First, factor the numerator: .  

Now your expression looks like 

Second, cancel the "like" terms -  - which leaves us with .  

Third, solve for , which leaves you with . 

Because the  term has a coefficient, you begin by multiplying the  and the  terms () together: . 

Find the factors of  that when added together equal the second coefficient (the  term) of the polynomial: . 

There are four factors of : , and only two of those factors, , can be manipulated to equal  when added together and manipulated to equal  when multiplied together:  

Step 1: Subtract 30 from both sides of the equation in order to make the equation equal to 0.

Step 2: Factor out a 3.

Step 3: Factor the trinomial.

At this step you set both factors equal to 0 and solve for .

The method to factor is to find the roots of  of the polynomial in standard form.

The factors are:  

Select the set that will most likely add or subtract to achieve the coefficient of the middle term.

The set  will satisfy.  Since  is a positive number, the signs inside the binomials will be negative.

To factor this problem correctly you must recognize the similarity in all three terms. What is common to all three? They all have x yes. But is there a common factor amongst the constants? To factor correctly you factor out as much as you can from each term.

First factor out as much from each constant as you can. It is similar to dividing (at least that is a way to think about what you are doing). In essence you are just removing quantities.

Now factor out as many x's as you can from each term:

Now factor what is left over in the paranthesis. A way to think about this is what two x terms can I multiply to get the first x term and what two constants multiplied by each other would equal the third term and what sign must they have to add or subtract to get the middle term. You are just doing the reverse of the FOIL method.

See how the the x terms in the parentheses multiply to equal X^2? And how 2*1 gives the desired 2?

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:

=