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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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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Factoring Polynomials
Of all the topics covered in this chapter factoring polynomials is probably the most important
topic
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So, if you can’t factor the polynomial then you won’t be able to even start the
problem let alone finish it
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Factoring is the process by which
we go about determining what we multiplied to get the given quantity
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For instance, here are a variety of ways to factor 12
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A common method of factoring numbers is to completely factor the number into positive prime
factors
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For example
2, 3, 5, and 7 are all examples of prime numbers
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If we completely factor a number into positive prime factors there will only be one way of doing
it
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For our example above with 12 the
complete factorization is,

12 = ( 2 )( 2 )( 3)

Factoring polynomials is done in pretty much the same manner
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We then try to factor each of the terms we
found in the first step
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When we can’t do
any more factoring we will say that the polynomial is completely factored
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x 2 − 16 = ( x + 4 )( x − 4 )

This is completely factored since neither of the two factors on the right can be further factored
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Note that the first
factor is completely factored however
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x 4 − 16 =

(x

2

+ 4 ) ( x + 2 )( x − 2 )

The purpose of this section is to familiarize ourselves with many of the techniques for factoring
polynomials
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When factoring in general this will also be the first thing that we should try as it will often
simplify the problem
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If there is, we will factor it out of the polynomial
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Remember that the distributive
law states that

a ( b + c ) = ab + ac

In factoring out the greatest common factor we do this in reverse
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Example 1 Factor out the greatest common factor from each of the following polynomials
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Also note that we can factor an x2
out of every term
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8 x 4 − 4 x3 + 10 x 2 2 x 2 ( 4 x 2 − 2 x + 5 )
=

Note that we can always check our factoring by multiplying the terms back out to make sure we
get the original polynomial
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Each term contains and x3 and a y so we can factor both of those out
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Here is the work for this one
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To do this we
need the “+1” and notice that it is “+1” instead of “-1” because the term was originally a positive
term
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One of the more common mistakes with these types of factoring problems is to forget this “1”
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To check that the “+1” is required, let’s drop it and then multiply out to see what we
get
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It is easy to get
in a hurry and forget to add a “+1” or “-1” as required when factoring out a complete term
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However, it works the same way
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Doing the factoring for this problem gives,

9 x 2 ( 2 x + 7 ) − 12 x ( 2 x + = 3 x ( 2 x + 7 )( 3 x − 4 )
7)

[Return to Problems]

Factoring By Grouping
This is a method that isn’t used all that often, but when it can be used it can be somewhat useful
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Example 2 Factor by grouping each of the following
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Doing this gives,

3 x 2 − 2 x + 12 x −= x ( 3 x − 2 ) + 4 ( 3 x − 2 )
8
We can now see that we can factor out a common factor of 3 x − 2 so let’s do that to the final

factored form
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That’s all that there is to factoring by grouping
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[Return to Problems]

(b) x 5 + x − 2 x 4 − 2
In this case we will do the same initial step, but this time notice that both of the final two terms
are negative so we’ll factor out a “-” as well when we group them
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At this point we can see that we can factor an x out of the first term and a 2 out of the second
term
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x 5 + x − 2 x 4 − 2=

(x

4

+ 1) ( x − 2 )

[Return to Problems]

(c) x 5 − 3 x 3 − 2 x 2 + 6
This one also has a “-” in front of the third term as we saw in the last part
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We will still factor a “-” out when we
group however to make sure that we don’t lose track of it
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Again, you can always check that this
was done correctly by multiplying the “-” back through the parenthsis
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x 5 − 3 x 3 − 2 x 2 + 6= x 3 ( x 2 − 3) − 2 ( x 2 − 3)=

(x

2

− 3)( x3 − 2 )
[Return to Problems]

Factoring by grouping can be nice, but it doesn’t work all that often
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Factoring Quadratic Polynomials
First, let’s note that quadratic is another term for second degree polynomial
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In these problems we will be attempting
to factor quadratic polynomials into two first degree (hence forth linear) polynomials
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Let’s take a look at some examples
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(a) x 2 + 2 x − 15 [Solution]
(b) x 2 − 10 x + 24 [Solution]
(c) x 2 + 6 x + 9 [Solution]
(d) x 2 + 5 x + 1 [Solution]
(e) 3 x 2 + 2 x − 8 [Solution]
(f) 5 x 2 − 17 x + 6 [Solution]
(g) 4 x 2 + 10 x − 6 [Solution]

Solution
(a) x 2 + 2 x − 15
Okay since the first term is x2 we know that the factoring must take the form
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Therefore, the first term
in each factor must be an x
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We can narrow down the possibilities considerably
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In other words these two numbers must be
factors of -15
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( −1)(15)

(1)( −15)

( −3)( 5)

( 3)( −5)

Now, we can just plug these in one after another and multiply out until we get the correct pair
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The correct pair of numbers
must add to get the coefficient of the x term
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Here is the factored form of the polynomial
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Note that the method we used here will only work if the coefficient of the x2 term is one
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[Return to Problems]

(b) x 2 − 10 x + 24
Let’s write down the initial form again,

(

x 2 − 10 x + 24 = x +

)( x + )

Now, we need two numbers that multiply to get 24 and add to get -10
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In this case 3 and 3 will
be the correct pair of numbers
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Here is the factored form for this polynomial
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You should always do this when it happens
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There aren’t two
integers that will do this and so this quadratic doesn’t factor
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[Return to Problems]

(e) 3 x 2 + 2 x − 8
Okay, we no longer have a coefficient of 1 on the x2 term
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Since the coefficient of the x2 term is a 3 and there are only
two positive factors of 3 there is really only one possibility for the initial form of the factoring
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However, finding the numbers for the two blanks will not be as easy as the previous examples
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( −1)(8)

(1)( −8)

( −2 )( 4 )

( 2 )( −4 )

At this point the only option is to pick a pair plug them in and see what happens when we
multiply the terms out
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Let’s plug the numbers in and see what
we get
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However, since the middle term isn’t correct this isn’t the
correct factoring of the polynomial
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With the previous parts of this example it
didn’t matter which blank got which number
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Let’s flip the order and see what
we get
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We did guess correctly the first time we just put them into the wrong spot
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[Return to Problems]

(f) 5 x 2 − 17 x + 6
Again the coefficient of the x2 term has only two positive factors so we’ve only got one possible

initial form
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Here they are
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They are often the ones that we want
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With
some trial and error we can get that the factoring of this polynomial is,

5 x 2 − 17 x + 6=

( 5 x − 2 )( x − 3)

[Return to Problems]

(g) 4 x 2 + 10 x − 6
In this final step we’ve got a harder problem here
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This means that the initial form must be one of the following
possibilities
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Here they are,

( −1)( 6 )

(1)( −6 )

( −2 )( 3)

( 2 )( −3)

With some trial and error we can find that the correct factoring of this polynomial is,

4 x 2 + 10 x − 6=

( 2 x − 1)( 2 x + 6 )

Note as well that in the trial and error phase we need to make sure and plug each pair into both
possible forms and in both possible orderings to correctly determine if it is the correct pair of
factors or not
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This
gives,

4 x 2 + 10 x −= 2 ( 2 x − 1)( x + 3)
6

This is important because we could also have factored this as,

4 x 2 + 10 x − 6=

( 4 x − 2 )( x + 3)

which, on the surface, appears to be different from the first form given above
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[Return to Problems]

Special Forms
There are some nice special forms of some polynomials that can make factoring easier for us on
occasion
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a 2 + 2ab + b 2 = ( a + b )

2

a 2 − 2ab + b 2 = ( a − b )

2

a 2 − b 2 = ( a + b )( a − b )

a 3 + b3 = ( a + b ) ( a 2 − ab + b 2 )
a 3 − b3 = ( a − b ) ( a 2 + ab + b 2 )

Let’s work some examples with these
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(a) x 2 − 20 x + 100 [Solution]
(b) 25 x 2 − 9 [Solution]
(c) 8 x 3 + 1 [Solution]
Solution
(a) x 2 − 20 x + 100
In this case we’ve got three terms and it’s a quadratic polynomial
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Notice as well that 2(10)=20 and this is the
coefficient of the x term
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The correct
factoring of this polynomial is,

x 2 − 20 x + 100 = x − 10 )
(

2

To be honest, it might have been easier to just use the general process for factoring quadratic
polynomials in this case rather than checking that it was one of the special forms, but we did need
to see one of them worked
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Here is the correct factoring for this polynomial
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Here is the factoring for this
polynomial
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There are rare cases where this can be done, but none of those special
cases will be seen here
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However, there are some that we can do so
let’s take a look at a couple of examples
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(a) 3 x 4 − 3 x 3 − 36 x 2 [Solution]
(b) x 4 − 25 [Solution]
(c) x 4 + x 2 − 20 [Solution]
Solution
(a) 3 x 4 − 3 x 3 − 36 x 2
In this case let’s notice that we can factor out a common factor of 3x2 from all the terms so let’s
do that first
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Doing this gives
us,

3 x 4 − 3 x3 − 36 x 2 3 x 2 ( x − 4 )( x + 3)
=

Don’t forget that the FIRST step to factoring should always be to factor out the greatest common
factor
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[Return to Problems]

(b) x 4 − 25
There is no greatest common factor here
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=
x 4 − 25

( x ) − ( 5)
2 2

2

So, we can use the third special form from above
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Note however, that often we will
need to do some further factoring at this stage
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u 2 + u − 20 =

( u − 4 )( u + 5)

We used a different variable here since we’d already used x’s for the original polynomial
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We can then

rewrite the original polynomial in terms of u’s as follows,

x 4 + x 2 − 20 = u 2 + u − 20
and we know how to factor this! So factor the polynomial in u’s then back substitute using the
fact that we know u = x 2
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The correct factoring of this polynomial is then,

x 4 + x 2 − 20 = ( x − 2 )( x + 2 ) ( x 2 + 5 )
Note that this converting to u first can be useful on occasion, however once you get used to these
this is usually done in our heads
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However, we did
cover some of the most common techniques that we are liable to run into in the other chapters of
this work

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