You need to use the sum of two cubes equation

To factor a polynomial that has a coefficient in front of the  term, follow the steps below;

1) Once the equation is in standard form () , multiply the  term by the  term

2) Find two factors of this term that give you the  term

3) Re-write the polynomial with the original  term expanded into the two factors

4) Factor by grouping

5) Distribute to check that the factorization is correct

You need to factor by grouping but the important step is to remember the difference of squares.

To factor this, use trial and error to see what works. Since we have a  as the leading coefficient, it's helpful to remember that there's only one way to get  . Same with  as our third term--there's only one way to get  . Use these facts as you try to factor. Remember that signs matter. Therefore, your answer is: .

The best method for factoring this polynomial is by grouping:

The expression is in quadratic form, so it can be factored as though it is quadratic.

The resulting expression can be factored further, as there are two difference of squares quadratic expressions.

In order to factor a polynomial, you have to first find the greatest common factor. In this case, each term has an x in it, so you can easily factor that out. Also, the greatest common factor for the three terms is 3. After you take both the 3 and the x out you are left with  

The next step is to figure out which two numbers multiply together to equal , but also add together to equal . The factors of 52 are . The only ones that add to equal are , and in order to equal a positive 52, and , both of the numbers need to be negative. 

We need two numbers that add to get , and multiply to get .  It looks like  and  would do the trick.  To double check our answer, we can use the quadratic formula.

Knowing our solutions are  and , we can insert them into an equation:

and set each equal to .

so:

Because of the leading  in the quadratic term of the function, it's hard to think of a clever combination of solutions, but we can just rely on the quadratic formula to help us out:

Setting each equation equal to  yields:

If we were to expand this function we would find that it's wrong by a factor of . To correct this, we multiply one of the terms by .  In this case we can multiply the  term by 2, which also serves to clean up the fraction.  This gives us an end result of:

We need to find two numbers that multiply to get , but we don't have to worry about them adding at all because there's a quadratic term and a linear term. If we check, we can see that the quadratic and linear terms multiply to get , which is very convenient for us.  Because the quadratic term is negative, we pair that sign with the linear term when factored: