Rachel,
When you solve problems like this factoring one, you are looking for factors (expressions) with whole number coefficients.
So here, you have what's called a cubic expression (the highest power on x is 3).
You solve a problem like this by working from both ends! Consider the general product:
(ax+b)(cx+d)(ex+f) which will give you obviously a cubic polynomial in general (though in certain circumstances some of the terms may be missing, if the factors match up just so).
What can you say about the cubic polynomial you get?
Definitely, the coefficient on the x^3 term of the product is going to be the product (a*c*e) (why? -- it's the "head end" of the three term multiplication).
And the "coefficient" on the x^0 term of the product (i.e. the constant +140) is going to be the product (b*d*f) (why? -- it's the "tail end" of the three term multiplication). If you don't see this immediately, stop now and multiply the product out, stage by stage, until you do see. It's very important that you do see, before you go on.)
So, if you know that the "6" is a product of a*c*e and that these are all integers (whole numbers), you really have only a few choices to try -- something is a 2, something is a 3, and something is a 1 (or maybe a 1, 1, and 6, possibly).
At the tail end, you have the 140 -- it could be 14 * 5 *2, or 7*5*4, or so on. Many more choices here -- factor it completely into 7*5*2*2. Now, obviously, since you have 4 values, and you need to combine them into three, there's a few ways of doing this: 7*4*5, 28*5*1, 35*2*2, etc. This end is going to be a little more trial and error. These combinations will give you the "tail end" of your factors.
Now, there's something else to notice: some of the interior terms are (-) signed. That can only be if some of the factors are differences. How many differences (i.e. b, d, or f are negative) can there be? Obviously, since b*d*f is positive, two are differences and one is a sum: (-)*(-)*(+) = + (make sure you see this, too).
From this point, you're ready to try synthetic division (or trial division, whatever you call it).
What's the first type of term to try? The easiest is the one with a = 1 -- make your calculations simple! So, you have to divide the original expression by a whole list of possible factors, with a=1 : (x+1), (x+2), (x+4), (x+5), (x+7), (x+10), (x+14), (x+20), (x+35), (x+70); then (x-1), (x-2), and so on to (x-70).
But, you don't need to try ALL of these! Right away, you will see from your long division terms that some of these are going WAY off from coming out "even", i.e. with no remainder at the end. A strategy might be to try one at each end of each of the positive and negative lists, and one in the middle. Remember, IF the expression is factorable with integer coefficients, one factor HAS to be from your list above. So, zero in on it, being clever as to what to try "next".
When you hit the factor, and it divides to give you a quadratic (fx^2 + gx +h) as the dividend, then you just repeat the whole analysis process above, with the new coefficients "f" at the head, and "h" at the tail. This round might go faster -- but remember that you will probably now have TWO different non-unity numbers to try for factors of f (2 and 3, but maybe could also be 1 and 6) -- but in exchange, you also KNOW that each of 2 and 3 are the leading coefficients for very simple terms: (2x +- n) and (3x +- m), so you may be able to start doing some of the cross multiplications in your head (you should try to work up to this anyway).
Altogether, *when you get up to speed*, with many problems worked, you should be able to crack a cubic like this in no more than 2 minutes, it really gets that easy!
