Factoring Questions on the PSAT/NMSQT
When you factor, you break down an expression into the original terms that were (or could have been) multiplied to get that expression. When you take the PSAT/NMSQT, the factoring you have to do is pretty basic. You’ll likely see an expression with three terms that were calculated using FOIL. For example, you may have to factor x^{2} + 10x + 24 or something similar.
Here are the steps for factoring the example (x^{2} + 10x + 24):

Draw two sets of parentheses. Now you have a “home” ready for two factors.

Look at the first term and decide how you get there. The first term is x^{2}, so you must have multiplied x by x. Place those values in the First spot in each parenthesis:
(x ) (x ).

Examine the last term and decide how you get there. The last term is 24. To arrive at 24, you may have multiplied 24 by 1, 12 by 2, 8 by 3, or 6 by 4. Even two negative numbers work, because when you multiply two negatives, you get a positive number. Because you can’t tell, for the moment leave the Last spot in each parenthesis open.

Look at the middle term. The numbers that you’re looking for to put into the parentheses along with the x’s need to add up to the value of the middle term. Of the aforementioned pairs, 24 and 1 give you 25 (too big), 12 and 2 give you 14 (too big), 8 and 3 give you 11 (still too big), and 6 and 4 give you 10 (perfect!).
Now you know that you need 6 and 4 in the spots: x^{2} + 10x + 24 = (x + 6)(x + 4). Keep in mind that if the numbers in the parentheses are negative, adding up may actually mean subtracting!
Just for practice, try this question:

The expression x^{2} – 2x – 15 is equivalent to
(A) x(x – 2)
(B) (x – 2)(x – 15)
(C) (x + 3)(x – 5)
(D) (x – 3)(x + 5)
(E) (x + 15)(x – 1)
Now check your answer:

C. (x + 3)(x – 5)
You know that the first term is x^{2}, so you may as well set up your parentheses for antiFOILing with x’s in them: (x + ?)(x + ?). You know that the last term is –15, so the second numbers in the parentheses must multiply to –15. Your choices are 1 and –15, –1 and 15, 3 and –5, or –3 and 5.
Already, Choices (A) and (B) are out of contention. You can quickly check the outer and inner terms of the remaining choices to determine the answer or look at the middle term (–2x) and see that the two numbers you’re looking for must add to –2. The only numbers that fit that condition are 3 and –5, so Choice (C) is the answer.