How To Solve Binomials And Trinomials
College Algebra
Tutorial 7: Factoring Polynomials
Learning Objectives
After completing this tutorial, you should be able to:
- Find the Greatest Common Factor (GCF) of a polynomial.
- Factor out the GCF of a polynomial.
- Factor a polynomial with four terms by grouping.
- Factor a trinomial of the form
. - Factor a trinomial of the form
. - Indicate if a polynomial is a prime polynomial.
- Factor a perfect square trinomial.
- Factor a difference of squares.
- Factor a sum or difference of cubes.
- Apply the factoring strategy to factor a polynomial completely.
Introduction
Factoring is to write an expression as a product of factors. For example, we can write 10 as (5)(2), where 5 and 2 are called factors of 10. We can also do this with polynomial expressions. In this tutorial we are going to look at several ways to factor polynomial expressions. By the time I'm through with you, you will be a factoring machine.
Basically, when we factor, we reverse the process of multiplying the polynomial which was covered in Tutorial 6: Polynomials.
Tutorial
The GCF for a polynomial is the largest monomial that divides (is a factor of) each term of the polynomial.
Step 1: Identify the GCF of the polynomial.
Step 2: Divide the GCF out of every term of the polynomial.
This process is basically the reverse of the distributive property.
Step 1: Identify the GCF of the polynomial.
The largest monomial that we can factor out of each term is 2 y .
Step 2: Divide the GCF out of every term of the polynomial.
| | *Divide 2 y out of every term of the poly. |
Be careful. If a term of the polynomial is exactly the same as the GCF, when you divide it by the GCF you are left with 1, NOT 0. Don't think, 'oh I have nothing left', there is actually a 1. As shown above when we divide 2 y by 2 y we get 1, so we need a 1 as the third term inside of the ( ).
Note that if we multiply our answer out, we should get the original polynomial. In this case, it does check out. Factoring gives you another way to write the expression so it will be equivalent to the original problem.
This problem looks a little different, because now our GCF is a binomial. That is ok, we treat it in the same manner that we do when we have a monomial GCF.
Note that this is not in factored form because of the plus sign we have before the 5 in the problem. To be in factored form, it must be written as a product of factors.
Step 1: Identify the GCF of the polynomial.
This time it isn't a monomial but a binomial that we have in common.
Our GCF is (3 x -1).
Step 2: Divide the GCF out of every term of the polynomial.
| | *Divide (3 x - 1) out of both parts |
When we divide out the (3 x - 1) out of the first term, we are left with x . When we divide it out of the second term, we are left with 5.
That is how we get the ( x + 5) for our second ( ).
Four Terms by Grouping
In some cases there is not a GCF for ALL the terms in a polynomial. If you have four terms with no GCF, then try factoring by grouping.
Step 1: Group the first two terms together and then the last two terms together.
Step 2: Factor out a GCF from each separate binomial.
Step 3: Factor out the common binomial.
Note how there is not a GCF for ALL the terms. So let's go ahead and factor this by grouping.
Step 1: Group the first two terms together and then the last two terms together.
| |
Step 2: Factor out a GCF from each separate binomial.
| | *Factor out an x squared from the 1st ( ) |
Step 3: Factor out the common binomial.
| | *Divide ( x + 3) out of both parts |
Note that if we multiply our answer out, we do get the original polynomial.
Note how there is not a GCF for ALL the terms. So let's go ahead and factor this by grouping.
Step 1: Group the first two terms together and then the last two terms together.
| |
Be careful. When the first term of the second group of two has a minus sign in front of it, you want to put the minus in front of the second ( ). When you do this you need to change the sign of BOTH terms of the second ( ) as shown above.
Step 2: Factor out a GCF from each separate binomial.
| | *Factor out a 7 x squared from the 1st ( ) |
Step 3: Factor out the common binomial.
| | *Divide ( x - 2) out of both parts |
Note that if we multiply our answer out that we do get the original polynomial.
(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.
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 a squared as our first term, we will need the following:
( a )( a )
Step 3: Find the factors that go in the last positions.
We need two numbers whose product is -14 and sum is -5. That would have to be -7 and 2.
Putting that into our factors we get:
| | *-7 and 2 are two numbers whose prod. is -14 |
Note that if we would multiply this out, we would get the original trinomial.
Note that this trinomial does have a GCF of 2 y .
We need to factor out the GCF before we tackle the trinomial part of this.
| | *Factor out the GCF of 2 y |
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: 2 y ( )( )
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:
2 y ( x )( x )
Step 3 (trinomial): Find the factors that go in the last positions.
We need two numbers whose product is 15 and sum is 8. That would have to be 5 and 3.
Putting that into our factors we get:
| | *5 and 3 are two numbers whose prod. is 15 |
Note that if we would multiply this out, we would get the original trinomial.
(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.
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 3 x and x.
In the second terms of the binomials, we need factors of 2. This would have to be 2 and 1. I used positives here because the middle term is positive.
Also, we need to make sure that we get the right combination of these factors so that when we multiply them out we get
.
| | |
 |  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.
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 x squared. This would have to be 5 x and x.
In the second terms of the binomials, we need factors of -8. This would have to be -8 and 1, 8 and -1, 2 and -4, or -2 and 4. 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
.
| | |
 |  This is our original polynomial. So this is the correct combination of factors for this polynomial. |
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.
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 12 and sum is 5.
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 12 is 3 and 4, which does not add up to be 5.
Another pair of factors are 2 and 6, which also does not add up to 5.
Another pair of factors are 1 and 12, which also does not add up to 5.
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.
OR
It has to be exactly in this form to use this rule. When you have a base being squared plus or minus twice the product of the two bases plus another base squared, it factors as the sum (or difference) of the bases being squared.
This is the reverse of the binomial squared found in Tutorial 6: Polynomials. Recall that factoring is the reverse of multiplication.
First note that there is no GCF to factor out of this polynomial.
Since it is a trinomial, you can try factoring this by trial and error shown above. But if you can recognize that it fits the form of a perfect square trinomial, you can save yourself some time.
| | *Fits the form of a perfect sq. trinomial |
Note that if we would multiply this out, we would get the original polynomial.
First note that there is no GCF to factor out of this polynomial.
Since it is a trinomial, you can try factoring this by trial and error shown above. But if you can recognize that it fits the form of a perfect square trinomial, you can save yourself some time.
| | *Fits the form of a perfect sq. trinomial |
Note that if we would multiply this out, we would get the original polynomial.
Note that the sum of two squares DOES NOT factor.
Just like the perfect square trinomial, the difference of two squares has to be exactly in this form to use this rule. When you have the difference of two bases being squared, it factors as the product of the sum and difference of the bases that are being squared.
This is the reverse of the product of the sum and difference of two terms found in Tutorial 6: Polynomials. Recall that factoring is the reverse of multiplication.
First note that there is no GCF to factor out of this polynomial.
This fits the form of a the difference of two squares. So we will factor using that rule:
| | *Fits the form of a diff. of two squares |
Note that if we would multiply this out, we would get the original polynomial.
First note that there is no GCF to factor out of this polynomial.
This fits the form of the difference of two squares. So we will factor using that rule:
| | *Fits the form of a diff. of two squares |
Note that if we would multiply this out and the original expression out we would get the same polynomial.
The sum of two cubes has to be exactly in this form to use this rule. When you have the sum of two cubes, you have a product of a binomial and a trinomial. The binomial is the sum of the bases that are being cubed. The trinomial is the first base squared, the second term is the opposite of the product of the two bases found, and the third term is the second base squared.
First note that there is no GCF to factor out of this polynomial.
This fits the form of the sum of cubes. So we will factor using that rule:
| | *Fits the form of a sum of two cubes |
Note that if we would multiply this out, we would get the original polynomial.
This is factored in a similar fashion to the sum of two cubes. Note the only difference is that the sign in the binomial is a - which matches the original sign, and the sign in front of ax is positive, which is the opposite sign.
The difference of two cubes has to be exactly in this form to use this rule. When you have the difference of two cubes, you have a product of a binomial and a trinomial. The binomial is the difference of the bases that are being cubed. The trinomial is the first base squared, the second term is the opposite of the product of the two bases found, and the third term is the second base squared.
First note that there is no GCF to factor out of this polynomial.
This fits the form of the difference of cubes. So we will factor using that rule:
| | *Fits the form of a diff. of two cubes |
Note that if we would multiply this out, we would get the original polynomial.
Now that you have a list of different factoring rules, let's put it all together. The following is a checklist of the factoring rules that we have covered in our tutorials.
When you need to factor, you ALWAYS look for the GCF first . Whether you have a GCF or not, then you continue looking to see if you have anything else that factors.
Below is a checklist to make sure you do not miss anything. Always factor until you can not factor any further.
I. GCF:
Always check for the GCF first, no matter what.
II. Binomials:
a.
b. Trial and error:
c. Perfect square trinomial:
IV. Polynomials with four terms:
The first thing that we always check when we are factoring is WHAT?
The GCF. In this case, there is one.
Factoring out the GCF of 3 we get:
| | *Factor a 3 out of every term |
Next, we assess to see if there is anything else that we can factor. We have a trinomial inside the ( ). It fits the form of a perfect square trinomial, so we will factor it accordingly:
| | *Fits the form of a perfect sq. trinomial |
There is no more factoring that we can do in this problem.
Note that if we would multiply this out, we would get the original polynomial.
The first thing that we always check when we are factoring is WHAT?
The GCF. In this case, there is not one.
So we assess what we have. It fits the form of a difference of two squares, so we will factor it accordingly:
| | *Fits the form of a diff. of two squares |
Next we assess to see if there is anything else that we can factor. Note how the second binomial is another difference of two squares. That means we have to continue factoring this problem.
| | *Fits the form of a diff. of two squares |
There is no more factoring that we can do in this problem.
Note that if we would multiply this out, we would get the original polynomial.
The first thing that we always check when we are factoring is WHAT?
The GCF. In this case, there is not one.
So we assess what we have. It fits the form of a sum of two cubes, so we will factor it accordingly:
| | *Fits the form of a sum of two cubes |
There is no more factoring that we can do in this problem.
Note that if we would multiply this out, we would get the original polynomial.
The first thing that we always check when we are factoring is WHAT?
The GCF. In this case, there is not one.
So we assess what we have. This is a trinomial that does not fit the form of a perfect square trinomial. Looks like we will have to use trial and error:
| | *Factor by trial and error |
There is no more factoring that we can do in this problem.
Note that if we would multiply this out, we would get the original polynomial.
The first thing that we always check when we are factoring is WHAT?
The GCF. In this case, there is not one.
So we assess what we have. This is a polynomial with four terms. Looks like we will have to try factoring it by grouping:
| | *Group in two's |
There is no more factoring that we can do in this problem.
Note that if we would multiply this out, we would get the original polynomial.
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 - 1f: Factor completely.
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Last revised on Dec. 13, 2009 by Kim Seward.
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How To Solve Binomials And Trinomials
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