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Of all the topics covered in this chapter factoring
polynomials is probably the most important topic. There are many sections in later chapters
where the first step will be to factor a polynomial. So, if you can’t factor the polynomial then
you won’t be able to even start the problem let alone finish it.
Let’s start out by talking a little bit about just what
factoring is. Factoring is the process
by which we go about determining what we multiplied to get the given quantity. We do this all the time with numbers. For instance, here are a variety of ways to
factor 12.
There are many more possible ways to factor 12, but these
are representative of many of them.
A common method of factoring numbers is to completely factor the number into
positive prime factors. A prime number is a number whose only
positive factors are 1 and itself. For
example 2, 3, 5, and 7 are all examples of prime numbers. Examples of numbers that aren’t prime are 4,
6, and 12 to pick a few.
If we completely factor a number into positive prime factors
there will only be one way of doing it.
That is the reason for factoring things in this way. For our example above with 12 the complete
factorization is,
Factoring polynomials is done in pretty much the same
manner. We determine all the terms that
were multiplied together to get the given polynomial. We then try to factor each of the terms we
found in the first step. This continues
until we simply can’t factor anymore.
When we can’t do any more factoring we will say that the polynomial is completely factored.
Here are a couple of examples.
This is completely factored since neither of the two factors
on the right can be further factored.
Likewise,
is not completely factored because the second factor can be
further factored. Note that the first
factor is completely factored however.
Here is the complete factorization of this polynomial.
The purpose of this section is to familiarize ourselves with
many of the techniques for factoring polynomials.
Greatest Common
Factor
The first method for factoring polynomials will be factoring
out the greatest common factor. When factoring in general this will also be
the first thing that we should try as it will often simplify the problem.
To use this method all that we do is look at all the terms
and determine if there is a factor that is in common to all the terms. If there is, we will factor it out of the
polynomial. Also note that in this case
we are really only using the distributive law in reverse. Remember that the distributive law states
that
In factoring out the greatest common denominator we do this
in reverse. We notice that each term has
an a in it and so we “factor” it out
using the distributive law in reverse as follows,
Let’s take a look at some examples.
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Example 1 Factor
out the greatest common factor from each of the following polynomials.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
(a)
First we will notice that we can factor a 2 out of every
term. Also note that we can factor an x2 out of every term. Here then is the factoring for this
problem.
Note that we can always check our factoring by multiplying
the terms back out to make sure we get the original polynomial.
[Return to Problems]
(b)
In this case we have both x’s and y’s in the
terms but that doesn’t change how the process works. Each term contains and x3 and a y
so we can factor both of those out.
Doing this gives,
[Return to Problems]
(c)
In this case we can factor a 3x out of every term. Here
is the work for this one.
Notice the “+1” where the 3x originally was in the final term, since the final term was the
term we factored out we needed to remind ourselves that there was a term
there originally. To do this we need
the “+1” and notice that it is “+1” instead of “-1” because the term was
originally a positive term. If it had
been a negative term originally we would have had to use “-1”.
One of the more common mistakes with these types of
factoring problems is to forget this “1”.
Remember that we can always check by multiplying the two back out to
make sure we get the original. To
check that the “+1” is required, let’s drop it and then multiply out to see
what we get.
So, without the “+1” we don’t get the original
polynomial! Be careful with this. It is easy to get in a hurry and forget to
add a “+1” or “-1” as required when factoring out a complete term.
[Return to Problems]
(d)
This one looks a little odd in comparison to the
others. However, it works the same
way. There is a 3x in each term and there is also a in each term and so that can also be
factored out. Doing the factoring for
this problem gives,
[Return to Problems]
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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.
This method is best illustrated with an example or two.
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Example 2 Factor
by grouping each of the following.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
(a)
In this case we group
the first two terms and the final two terms as shown here,
Now, notice that we can factor an x out of the first grouping and a 4 out of the second grouping.
Doing this gives,
We can now see that we can factor out a common factor of so let’s do that to the final factored form.
And we’re done.
That’s all that there is to factoring by grouping. Note again that this will not always work
and sometimes the only way to know if it will work or not is to try it and
see what you get.
[Return to Problems]
(b)
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.
Doing this gives,
Again, we can always distribute the “-” back through the
parenthesis to make sure we get the original polynomial.
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. This gives,
We now have a common factor that we can factor out to
complete the problem.
[Return to Problems]
(c)
This one also has a “-” in front of the third term as we
saw in the last part. However, this
time the fourth term has a “+” in front of it unlike the last part. We will still factor a “-” out when we
group however to make sure that we don’t lose track of it. When we factor the “-” out notice that we
needed to change the “+” on the fourth term to a “-”. Again, you can always check that this was
done correctly by multiplying the “-” back through the parenthsis.
Now that we’ve done a couple of these we won’t put the
remaining details in and we’ll go straight to the final factoring.
[Return to Problems]
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Factoring by grouping can be nice, but it doesn’t work all
that often. Notice that as we saw in the
last two parts of this example if there is a “-” in front of the third term we
will often also factor that out of the third and fourth terms when we group
them.
Factoring Quadratic
Polynomials
First, let’s note that quadratic is another term for second
degree polynomial. So we know that the
largest exponent in a quadratic polynomial will be a 2. In these problems we will be attempting to factor
quadratic polynomials into two first degree (hence forth linear)
polynomials. Until you become good at
these, we usually end up doing these by trial and error although there are a
couple of processes that can make them somewhat easier.
Let’s take a look at some examples.
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Example 3 Factor
each of the following polynomials.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
(f) [Solution]
(g) [Solution]
Solution
(a)
Okay since the first term is x2 we know that the factoring must take the form.
We know that it will take this form because when we
multiply the two linear terms the first term must be x2 and the only way to get that to show up is to
multiply x by x. Therefore, the first
term in each factor must be an x. To finish this we just need to determine
the two numbers that need to go in the blank spots.
We can narrow down the possibilities considerably. Upon multiplying the two factors out these
two numbers will need to multiply out to get -15. In other words these two numbers must be factors
of -15. Here are all the possible ways
to factor -15 using only integers.
Now, we can just plug these in one after another and
multiply out until we get the correct pair.
However, there is another trick that we can use here to help us
out. The correct pair of numbers must
add to get the coefficient of the x
term. So, in this case the third pair
of factors will add to “+2” and so that is the pair we are after.
Here is the factored form of the polynomial.
Again, we can always check that we got the correct answer
my doing a quick multiplication.
Note that the method we used here will only work if the
coefficient of the x2
term is one. If it is anything else
this won’t work and we really will be back to trial and error to get the
correct factoring form.
[Return to Problems]
(b)
Let’s write down the initial form again,
Now, we need two numbers that multiply to get 24 and add
to get -10. It looks like -6 and -4
will do the trick and so the factored form of this polynomial is,
[Return to Problems]
(c)
Again, let’s start with the initial form,
This time we need two numbers that multiply to get 9 and
add to get 6. In this case 3 and 3
will be the correct pair of numbers.
Don’t forget that the two numbers can be the same number on occasion
as they are here.
Here is the factored form for this polynomial.
Note as well that we further simplified the factoring to
acknowledge that it is a perfect square.
You should always do this when it happens.
[Return to Problems]
(d)
Once again, here is the initial form,
Okay, this time we need two numbers that multiply to get 1
and add to get 5. There aren’t two
integers that will do this and so this quadratic doesn’t factor.
This will happen on occasion so don’t get excited about it
when it does.
[Return to Problems]
(e)
Okay, we no longer have a coefficient of 1 on the x2 term. However we can still make a guess as to the
initial form of the factoring. 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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