We can split this problem into two different parts.  The first part is to find the greatest common constant (the numbers our in front of each term), and the greatest common variable (the ).  The variable part is much easier; each term has an  in it, and the limiting term is the last one (it only has 1 ), so we can factor it out.  That leaves us with:

Looking at the numbers, we can see that each has  as a factor, so we can also factor that out:

At this point, there's nothing else that we can factor out, so the largest common factor (that we were able to remove from each term) was:

We can start by only worrying about the numerator, and grouping the first two and the last two terms together:

From here we can factor out a  from the first term, and a  from the second term:

We can now factor out the  from each term:

While we could continue to factor the  term, let's see what the problem looks like overall for now:

We can cancel the  from the numerator and the denominator to save some time.  If we were to factor everything out and then cancel, we would still get the same answer:

Check if the problem factors using the following property,

where c will be the sum of a and b, and d will be the product of them. We can prove this by expanding out the right hand side (Using FOIL!):

Therefore:     and:  

In our problem, 7 is the sum of a and b, and 12 is the product of a and b. What numbers for a and b can we use to satisfy our equation? Well, the numbers will have to be factors of 12, so we can start by writing these down: 1,12  2,6, and 3,4. Trying a few of these out we can clearly see that the proper numbers are 3, and 4, since they add up to 7! Now that you have a and b, plug these back into the original relationship to find:

Expand this out for yourself to see that the property holds true!

In order to factor, we need to find two numbers that add up to , and multiply together to get .  Because the signs of those numbers are both negative, one of the numbers we're looking for will be positive, and one will be negative. It helps if we first look at all of the pairs of the factors of :

We know that these pairs will multiply to get , we just need to narrow it down to the pair that, when added, will give .  We find that the  pair does that.  We plug those numbers into our answer:

The first thing we can do is factor a  out of each term:

Now we need to find two numbers that add up to , and when multiplied make .  We can look at the pairs of factors of :

Of these pairs, only one will equal  when added together, so that's the one we're looking for.  To finish, we plug those numbers into our function:

Using the AC method, A=8,B=-2 C=-6, therefore A*C=-48. So you must find 2 factors of -48 which multiply and add/subtract to get the B term which is -2. So going through the factors of -48 you come to the factors of -8 and +6, which multiply together to give you -48 and add together to give you -2.

Then you insert factors as follows:

Factor by grouping which is as follows:

Then factor

Then

To solve this, first find what numbers give a product of 20 and when added yield -12. 

 Since the latter is negative, both numbers are negative. 

    and   

Therefore 

is the solution and can be verified by distributing:

To simplify the factorization pull out an x term

Find two numbers that add to 16 and multiply to 60.

To begin, let's clear up the  from the quadratic term by multiplying the entire numerator and denominator by :

We can distribute the  through the numerator:

Let's factor the  out of the denominator now, just to clear the problem up a bit:

Now that we have the equation in a more recognizable form, we need to find two numbers that add up to , and multiply to get .  Let's look at the factor pairs of :

We know that these pairs multiply to get  (if we give one of the numbers a negative sign), so we need to find the pair that adds to .  Because the three is negative, we know that the larger number will have the negative sign.  We find that:

So our numbers are  and .  Now we can plug them back into our equation to get a final answer of:

The polynomial will be easier to factor with an x pulled out

Two numbers are needed that add to 15 and multiply to 56. Trial and error is show that those two numbers are 7 and 8.