Intermediate Algebra
Tutorial 28:
Factoring Trinomials
 Learning Objectives
|
After completing this tutorial, you should be able to:
-
Factor a trinomial of the form
 .
-
Factor a trinomial of the form
 .
-
Factor using substitution.
-
Indicate if a polynomial is a prime polynomial.
|
Introduction
|
| In this tutorial we add on to your factoring repertoire
by talking
about factoring trinomials. Basically, you will be doing the FOIL
method backwards. This is one of those things that just takes
practice
to master. Make sure that you work through the
problems
on this page as well as any that you teacher may have assigned you. You
never know when your math skills will be put to the test. Now
let's
get to factoring these trinomials. |
Tutorial
|
|
Factoring
Trinomials of
the Form
(Where the number in front of x
squared is 1)
|
| Basically, we are reversing the FOIL method to get
our factored
form. We are looking for two binomials that when you multiply
them
you get the given trinomial.
Step 1: Set up a product of two ( )
where each will
hold two terms.
| It will look like this:
( )(
). |
Step 2: Find the factors that go in
the first positions.
| To get the x
squared (which is the F in
FOIL), we would have to have an x in
the first
positions in each ( ).
So it would look like this: (x
)(x ).
|
Step 3: Find the factors that
go in the last
positions.
| The factors that would go in the last position
would have to be two
expressions such that their product equals c
(the constant) and at the same time their sum equals b
(number in front of x term).
As you are finding these factors, you have to
consider the sign of the
expressions:
| If c is positive,
your factors are going to both have the same sign depending on b’s
sign.
If c is negative,
your factors are going to have opposite signs depending on b’s
sign.
|
|
|
 Example
1: Factor the trinomial  . |
| Note that this trinomial does not have a GCF.
So we go right into factoring the trinomial of the
form .
|
| Step 1: Set up a product of two ( )
where each will
hold two terms. |
| It will look like
this:
(
)( ) |
| Step 2: Find the factors that go in
the first positions. |
| Since we have y squared as our first term, we will need
the following:
(y
)(y )
|
| Step 3: Find the factors that
go in the last
positions. |
| We need two numbers whose product is 6 and sum is
-5. That
would have to be -2 and -3.
Putting that into our factors we get:
|
 |
*-2 and -3 are two numbers whose
prod. is
6
and sum is -5 |
| Note that if we would multiply this out, we would get
the original
trinomial. |
Example
2: Factor the trinomial  |
 |
*Factor out the GCF of 2y |
| We are not finished, we can still factor the
trinomial. It is
of the form .
Anytime you are factoring, you need to make sure that
you factor everything
that is factorable. Sometimes you end up having to do several
steps
of factoring before you are done.
|
| Step 1 (trinomial): Set up a product
of two ( ) where
each will hold two terms. |
| It will look like this: 2y(
)( ) |
| Step 2 (trinomial): Find the factors
that go in the
first positions. |
| Since we have x squared as our first term, we will need
the following:
2y(x
)(x )
|
| Step 3 (trinomial): Find the
factors that go
in the last positions. |
| We need two numbers whose product is -20 and sum is
1. That
would have to be 5 and -4.
Putting that into our factors we get:
|
 |
*5 and -4 are two numbers whose
prod. is -20
*and sum is 1 |
| Note that if we would multiply this out, we would get
the original
trinomial. |
|
Factoring Trinomials of the Form
(where a does not
equal 1)
|
| Again, this is the reverse of the FOIL method.
The difference between this trinomial and the one
discussed above, is
there is a number other than 1 in front of the x
squared. This means, that not only do
you
need to find factors of c, but also a.
Step 1: Set up a product of two
( ) where each
will hold two terms.
| It will look like this
( )(
) |
Step 2: Use trial and error to find
the factors needed.
| The factors of a will
go in the first
terms of the binomials and the factors of c
will go in the last terms of the binomials.
The trick is to get the right combination of these
factors. You
can check this by applying the FOIL method. If your product comes
out to be the trinomial you started with, you have the right
combination
of factors. If the product does not come out to be the given
trinomial,
then you need to try again.
|
|
Example
3: Factor the trinomial  . |
| Note that this trinomial does not have a GCF.
So we go right into factoring the trinomial of the
form .
|
| Step 1: Set up a product of
two ( ) where each
will hold two terms. |
| It will look like this:
(
)( ) |
| Step 2: Use trial and error to find
the factors needed. |
| In the first terms of the binomials, we need factors
of 3 x
squared. This would have to be 3x and x.
In the second terms of the binomials, we need factors
of 2.
This would have to be -2 and -1. I used negatives here because
the
middle term is negative.
Also, we need to make sure that we get the right
combination of these
factors so that when we multiply them out we get .
|
|
Possible Factors
|
Check using the FOIL
method
Tutorial 26:
Multiplying
Polynomials
|
First try:
|
This is not our original polynomial.
So we need to try again.
|
Second try:
|
This is our original polynomial.
So this is the correct combination of factors for this polynomial.
|
| This process takes some
practice. After
a while you will get used to it and be able to come up with the right
factor
on the first try. |
Example
4: Factor the trinomial  |
| Note that this trinomial does not have a GCF.
So we go right into factoring the trinomial of the
form
|
| Step 1: Set up a product of
two ( ) where each
will hold two terms. |
| It will look like this:
(
)( ) |
| Step 2: Use trial and error to find
the factors needed. |
| In the first terms of the binomials, we need factors
of 5 a
squared.
This would have to be 5a and a.
In the second terms of the binomials, we need factors
of -6.
This would have to be -6 and 1, 6 and -1, 3 and -2, or -3 and 2.
Since the product of these factors has to be a negative number, we need
one positive factor and one negative factor.
Also we need to make sure that we get the right
combination of these
factors so that when we multiply them out we get .
|
|
Factor a Trinomial by
Substitution
|
| Sometimes a trinomial does not exactly fit the
form
OR ,
but with
the use of substitution it can be written in that form.
Step 1: Substitute x for the
non-coefficient
part of the middle term.
| Note that if x is
already being used in
the polynomial, you want to use a different variable like y
to
avoid confusion.
When you substitute, the trinomial should be in
the form
OR .
If it isn't, then you may have to seek a different
method than what
we are covering here. All of the ones we are using
substitution
with on this web page will fit this form.
|
Step 2: If the trinomial
is in the form OR ,
factor it accordingly.
Step 3: Substitute back in what you
replaced x
with in step 1.
|
Example
5: Factor the trinomial  |
| Note that this trinomial does not have a GCF. |
| Step 1: Substitute x for the
non-coefficient
part of the middle term.
Since x is already being
used in this problem,
let's use y for our substitution.
|
Let 
|
*Substitute y
in
for x squared
|
| Step 2: If the trinomial
is in the form OR ,
factor it accordingly |
| Now it is in a form that we do know how to
factor.
Now we proceed as we did in examples 3 and 4 above.
|
| In the first terms of the binomials, we need factors
of 2 y
squared.
This would have to be 2y and y.
In the second terms of the binomials, we need factors
of -9.
This would have to be -3 and 3, 9 and -1, or -9 and 1. Since the
product of these factors has to be a negative number, we need one
positive
factor and one negative factor.
Also we need to make sure that we get the right
combination of these
factors so that when we multiply them out we get  .
|
| Step 3: Substitute back in what you
replaced x
with in step 1. |
 |
*Substitute x
squared
back in for y |
| Note that if we would multiply this out, we would get
the original
trinomial. |
Example
6: Factor the trinomial  . |
| Note that this trinomial does not have a GCF. |
| Step 1: Substitute x for the
non-coefficient
part of the middle term. |
Let 
|
*Substitute x
in
for (a + 2)
|
| Step 2: If the trinomial
is in the form OR ,
factor it accordingly. |
| Now it is in a form that we do know how to
factor.
Now we proceed as we did in examples 1 and 2 above.
We need two numbers whose product is 32 and sum is
-12. That
would have to be -8 and -4.
Putting that into our factors we get:
|
 |
*-8 and -4 are two numbers whose
prod. is
32
and sum is -12 |
| Step 3: Substitute back in what you
replaced x
with in step 1. |
 |
*Substitute (a + 2) back
in for x
*Combine like terms in
( )
|
| Note that if we would multiply this out and the
original expression
out we would get the same trinomial. |
| Not every polynomial is factorable. Just like not
every number
has a factor other than 1 or itself. A prime number is a number
that
has exactly two factors, 1 and itself. 2, 3, and 5 are examples
of
prime numbers.
The same thing can occur with polynomials. If
a polynomial
is not factorable we say that it is a prime polynomial.
Sometimes you will not know it is prime until you start
looking for
factors of it. Once you have exhausted all possibilities, then
you
can call it prime. Be careful. Do not think because you
could not factor it on the first try that it is prime. You must
go
through ALL possibilities first before declaring it prime.
|
Example
7: Factor the trinomial  . |
| Note that this trinomial does not have a GCF.
So we go right into factoring the trinomial of the
form
|
| Step 1: Set up a product of two ( )
where each will
hold two terms. |
| It will look like
this:
(
)( ) |
| Step 2: Find the factors that go in
the first positions. |
| Since we have x squared
as our first term,
we will need the following:
(x
)(x )
|
| Step 3: Find the factors that
go in the last
positions. |
| We need two numbers whose product is 9 and sum is 2.
Can you think of any????
Since the product is a positive number and the sum is
a positive
number, we only need to consider pairs of numbers where both signs are
positive.
One pair of factors of 9 is 3 and 3, which does not
add up to be
2.
Another pair of factors are 1 and 9, which also does not
add up
to 2.
Since we have looked at ALL the possible factors, and
none of them
worked, we can say that this polynomial is prime.
In other words, it does not factor.
|
Practice Problems
|
| These are practice problems to help bring you to the
next level.
It will allow you to check and see if you have an understanding of
these
types of problems. Math works just like
anything
else, if you want to get good at it, then you need to practice
it.
Even the best athletes and musicians had help along the way and lots of
practice, practice, practice, to get good at their sport or instrument.
In fact there is no such thing as too much practice.
To get the most out of these, you should work the
problem out on
your own and then check your answer by clicking on the link for the
answer/discussion
for that problem. At the link you will find the answer
as well as any steps that went into finding that answer.
|
 Practice
Problems 1a - 1d:
Factor Completely.
|
Need Extra Help on These Topics?
|
All contents copyright (C) 2001 - 2008, WTAMU and Kim Seward. All rights reserved.
Last revised on June 22, 2003 by Kim Seward. |
(Back
to the THEA Math Help Page)
