Lesson 3.2 Factoring Polynomials with 4 terms
Factoring polynomials with more than 3 terms can sometimes be done by grouping some of the terms together. There are several types of problems with 4 terms, but I am just going to teach you about one type that has been seen a lot in TEXTBOOKS lately!!
First look at these that have been "grouped" for you.
3(x5) +4(x5). Notice the group formed, x 5, occurs twice.
That is our first goal (step).
Try this one: 4x8 + 3xy 6y. This will become 4(x2) + 3y(x2).
I did this by common factoring the 4 from the 4x  8
and then common factoring the 3y from the 3xy  6y.t
Notice that the group (x2) occurs twice.
I, of course, was careful in creating the 4 terms of the polynomial!
I wanted there to be an expression that was repeated.
Try these 2 problems.
1. 5x20+ 2x8
2. xy7y + x 7
Hint: for the last 2 terms common factor a 1 from each!
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Example A 

Example B

5(x4) + 2(x4)


y(x7) + 1(x7)

You have learned the first step. We can factor now.
Now you are ready for the second step.
4x 8 + 3xy 6y =

Notice the 2 expressions inside

4(x2) + 3y(x2)

the parenthesis are EQUAL.


I want to temporarily


call the (x2) BIG P

4P + 3yP

Replace each (x2) with P.


Now this is an easy problem to factor. As we did in Lesson 3.1. Do so.

or P(4 +3y)

The 2 terms have a BIG P in common.

= (x  2)(4 + 3y)

Now replace the P with its equal (x2)

THUS 4x  8 + 3xy  6y factors to (x2)(4+3y).
We have completely factored a 4 term polynomial by grouping,
BUT the factoring depended upon us having ,at least twice,
the SAME expression inside parenthesis!
Be sure you see that we had exactly the same expression in each parenthesis.
4(x2)+ 3y(x2) I call this the BIG P Principle.
We temporarily replace all of the identical parenthesis with P
and then see we can factor with common factoring of the P.
It is very useful at other times in algebra.
I created the BIG P Principle for my Algebra 1 classes.
Now finish the problems we started above.
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Example A 

Example B 
5x 20 + 2x 8


yx7y + x  7

5(x4) + 2(x4)


y(x7) + 1(x7)
sometimes we have to use a 1 for the common factor.

5P + 2P


yP + 1P

P(5 + 2)


P(y+1)

P7 = (x4)7


(x 7)(y+1)

Remember this BIG P Principle of Factoring.
It would be a good idea for you to copy these to your notebook. We will use this again.
NOW watch this one.
It is a little different. Do you see the difference? Look at the signs.
When the signs of the first two terms are not the same as the signs of the last two terms
then use a negative common factor on the last two terms. HOPE YOU wrote this down.
4x+8 3xy 6y
LOOK HERE FIRST. Do you see the signs are not the same for our groups? The first 2 terms have positives and the last two have negatives. When the signs of the first 2 terms are not the same as the signs of the last two terms then that is a BIG SIGNAL to you. You must factor out a negative from the last two terms and that will change the last 2 signs to the opposite of what they are. Read this again. Then read the box to the right starting at the top.

Factor a 4 from the 4x+8 and
then factor a 3y from the 3xy 6y. I did that not because they are both negative, but because they are not the same as the signs of the first 2 terms, 4x+8. THEY ARE OPPOSITES!
4x+8= 4(x + 2)
and 3xy  6y = 3y(x + 2) We have the same terms in both parenthesis. YEA!
Again, if you do not understand this
then go back to common factoring.
When the signs of the first two terms are not the same as the signs of the last two terms then use a negative common factor, see the 3y in this problem.

4(x+2)  3y(x+2)
Notice the 2 expressions inside the parenthesis are EQUAL.

See the  3y.
When I factor it out of the 3xy 6y it will make both of these positive. Let's check that part.
3xy6y = 3y(x+2) Right?
Multiply the 3y(x+2) back out and you should get the
3xy 6y. Be sure you see this.
Notice the 2 expressions inside the parentheses are EQUAL. In this problem they are both x+2. They MUST be equal or something is wrong or it is not this type of factoring problem!!! There are other kinds!! Be sure you have not made a mistake.


I want to temporarily call (x+2) a BIG P.

4P  3yP

Replace each (x+2) with P.


Now this is an easy problem to factor, that is, if you know common factoring.

= P(4  3y)

The 2 terms have a BIG P in common.

becomes
(x + 2)(4  3y)

Now replace the P with its equal (x+2)

Now remember if the signs of the first two terms are not the same
(same order also) as the last two terms then that is your BIG SIGNAL to
common factor out a negative and change the signs of the last two terms.
Another similar, but different!, Example:
Are the signs of the first two terms similar to the last two terms in this one?
5xv  20v  2x + 8
No. The first two terms: positive and then a negative.
The last two terms: negative and then a positive.
This is our BIG SIGNAL to factor out a negative from the last two terms and that will change their signs.
5xv  20v  2x + 8 = 5v(x  4)  2(x  4) See the 2 factor.
See the signs inside the parenthesis.
The x is now positive and the 4 is negative.
Just to check it, does 2(x  4) = 2x + 8 ?
So here are the steps for you to write in your notebook:
5vx  20v  2x+ 8=
5v(x 4)  2(x 4)= NOTE: x 4 is inside both parenthesis
5vP2P=
P(5v 2) Replace the P.
( x 4)(5v2)

The next example requires more factoring after the P principle factoring is completed.
x^{3}+8x^{2} 3x^{2} 24x
= ????
= x^{2}(x+8)  3x(x+8) .
You finish then find answer below.
No peeking.
Write this one down and factor it completely.

Do this problem then look at the answer below.
What do the first two terms have in common??? x^{2}
What do the last two terms have in common???  3x
x^{3}+ 8x^{2} 3x^{2} 24x =
x^{2}(x+8)  3x(x+8) Think: x^{2}P 3xP = P(x^{2} 3x)
(x+8)(x^{2}3x) = But we need to look more at the x^{2}3x.
(x+8)x(x3)
or x(x+8)(x  3)

Do Assignment 3.2a. Look in the Section 1.
Can you common factor when the exponent is a variable too!
x^{2n} x^{n+1}+ x^{n} do you see all have x^{n} in common.
Factor x^{n }out!
BUT first do you recall that when multiplying we add the exponents.
Thus x^{n} times x^{n} is x^{2n} ( I added n+n).
So what is x^{n} times x. What are the exponents?
n and 1 so we get x^{n}^{+1}.

x^{2n} x^{n+1}+ x^{n} You may want to think x^{n}x^{n}^{ }  x^{n}x^{1}+ 1x^{n }.
factors to x^{n}(x^{n} x^{1}+ 1)
OR x^{n}(x^{n} x + 1)

You will need the above in coming lessons!
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