Notesale: Turn your study into money

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Preface
Here are my online notes for my Algebra course that I teach here at Lamar University, although I
have to admit that it’s been years since I last taught this course
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Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to
learn Algebra or needing a refresher for algebra
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However, they do assume that you’ve had some
exposure to the basics of algebra at some point prior to this
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Here are a couple of warnings to my students who may be here to get a copy of what happened on
a day that you missed
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Because I wanted to make this a fairly complete set of notes for anyone wanting to learn
algebra I have included some material that I do not usually have time to cover in class
and because this changes from semester to semester it is not noted here
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2
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Likewise, even
if I do work some of the problems in here I may work fewer problems in class than are
presented here
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Sometimes questions in class will lead down paths that are not covered here
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Sometimes a very good question gets asked in class
that leads to insights that I’ve not included here
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4
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THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!!
Using these notes as a substitute for class is liable to get you in trouble
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Polynomials
In this section we will start looking at polynomials
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We will start off with polynomials in one variable
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e
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The degree of a
polynomial in one variable is the largest exponent in the polynomial
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Here are examples of polynomials and their degrees
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Also,
polynomials can consist of a single term as we see in the third and fifth example
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This really is a polynomial even it
may not look like one
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Another way to write the last example is

−8x 0
Written in this way makes it clear that the exponent on the x is a zero (this also explains the
degree…) and so we can see that it really is a polynomial in one variable
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4 x 6 + 15 x −8 + 1
5 x − x + x2
2 3
+ x −2
x
The first one isn’t a polynomial because it has a negative exponent and all exponents in a
polynomial must be positive
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1
2

5 x − x + x = 5x − x + x2
2

By converting the root to exponent form we see that there is a rational root in the algebraic
expression
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As a general rule of thumb if an algebraic
expression has a radical in it then it isn’t a polynomial
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2 3
+ x − 2= 2 x −1 + x 3 − 2
x
So, this algebraic expression really has a negative exponent in it and we know that isn’t allowed
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Note that this doesn’t mean that radicals and fractions aren’t allowed in polynomials
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For instance, the following is a polynomial
3

5 x4 −

7 2 1
x +
x − 5 14 113
12
8

There are lots of radicals and fractions in this algebraic expression, but the denominators of the
fractions are only numbers and the radicands of each radical are only a numbers
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Therefore this is a polynomial
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Polynomials in two variables are
algebraic expressions consisting of terms in the form ax n y m
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Here are some examples of polynomials in two variables and their degrees
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Also, the
degree of the polynomial may come from terms involving only one variable
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We can also talk about polynomials in three variables, or four variables or as many variables as
we need
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Next we need to get some terminology out of the way
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A binomial is a polynomial that consists of exactly two terms
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We will use these terms off and on
so you should probably be at least somewhat familiar with them
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You’ll note that we
left out division of polynomials
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Before actually starting this discussion we need to recall the distributive law
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Here is the distributive law
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This is probably best done with a couple
of examples
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(a) Add 6 x 5 − 10 x 2 + x − 45 to 13 x 2 − 9 x + 4
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[Solution]
Solution
(a) Add 6 x 5 − 10 x 2 + x − 45 to 13 x 2 − 9 x + 4
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(6x

5

− 10 x 2 + x − 45 ) + (13 x 2 − 9 x + 4 )

In this case the parenthesis are not required since we are adding the two polynomials
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To add two polynomials all that
we do is combine like terms
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In this case this is,

(6x

5

− 10 x 2 + x − 45 ) + (13 x 2 − 9 x + 4= 6 x5 + ( −10 + 13) x 2 + (1 − 9 ) x − 45 + 4
)
= 6 x5 + 3 x 2 − 8 x − 41
[Return to Problems]

(b) Subtract 5 x 3 − 9 x 2 + x − 3 from x 2 + x + 1
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We will also need to be very careful
with the order that we write things down in
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We are subtracting the
whole polynomial and the parenthesis must be there to make sure we are in fact subtracting the
whole polynomial
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This means that we will change the sign on every term in the second polynomial
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After distributing the minus through the parenthesis we again combine like
terms
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x 2 + x + 1 − ( 5 x 3 − 9 x 2 + x − 3) = x 2 + x + 1 − 5 x 3 + 9 x 2 − x + 3
=x3 + 10 x 2 + 4
−5

Note that sometimes a term will completely drop out after combing like terms as the x did here
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[Return to Problems]

Now let’s move onto multiplying polynomials
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Example 2 Multiply each of the following
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4 x 2 ( x 2 − 6 x + 2 ) = 4 x 4 − 24 x3 + 8 x 2

[Return to Problems]

(b)

( 3x + 5)( x − 10 ) This one will use the FOIL method for multiplying these two binomials
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If either of the
polynomials isn’t a binomial then the FOIL method won’t work
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The FOIL acronym is simply a convenient way to remember
this
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( 4x

2

− x ) ( 6 − 3 x ) = x 2 − 12 x 3 − 6 x + 3 x 2 = x 3 + 27 x 2 − 6 x
24
−12

(d) ( 3 x + 7 y )( x − 2 y )

[Return to Problems]

We can still FOIL binomials that involve more than one variable so don’t get excited about these
kinds of problems when they arise
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Recall
however that the FOIL acronym was just a way to remember that we multiply every term in the
second polynomial by every term in the first polynomial
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( 2 x + 3) ( x 2 − x + 1)=

2 x3 − 2 x 2 + 2 x + 3x 2 − 3x + 3 2 x3 + x 2 − x + 3
=
[Return to Problems]

Let’s work another set of examples that will illustrate some nice formulas for some special
products
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Example 3 Multiply each of the following
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( 3x + 5)( 3x − 5) =

9 x 2 − 15 x + 15 x − 25 = 9 x 2 − 25

In this case the middle terms drop out
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Squaring with polynomials works the same way
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(1 − 7 x )

2

=1 − 7 x )(1 − 7 x ) =− 7 x − 7 x + 49 x 2 =− 14 x + 49 x 2
1
1
(
[Return to Problems]

(d) 4 ( x + 3)

2

This part is here to remind us that we need to be careful with coefficients
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4 ( x + 3) = 4 ( x + 3)( x + 3) = 4 ( x 2 + 6 x + 9 ) = 4 x 2 + 24 x + 36
2

You can only multiply a coefficient through a set of parenthesis if there is an exponent of “1” on
the parenthesis
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Just to illustrate the point
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( a + b )( a − b ) = a 2 − b2
2
( a + b ) =a 2 + 2ab + b2
2
( a − b ) =a 2 − 2ab + b2

Be careful to not make the following mistakes!

( a + b ) ≠ a 2 + b2
2
( a − b ) ≠ a 2 − b2
2

These are very common mistakes that students often make when they first start learning how to
multiply polynomials

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