Factoring can be quite difficult for those who are new to the concept. There are many ways to go about it. The guess and check way seems to be the most common, and in my mind, it is the best, especially if one wants to go further into mathematics, than Calculus 1. But for those just getting through a required algebra course, here is another way to consider, that I picked up while tutoring some time ago:
If you have heard of factor by grouping, then this concept will make some sense to you. Let's use an example to demenstrate how to do this operation:
Ex| x2 + x - 2
With this guess and check method, we would use (x + 1)(x - 2) or (x + 2)(x - 1). When we "foil" this out, we see that the second choice is the correct factorization. But, instead of just using these guesses, why not have a concrete way to do this.
Let's redo the example, with another method.
Ex| x2 + x - 2
First notice the -2, the negative shows us that the only way to this is (x - ?)(x + ?)
Now we look for the factors of 2. The only factors are 1,2 and 2, 1.
(In another case in which there are more factors, the difference or addition of these two factor must equal the middle term. That determines the correct pair of factors.
Now notice the middle term is positive. That means the larger of the two factors we choose must be positive also.
So, we can now write this as:
x2 + 2x - 1x - 2
What we just did was replace the original "+x" with "2x - 1x". These two statements are equivalent, so this is fair game.
Now, we group the left and right sides together. Note that when they are grouped, the negative stays with the 1, as shown below:
(x2 + 2x) + (-1x - 2)
Now we factor out common terms from each:
x is common on the left, and -1 is common on the right, so we factor out each of these:
x(x + 2) - 1(x + 2)
Now we check to see if our terms in parenthesis are equal. Since they are, we can group them into one term, and and the outer terms together. These two expressions are then multiplied:
(x + -1)(x + 2)
(x - 1)(x + 2)
If we foil this out, we get:
x2 + x - 2
Which is our original problem. Therefore we have correctly factored this. And in so doing, we have learned a new way to factor that requires no guessing.