The simplest way to solve this problem is to recognize that it is a difference of two perfect squares. Recall that the factors of  are  and . Looking back at our original function, we can see that both of our terms are perfect squares -  is a perfect square because  and  is a perfect square because . Therefore, we can treat  as , meaning that , and  as , meaning that , resulting in the factored form . From here, we solve for the zeros like we would with any factoring problem, by setting the right side of our function equal to zero and solving for x. 

Remember that because if either of these two factors is equal to zero, the entire side is equal to zero (as anything multiplied by zero is still zero), we can evaluate each factor separately and solve for x. 

Therefore...

The first step in solving this problem is to factor as many  as possible from the original expression. This can be seen below.

Next, we can see that inside of the parentheses we have a typical polynomial in the quadratic form that can be factored just like any quadratic. This step is seen below.

This is our final answer, because there is nothing left to factor out. 

In order to factor this, we will need to determine the multiples of the first and last terms.

Using trial and error, and the rule of the FOIL method:

Only  and  will provide us the middle term since:

Substitute the correct values into the binomials.

The answer is:  

In order to factor this, we will first need to pull a negative 1 out as the common factor.

Factor the term in the parentheses.

The factors of the first term are:

The factors of the last term are: 

Using trial and error, we can determine that the factors  for the  term and  for the number 14 will work since the numbers can be manipulated to achieve the middle term.

Remember to add the negative sign in front of the binomials.

The answer is:  

First, we notice that there's an  in both the first two terms, and a  in the last two terms, so let's factor those out:

Now we can see a  in both terms, so let's factor that out as well:

The first thing we can do is factor a  out of both terms:

Now we can see that each term is the square of something simpler:

We can use our squared terms formula to solve:

1) In order to start factoring we first want to see if the expression is divisible by any common factor. This allows to greatly simplify factoring in expressions and equations where polynomials have large coefficients.

In this case 

is divisible by two, so it can be simplified to 

 When pulling out a number/factoring you should make sure that the  simplified expression is equivalent to the unsimplified expression by doing the reverse of the operation given previously. In this case the  can be simplified to  by simply distributing the two.

2) With  the next step is to look at how you can simplify the expression such that  when multiplied out will given . Thinking about this you want a number that can multiply to 15 (the constant) but add up to 2 (the coefficient of the x-term).  and  will provide you this, so you can substitute them into the expression where the blanks are.

3) The expression can now be simplified to , which is your final answer. The equivalency to its unfactored form can be checked through multiplying everything out and if the factored answer is correct it'll given the initial expression, otherwise you did something wrong in your operations and should double check each step.

The simplest way to solve this problem is to identify that it is a difference of two perfect squares. Recall that the factors of  are  and .

Looking back at our original function, we can see that both of our terms are perfect squares -  is a perfect square because  and  is a perfect square because .

Therefore, we can treat  as , meaning that , and  as , meaning that , resulting in the factored form 

.

The correct answer is . Because the constant is negative, the answer must contain terms with positive and negative constants, thus ruling out , which would result in a positive coefficient. The   answer choice can be ruled out because the factored out polynomial results in a , not a . Therefore, the correct answer choice is, which can be confirmed if multiplied out. 

This problem is an example of a difference of squares. The formula for such a problem is . In this problem  corresponds to  and  corresponds to . Therefore, the solution is .