# FOIL for Binomials

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Algebra One $\longrightarrow$
Polynomials $\longrightarrow$

Objectives
• Understand why multiplying two binomials yields four terms
• Perform binomial multiplication using the FOIL method
• Learn two basic FOIL shortcuts
Lesson Description

For a few reasons, multiplying two binomials together is the most common form of polynomial multiplication we will encounter in our math careers. For this reason, it has its own named method, and we have an interest in knowing what usual outcome to expect. This lesson instructs us on how to multiply binomials efficiently and correctly.

Practice Problems

Practice problems and worksheet coming soon!

## Multiplying Binomials

Now that we're a little more familiar with what a binomial is, we need to know how to multiply them together. We recently defined a binomial as an object that is made up of a sum of two things. We commonly consider a binomial to be a single object despite its composition, because as a whole unit, it acts like its own factor in many Algebra topics. In fact, more often than not, we will find ourselves in the business of multiplying and dividing binomials - sometimes we will find it convenient to leave it as is, essentially taking no action:$$(3x + 5) \cdot \left(x^2 + 1\right) \longrightarrow (3x + 5)\left(x^2 + 1\right)$$Other times, however, we will be interested in multiplying out the pieces and seeing what polynomial result we get:$$(x+2)(x-1) = x^2 + x - 2$$This lesson will show us how to multiply out the pieces and get that polynomial result using a systematic approach that will minimize mistakes, called the FOIL method.

## Understanding Binomial Multiplication

When you multiply two binomials together, you're really multiplying each term of the first binomial with each of the terms in the second binomial. This is because the distributive property applies two times when multiplying two binomials together. Let's look at a simple case of distributive property:$$a \cdot (x+y) = ax + ay$$(1)All we did here was apply the distributive property in a way we've seen in the past. But what if $a$ was some other object? What if, instead of multiplying $a$ with $(x+y)$, we were to multiply $(a+b)$ with $(x+y)$? All that should be different is that everywhere we had an $a$ before, we will instead have $(a+b)$:$$(a+b) \cdot (x+y) = (a+b) \cdot x + (a+b) \cdot y$$(2)We followed the same distributive property step as before, but this time we are not quite done. In the top example (equation 1), we perform the distributing and that's that. In the second case (equation 2), notice that we're left with two pieces that can each be further distributed. Let's go ahead and finish the second example by distributing in each term, and thus finish the job overall.$$(a+b) \cdot x + (a+b) \cdot y$$(2)$$= [ax + bx] + [ay + by]$$(3)...and that's as far as we can take it, since there are no "like terms" that can be combined. So let's see what happened again, start to finish.$$(a+b)(x+y) = ax + ay + bx + by$$(3) (Re-written)This is a very important result, because we use this concept frequently in algebra and beyond. But just because we need to do it frequently and accurately doesn't mean we have to memorize the result as a bunch of symbols. Yet we also don't need to step through the derivation of the process every time by using the distributive property twice, as we did above. Instead, when we need to multiply two binomials, we get to the answer quickly and accurately by using a time-tested shortcut called the FOIL method.

## How FOIL Works

Multiplying binomials ultimately means that each term in each binomial gets multiplied with each term in the other binomial exactly once. The FOIL method takes each combination of terms that needs to be multiplied together and steps through them one-by-one to make sure we quickly and correctly get the right answer. If you can remember FOIL, you can multiply binomials. FOIL is an acronym that stands for "First, Outer, Inner, Last".Take another look at the two binomials next to each other from equation (2) above, before we multiply:$$(a+b)(x+y)$$The words "First, Outer, Inner, Last" refer to pairing up the terms as follows: Now that we know what the FOIL acronym stands for, let's get to the good part: how to use it! We'll use our $(a+b)(x+y)$ example one last time to apply the FOIL method before we practice other examples.First multiply the "First" terms; in this case, $a$ and $x$ are the "First" terms, and they multiply to $ax$. Next, multiply the "Outer" terms $a$ and $y$ to get $ay$. Next the "Inner" terms $b$ and $x$ multiply to $bx$, and finally, the "Last" terms, $b$ and $y$, multiply to $by$. Add up the result, and just like before, we get$$(a+b)(x+y) = ax + ay + bx + by$$(3) RepeatedThe biggest challenge to learning this process is memorizing what we mean when we say the "First" terms, or the "Outer" terms, etc. Look above at when we first told you which pairs had which name and see why the names help you remember which pair is which. The "First" terms are named as such because they are the first ones to appear in each binomial. The "Last" are named similarly. Personally, I like to take a visual approach and think of "Inner" as the terms that are stuck in the middle, and "Outer" as the ones on the "edge". How you choose to remember what is referred to as "Inner" or "Outer" is up to you. However you decide to think about these, you'll be using the FOIL method flawlessly once you remember what each of the four words "First", "Outer", "Inner", and "Last" belongs to.
Pro Tip Let's face it - you're going to have to multiply binomials frequently in math from here on out. It's also one of those skills that teachers expect you to remember even if you haven't used it in a year, say when you're working on Pre-Calc or Calc topics. Understand and practice the FOIL method. It's fast, effective, and it will never let you down!

## Using FOIL in Practice

Let's take a look now at a more common example that has some numeric coefficients, not just all variables.

Example 1$$(x+2)(2x-1)$$$\blacktriangleright$ We're going to multiply using FOIL. Some very common things will happen here. First we'll get some common terms which need to be combined in our final answer. Second, since we are multiplying two linear binomials of the same variable, we will end up getting a quadratic result.$$(x+2)(2x-1) \longrightarrow$$$$\mathrm{First:} \,\, (x) \cdot (2x) \Rightarrow 2x^2$$$$\mathrm{Outer:} \,\, (x) \cdot (-1) \Rightarrow -x$$$$\mathrm{Inner:} \,\, (2) \cdot (2x) \Rightarrow 4x$$$$\mathrm{Last:} \,\, (2) \cdot (-1) \Rightarrow -2$$The answer is the sum of these pieces. We also need to combine the inner and outer terms, which are indeed like terms.$$(x+2)(2x-1) = 2x^2 -x + 4x - 2$$$$= 2x^2 + 3x - 2$$
Remember! Clean up your result by combining like terms, if possible. Failure to do this will not only lose you points, but also make your life more difficult in situations were this FOIL process will be done as part of a bigger overall process.
Let's take a look at two more examples.

Example 2$$(x-3)\left(x^2 +2x \right)$$$\blacktriangleright$ Let's apply FOIL just like we did in the last example.$$(x-3)\left(x^2 +2x \right) \longrightarrow$$$$\mathrm{First:} \,\, (x) \cdot \left(x^2\right) \Rightarrow x^3$$$$\mathrm{Outer:} \,\, (x) \cdot (2x) \Rightarrow 2x^2$$$$\mathrm{Inner:} \,\, (-3) \cdot \left(x^2\right) \Rightarrow -3x^2$$$$\mathrm{Last:} \,\, (-3) \cdot (2x) \Rightarrow -6x$$$$\therefore (x-3)\left(x^2 +2x \right) = x^3 + 2x^2 -3x^2 -6x$$$$= x^3 - x^2 - 6x$$
Pro Tip Pay attention to $+$ and $-$ signs. Because we often omit writing the plus sign for positive numbers, forgetting to include a negative sign is a common mistake for students who rush. Additionally, you have two opportunities to make a mistake - first when you look at each term during the part where you identify first, outer, inner, and last terms, and second when you multiply and combine the results. Stay sharp and always mind your minuses!

Example 3$$(4x-3y)(9x+8y)$$Try this one for yourself before you look at the answer!
Show solution
$$\blacktriangleright \,\, (4x-3y)(9x+8y) \longrightarrow$$$$\mathrm{First:} \,\, (4x) \cdot (9x) \Rightarrow 36x^2$$$$\mathrm{Outer:} \,\, (4x) \cdot (8y) \Rightarrow 32xy$$$$\mathrm{Inner:} \,\, (-3y) \cdot (9x) \Rightarrow -27xy$$$$\mathrm{Last:} \,\, (-3y) \cdot (8y) \Rightarrow -24y^2$$$$\therefore (4x-3y)(9x+8y) = 36x^2 + 32xy -27xy -24y^2$$$$= 36x^2 + 5xy - 24y^2$$
You Should Know When we learn concepts and algebra manipulations, we're commonly asked to see and practice on expressions with multiple variables. However, when we apply knowledge and techniques to situations where we're actually trying to somehow solve equations, we nearly always use $x$ and no other variables. That's just how it is among teachers and textbooks, so be ready to work problems in concept lessons (like this one!) with multiple variables but don't let it worry you - it usually doesn't make the process that you're learning extra challenging, it just sometimes looks intimidating.

## Special Forms and Shortcuts

There are two binomial multiplication situations that show up relatively frequently, and have special patterns to them. It's worth pointing these out to at least recognize them, though the erudite student will often want to memorize the patterns as a nice shortcut.Squared BinomialsWhen a binomial is multiplied by itself, a special pattern is always present. Let's see what happens when we multiply $(a + b)$ by itself.$$(a+b)(a+b)$$Applying the FOIL method:$$(a+b)(a+b) = a^2 + ab + ab + b^2$$$$= a^2 + 2ab + b^2$$In words, when you square a binomial, the result will be the sum of the first term squared, two times each term, and the second term squared. Again, shrewd students often apply this knowledge as a shortcut, but you can always resort to simply using the FOIL method. It only saves you one step of work, anyway.
You Should Know A squared binomial creates a trinomial that we call a "perfect square trinomial", due to the fact that it is a result of a squared object. Additionally, look for squared binomials to be given to you not only in the explicitly written $(a+b)(a+b)$ form, but also in the concise $(a+b)^2$ form.

Example 4$$(5x - 2y)^2$$
Show solution
$\blacktriangleright$ We can either FOIL this or apply the direct result shortcut. Either way, you should end up with$$(5x - 2y)^2 = 25x^2 -10xy + 4y^2$$This is another chance to point out the need to pay attention to $+$ and $-$ signs. Note that since the second term was negative, the middle term of the result (obtained by doing two times each term) is negative. Also note that the squared pieces are positive because squaring a term will make it positive regardless.
Warning! If you are given a squared binomial in the form $(a+b)^2$, I very, very, very strongly recommend you write it out as $(a+b)(a+b)$ before you multiply, at least to start while you're new to this. Otherwise, you're in danger of falling into one of the most notoriously common and wrong traps in the history of Algebra.Please write down the following fact 10 times somewhere immediately, please and thank you.$$(a+b)^2 \neq a^2 + b^2$$
Difference of SquaresAnother special form that we see a lot is called a "Difference of Squares", which we'll be looking at in its own lesson » soon when we want to operate this process in reverse (and yes / sorry, we will need to do that a lot too).A Difference of Squares occurs when you have two binomials multiplied together with the same first term and opposite second terms, such as$$(a+b)(a-b)$$The name Difference of Squares won't seem very appropriate until we multiply this result out. Proceeding with FOIL:$$(a+b)(a-b)$$$$= a^2 - \cancel{ab} + \cancel{ab} -b^2$$$$= a^2 -b^2$$In words, when you have a Difference of Squares multiplication form, the result will be the first term squared minus the second term squared.

Example 5$$(x+6)(x-6)$$
Show solution
$\blacktriangleright$ Recognizing the Difference of Squares form, we have$$(x+6)(x-6) = x^2 - 36$$
You Should Know Recognizing the Difference of Squares form will be very important when we study factoring in the near future, so it is valuable to you now if you can understand and recognize it, regardless of whether you use the step-saving shortcut.

## Put It To The Test

We'll be encountering this skill explicitly and implicitly throughout several topics in the future. Make sure you can handle anything they can throw at you!

Example 6$$(2x-7)(5x+3)$$
Show solution
$$\blacktriangleright \,\, (2x-7)(5x+3)$$$$=10x +6x -35x -21$$$$=10x -29x -21$$

Example 7$$(3x + 4y)(2x - 6y)$$
Show solution
$$\blacktriangleright \,\, (3x + 4y)(2x - 6y)$$$$=6x^2 -18xy + 8xy -24y^2$$$$=6x^2 -10xy -24y^2$$

Example 8$$\left(x^2 + 1\right)\left(3-x^2\right)$$
Show solution
$$\blacktriangleright \,\, \left(x^2 + 1\right)\left(3-x^2\right)$$$$=3x^2 -x^4 + 3 -x^2$$$$=-x^4 +2x^2 + 3$$

Example 9$$\left(3x^2 + 2y\right)\left(7x^2 - 3y\right)$$
Show solution
$$\blacktriangleright \,\, \left(3x^2 + 2y\right)\left(7x^2 - 3y\right)$$$$=21x^4 -9x^2 y + 14x^2 y -6y^2$$$$=21x^4 +5x^2 y -6y^2$$

Example 10$$(2x-1)(2x+1)$$
Show solution
$$\blacktriangleright \,\, (2x-1)(2x+1)$$$$=4x^2 + 2x -2x -1$$$$=4x^2 - 1$$This problem is most efficiently done by recognizing the Difference of Squares form.

Example 11$$(5x - 2y)(5x - 2y)$$
Show solution
$$\blacktriangleright \,\, (5x - 2y)(5x - 2y)$$$$=25x^2 -10xy -10xy +4y^2$$$$=25x^2 -20xy +4y^2$$This problem is most efficiently done by recognizing the squared binomial form.

Lesson Takeaways
• Understand why multiplying two binomials gives four pieces
• Execute the FOIL method correctly and consistently
• Clean up all answers and always double check that like terms were combined, if possible
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