# 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 4

*x*^{2}(6*x*^{2}*–*2). First, multiply 4*x*^{2}by 6*x*^{2}, which gives you 24*x*^{4}. Now multiply 4*x*^{2}by –2, which gives you –8*x*^{2}. Put it all together and you have 24*x*^{4}– 8*x*^{2}.**To multiple two terms by two other terms, use FOIL.**The letters of FOIL stand for**F**irst,**O**uter,**I**nner,**L**ast. 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

**F**irst by multiplying*a*x*a,*which gives you*a*^{2}.Go to the

**O**uter limits and multiply*a*x –8, which gives you –8*a.*Work your way to the

**I**nner layer by multiplying –2 x*a,*which gives you –2*a.*Take the (almost)

**L**ast step and multiply –2 x –8, which gives you 16.Now put it together and you have

*a*^{2}*a*–2*a*+16.Combine like terms (–8

*a*– 2*a*) and you get –10*a.*Replace the separate terms (–8*a*and –2*a*) with –10*a.*There you go: Your answer is

*a*^{2}– 10*a*+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}+ 2*ab*+*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: (2

*a*+ 3)(*a*– 4)(A)

*a*^{2}–*a*– 12(B) 2

*a*^{2}– 11*a*– 12(C) 2

*a*^{2}– 5*a*– 12(D) 2

*a*^{2}–*a*– 12(E) 2

*a*^{2}+ 5*a*– 12The expression (

*x*+*y*)(2*x*– 3*y*) is equivalent to(A)

*x*^{2}– 3*y*^{2}(B)

*x*^{2}–*xy*– 3*y*^{2}(C) 2

*x*^{2}– 3*y*^{2}(D) 2

*x*^{2}–*xy*– 3*y*^{2}(E) 2

*x*^{2}+*xy*– 3*y*^{2}

Now check your answers:

**C.**2*a*^{2}– 5*a*– 12FOIL! First: (2

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

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