Question 31921
The goal in factoring trinomials/binomials, as you have in your problem, is to find the two binomials which, when multiplied together, will result in the given trinomial/binomial. Now not all trinomials/binomials are factorable, but in general, you can nearly always find two such binomials. The trick is to use the right technique for your particular trinomial/binomial.

In the first one {{{x^2-5x-14}}}, for example, you have have divided the constant term (14) by 2. Why did you do this?

Here's an approach you might try:
Factor:
{{{x^2-5x-14}}} You want to find two binomials of the form{{{(x+m)(x-n)}}} such that:
m X n = -14 and
m + n = -5

Using a bit of trial & error:
2 X (-7) = -14
(-2) X 7 = -14

2 + (-7) = -5 This works!
(-2) + 7 = 5 This doesn't work!

So from the above, m = 2 and n = -7

The factors are:
{{{x^2-5x-14 = (x+2)(x-7)}}}

For you second problem, notice that your binomial is "the difference of two squares". There is a formula for factoring these:

{{{A^2 - B^2 = (A-B)(A+B)}}}
To find the A and B in your binomial, you must take the square root of each term.
{{{A = sqrt(4x^2)}}} = 2x
{{{B = sqrt(36y^2)}}} = 6y
Now substitute these values for A (2x) and B (6y) into the formula:

{{{4x^2-36y^2 = (2x-6y)(2x+6y)}}}

I hope this helps!