FOIL: Multiplying Algebraic Terms on the PSAT/NMSQT
FOIL is a mnemonic (a memory aid) that helps you remember how to multiply in Algebra Land, which will help you on the PSAT/NMSQT. You’ll learn how to multiply multiple terms, with and without exponents. Before getting to FOIL, here’s some easy stuff:

To multiply two or more terms by one term, use the distributive property. What, you forgot the distributive property? Not to worry: It’s simple. Just multiply the single term by each of the terms in the parenthesis. Then recombine everything.
Here’s a sample: Imagine that you have to multiply 4x^{2}(6x^{2} – 2). First, multiply 4x^{2} by 6x^{2}, which gives you 24x^{4}. Now multiply 4x^{2} by –2, which gives you –8x^{2}. Put it all together and you have 24x^{4} – 8x^{2}.

To multiple two terms by two other terms, use FOIL. The letters of FOIL stand for First, Outer, Inner, Last. When you multiply two terms by two terms, you work in FOIL order. Take a look at this problem:
(a – 2) (a – 8)

Run for First by multiplying a x a, which gives you a^{2}.

Go to the Outer limits and multiply a x –8, which gives you –8a.

Work your way to the Inner layer by multiplying –2 x a, which gives you –2a.

Take the (almost) Last step and multiply –2 x –8, which gives you 16.

Now put it together and you have a^{2} – 8a –2a +16.

Combine like terms (–8a – 2a) and you get –10a. Replace the separate terms (–8a and –2a) with –10a.

There you go: Your answer is a^{2} – 10a +16.
The PSAT/NMSQT writers recommend that you memorize two FOIL problems that pop up all over the place. So memorize them!

(a + b) (a – b) = a^{2} – b^{2}. This shortcut works only when you’re multiplying terms that are exactly alike, except for their signs. You can use it for (b + 3) (b – 3), which equals b^{2} – 9. You can’t use it for (b + 3) (a – 15). This FOIL problem is known as the difference of two squares.

(a + b)^{2} = (a + b) (a + b) = a^{2} + 2ab + b^{2}. This is FOIL, plain and simple, already worked out for you. If you see a problem that looks like this, try backsolving for a and b.

See if you can FOIL all by yourself:

Simplify: (2a + 3)(a – 4)
(A) a^{2} – a – 12
(B) 2a^{2} – 11a – 12
(C) 2a^{2} – 5a – 12
(D) 2a^{2} – a – 12
(E) 2a^{2} + 5a – 12

The expression (x + y)(2x – 3y) is equivalent to
(A) x^{2} – 3y^{2}
(B) x^{2} – xy – 3y^{2}
(C) 2x^{2} – 3y^{2}
(D) 2x^{2} – xy – 3y^{2}
(E) 2x^{2} + xy – 3y^{2}
Now check your answers:

C. 2a^{2} – 5a – 12
FOIL! First: (2a)(a) = 2a^{2}. Outer: (2a)(–4) = –8a. Inner: (3)(a) = 3a. Last: (3)(–4) = –12. Add all those terms up and combine like terms: 2a^{2} – 8a + 3a – 12 = 2a^{2} – 5a –12, or Choice (C).

D. 2x^{2} – xy – 3y^{2}
FOIL again! First: (x)(2x) = 2x^{2}. Outer: (x)(–3y) = –3xy. Inner: (y)(2x) = 2xy. Last: (y)(–3y) = –3y^{2}. Now combine the terms: 2x^{2} – 3xy + 2xy – 3y^{2} = 2x^{2} – xy – 3y^{2}, or Choice (D).