Rewriting the equation as , we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is  and between the third and fourth terms, the GCF is 4. Thus, we obtain .   Setting each factor equal to zero, and solving for , we obtain  from the first factor and  from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept . 

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)

Add 8 to both sides to set the equation equal to 0:

To factor, find two integers that multiply to 24 and add to 10.  4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

Then factor by grouping:

Set each factor equal to 0 and solve:

and

First, find the factors of 90 and 315.  The greatest common factor is the largest factor shared by both of the numbers: 45.

Then, find the least common multiple of 5 and 15.  This will be the smallest number that can be divided by both 5 and 15: 15.

Finally, the greatest common factor (45) divided by the least common multiple (15) = 45 / 15 = 3.

The given expression is a special binomial, known as the "difference of squares". A difference of squares binomial has the given factorization: . Thus, we can rewrite  as  and it follows that 

The product of is .

For the equation , 

must equal and  must equal .

Thus  and must be and , making the answer  .

This is a quadratic equation. We can solve  for either by factoring or using the quadratic formula. Since this equation is factorable, I will present the factoring method here.

The factored form of our equation should be in the format .

To yield the first value in our original equation (),  and .

To yield the final term in our original equation (), we can set  and .

Now that the equation has been factored, we can evaluate . We set each factored term equal to zero and solve.

First factor the numerator. We need two numbers with a sum of 3 and a product of 2. The numbers 1 and 2 satisfy these conditions:

Now, look to see if there are any common factors that will cancel:

The  in the numerator and denominator cancel, leaving .

The general form for a factored expression of order 2 is 

, which, when FOILED, gives .

Comparing this generic expression to the one given in the probem, we can see that the  term should equal , and the  term should equal 2.

The values of  and  that satisfy the two equations are  and ,

so your factored expression is