Question 90319
Ah yes...factoring!!
1) Factor:
{{{x^2 - 25}}} Look at each term and notice that they are both "perfect squares".
{{{(x)^2 - (5)^2}}}
This suggests that the factors will be of the form:
{{{(x + n)(x - n)}}} which would get you {{{x^2 - n^2}}} when multiplied.
So, in this case, n = 5 so the factors are:
{{{x^2-25 = (x+5)(x-5)}}}

2) Factor:
{{{x^2-7x-12}}}
The factors will have the form: {{{(x+m)(x+n)}}} where:
{{{m+n = -7}}} and
{{{m*n = -12}}}
The two numbers come to mind are 3 and 4 becuase:
{{{3+4 = 7}}} and 
{{{3*4 = 12}}}
Now all you have to do is to make sure the signs of the two numbers, n and m, will give the correct signs (negative, in this case) for the sum and the product.
But, there is a problem!
To get a product of -12, the 3 and the 4 must have opposite signs, right?
But if the 3 and the 4 have opposite signs, then their sum can never be -7, can it?
So, scrap the 3 and the 4.
What other factors of 12 are there?
-1*12 or 1*-12
-2*6 or 2*-6
Looking at these factors, there's no way you are going to get a sum -7
So, you have to say that:
{{{x^2-7x-12}}} is prime! (Not factorable)
Now if your expression had been:
{{{x^2-7x+12}}} then you could factor this as:
{{{x^2-7x+12 = (x-3)(x-4)}}}
But it isn't, so you can't!