 1 − 4
1(5) − 32 =? 1 + 3 − 4 − 4
5 − 9 =? 4 − 4 − 4
−4 =? 0 − 4 YES!
Again, this is only a partial check.
Substitute 1 for x.
Advanced math fact: The general rule is that polynomials with highest power x3 are
identical if they agree at 4 different values of x. Therefore you would need to check
your answer for 3 additional values of x to be sure it is correct.
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Worked examples




3 − 4(x + 2)
= 3 − (4x + 8)
14)
= 3 − 4x − 8



= −4x − 5

x2 − x(x + 2)




 = x2 − (x2 + 2x)
= x2 − x2 − 2x
16)


= 0 − 2x



= −2x
(x)(x + 4) − (x + 7)2
= (x2 + 4x) − (x2 + 14x + 49)
15)
= x2 + 4x − x2 − 14x − 49



= −10x − 49

(x + 3)(x + 4) − 3(x + 7)2




 = (x2 + 7x + 12) − 3(x2 + 14x + 49)
= x2 + 7x + 12 − 3x2 − 42x − 147
17)


= x2 − 3x2 + 7x − 42x + 12 − 147



= −2x2 − 35x − 135




Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Quiz review
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Exercises
Click on Wolfram Calculator to find an answer checker. For example, type in “simplify
(2 − x)ˆ2.”
Click on
Click on
to see how to check various types of algebra problems.
to work on WebAssign HW problems.
Wolfram Algebra Examples
WA
1. Rewrite each of the following as a simplified sum:
a) (3 − x)2 − (4 − x)2
b) (x + 2)2 (x + 2)
d) (5x + 2)(2x − 1) − (3x + 1)2
c) 5x − (3x + 1)2
2. Substitute x + 2 for x in each of the following and rewrite the answer as a
simplified sum.
a) 4(3 − x)
b) 2 − x2
c) (2 − x)2
d) (x + 1)2 − x2
3. Substitute x − h for x in each of the following and rewrite the answer as a
simplified sum.
a) 4(3 − x)
b) 2 − x2
c) (2 − x)2
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d) (x + 1)2 − x2
M19500 Precalculus Chapter 0: Algebra preliminaries
Quiz review
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
4. Substitute −2 for h in each of your answers in question 3 above. Your answers
should match exactly your answers to question 2.
5. In
•
•
•
the function definition f (x) = x2 + 7
Find f (x + h) by substituting x + h for x . Simplify your answer.
Find f (x − h) by substituting x − h for x. Simplify your answer.
(x−h)
. Simplify your answer.
Use the last two results to simplify f (x+h)−f
h
6. Redo problem 5, this time using f (x) = 3 − 7x − 2x2 .
7. Redo problem 5, this time using f (x) = x3 .
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Quiz review
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Review exercises for Chapter 0
Example 1: In the expression A2 − B 2 substitute x + 3 for A and y − 2 for B.
Example 2: In the expression A2 − B 2 substitute A + 3 for A and B − 2 for B.
Example 3: In the statement f (x) = 3 − 7x − 2x3 substitute x + h for x.
Example 4: Substitute x + h for x in 3 − x2 .
Example 5: Substitute x + y for A and z for B in (A + B)(A − B) = A2 − B 2
Example 6. Substitute z for A and x + y for B in (A + B)(A − B) = A2 − B 2
Example 7: Substitute x for A, x + 1 for B, and x + 1 for C in
A(B + C) = AB + AC.
Example 8: Substitute 3x for A and 5 for B in the identity:
(A + B)(A − B) = A2 − B 2 . Then remove unnecessary parentheses:
Example 9: Check that the new identities obtained in Examples 7 and 8 yield true
statements when you substitute a) −5 for x
Stanley Ocken
b) 7 for x.
M19500 Precalculus Chapter 0: Algebra preliminaries
Quiz review
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Example 10:Find each of the following:
a) 2 + 3 · 4 b) 3 · 2ˆ2 c) 3 · 22 d) 100 − 2 + 1
e) (3 + 2 · 4)2 + (8/2 · 4 − 3)3 .
Order of operations
Exercises
e) 8/2 · 4
Rewrite each of the following as a simplified sum:
Example 11: (x + 2)(x + 3) − x2
Example 12: x2 − (x + 2)(x + 3)
Example 13: x2 (x + 4) − (x + 2)2
Example 14: 3 − 4(x + 2)
Example 15: (x)(x + 4) − (x + 7)2
Example 16: x2 − x(x + 2)
Example 17: (x + 3)(x + 4) − 3(x + 7)2
Stanley Ocken
M19500 Precalculus Chapter 0: Algebra preliminaries
Quiz review