To factor trinomials like this one, we need to do a reverse FOIL. In other words, we need to find two binomials that multiply together to yield .

Finding the "first" terms is relatively easy; they need to multiply together to give us , and since  only has two factors, we know the terms must be  and . We now have         , and this is where it gets tricky.

The second terms must multiply together to give us , and they must also multiply with the first terms to give us a total result of . Many terms fit the first criterion. , ,  and  all multiply to yield . But the only way to also get the "" terms to sum to  is to use . It's just like a puzzle!

To find the greatest common factor, we must break each term into its prime factors:

The terms have , , and  in common; thus, the GCF is .

Pull this out of the expression to find the answer: . 

Use the -method to split the middle term into the sum of two terms whose coefficients have sum  and product . These two numbers can be found, using trial and error, to be  and .

and

Now we know that is equal to .

Factor by grouping.

First, we note that the coefficients have an LCD of 3, so we can factor that out:

We try to factor further by factoring quadratic trinomial . We are looking to factor it into two factors, where the question marks are to be replaced by two integers whose product is  and whose sum is . 

We need to look at the factor pairs of  in which the negative number has the greater absolute value, and see which one has sum :

None of these pairs have the desired sum, so  is prime.  is the complete factorization.

When working with a rational expression, you want to first put your monomials in standard format.

Re-order the bottom expression, so it is now reads . 

Then factor a  out of the expression, giving you .

 The new fraction is  .

Divide out the like term, , leaving , or .

First pull out 3u from both terms.

3u⁴ – 24uv³ = 3u(u³ – 8v³) = 3u[u³ – (2v)³]

This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember, but pat yourself on the back for getting to such hard questions! The difference of cubes formula is a³ – b³ = (a – b)(a² + ab + b²). In our problem, a = u and b = 2v:

3u⁴ – 24uv³ = 3u(u³ – 8v³) = 3u[u³ – (2v)³]

                = 3u(u – 2v)(u² + 2uv + 4v²)

This is a difference of squares. The difference of squares formula is a² – b² = (a + b)(a – b).

In this problem, a = 6x and b = 7y:

36x² – 49y² = (6x + 7y)(6x – 7y)

To factor, we are looking for two terms that multiply to give  and add together to get .

Possible factors of :

Based on these options, it is clear our factors are  and .

Our final answer will be:

To factor, we are looking for two terms that multiply to give  and add together to get .

Possible factors of :

Based on these options, it is clear our factors are  and .

Our final answer will be: