Before we do anything, we notice that both terms in the expression have a common factor of 4. Thus, we can factor it out, leaving us with: . We recognize that the expression inside the parentheses is a difference of squares, and factors as such: . Finally, we are done. 

We notice that the polynomial  has a greatest common factor (i.e. the biggest multiplicatve "part") of .

First , we put the terms in ascending order of degree:

Since the number  is the biggest number that can fit into all of the terms, we can pull it out front of the parentheses.

This results in  times a reducible quadratic factor. 

  is of a special class of polynomial in which  the middle term squared is equal to the last term. in this case, it can be factored into the form 

consider:

so, the final, simplest factorized form of the polynomial is:

Note:In the original form of this polynomial there are other common factors. One could factor out  or  but the greatest common factor is  because it reduces 

The first thing to notice is that each term in the polynomial has a common factor of . We can pull that out and see what we end up with.

Now, in the parentheses, we have a reducible quadratic factor. We see that it fits a special class of quadratic since half the middle number  squared  is equal to the last number. Hence, it can be factored as such:

Which is our answer because there are no more common terms and the contents of the parentheses can't be simplified anymore.

To check if this is true, we can expand, using FOIL, the answer to see if we get the original polynomial.

Which is our original polynomial.

The first thing we notice is that there's a common factor of  in both terms. We can factor it out in front.

Next, we notice that the polynomial in the parentheses can be factored, since it is the difference of two squared numbers:  and .  Hence it can be factored in the following way:

which is our answer.

To see that this is the right answer, we can re-expand what we came up with to see if the result is the original polynomial.

expanding the two linear binomial factors using FOIL:

the s cancel out and we're left with

Destributing the 3 out, we end up with the original polynomial.

We notice that this is the difference of two squared numbers:  and .

Hence, we can follow the rule that the difference of two perfect squares  is equal to

To see this a little better, we can FOIL out the answer:

the s cancel out and we're left with the original expression:

=

We look for the comon factors of each term in the polynomial (that means, the things that each term has in common).

It turns out that each term in the polinomial has a factor (multiplicative "part") of . In this case  could be a number or a variable but it doesn't matter in this case. To factor, we need to synthisize what's common with all the factors and put it in front of parentheses. Factoring like this helps to simplify equations and expressions down the road. 

Noticing that  is the common factor, we can take it outside of the parentheses, writing within what's left over after we take out the .

The polynomial  fits a special class of polynomials because the last number, , is the square of half the middle number, . i.e.  the middle number is the last number squared.

to see if it's right, we can expand it all over again:

usinig FOIL to expand the expression,

Which was our original polynomial.

We start with the polynomial

and, putting the summands in ascending order of degree, we have

We notice that the comon factor is  because it "fits" in each of the terms.

Notice

so, we factor  out and we have

.

 fits a common model of a special class of polynomial because the last number, , is half the middle term, squared (i.e. take  the middle number and square it and you have the last term in the polinomial.) This is very similar to the process of "completing the square" of a quadratic.

The goal is to factor out the greatest common factor to leave the polynomial in a much cleaner state. We notice that there is a factor of  and  (it could be that  is a number or a variable but, in this case, it doesn't matter). We can pull out  from all of the terms and put it in front.

Factoring out the 2b and leaving what's left inside of the parentheses, we get:

Note that this can't be simplified or factored anymore because there are no more common factors within the parentheses.