SOLUTION: The problem says factor: x^2-25 x^2-7x-12 I forgot how to factor these so maybe if you just explain it briefly I might remember,thank you.

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Question 90319: The problem says factor:
x^2-25
x^2-7x-12
I forgot how to factor these so maybe if you just explain it briefly I might remember,thank you.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Ah yes...factoring!!
1) Factor:
x%5E2+-+25 Look at each term and notice that they are both "perfect squares".
%28x%29%5E2+-+%285%29%5E2
This suggests that the factors will be of the form:
%28x+%2B+n%29%28x+-+n%29 which would get you x%5E2+-+n%5E2 when multiplied.
So, in this case, n = 5 so the factors are:
x%5E2-25+=+%28x%2B5%29%28x-5%29
2) Factor:
x%5E2-7x-12
The factors will have the form: %28x%2Bm%29%28x%2Bn%29 where:
m%2Bn+=+-7 and
m%2An+=+-12
The two numbers come to mind are 3 and 4 becuase:
3%2B4+=+7 and
3%2A4+=+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%5E2-7x-12 is prime! (Not factorable)
Now if your expression had been:
x%5E2-7x%2B12 then you could factor this as:
x%5E2-7x%2B12+=+%28x-3%29%28x-4%29
But it isn't, so you can't!