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# Foil Method|Definition & Meaning

## Definition

The **FOIL Method** is used to calculate the **product** of binomials. The acronym **‘FOIL’** stands for **First, Outside, Inside,** and **Last** which refers to the arrangement of multiplying terms. We multiply the **first** element, then **the outside** element, then **the inside** element, then the **last** element, and then combine like terms for our answer.

The **general form** of the above explanation can be **depicted** below:

**FOIL** is an abbreviation for the classic method of multiplying the two **binomials,** therefore named **‘FOIL.’** The word FOIL represents the four terms of the product, as explained next.

**‘F’ for “first”** represents that the first terms of each binomial are multiplied together. **‘O’ for “outer”** means that the first element of the left binomial and the last term of the right side binomial are multiplied. **‘I’ for “inner”** represents that the last element of the left binomial and the first element of the right side binomial are **multiplied. **

Finally, **‘L’ for “last”** represents that the last terms of each binomial are multiplied.

## FOIL Formula

The **FOIL** Method utilizes the distributive **property** to expand the product of two **polynomials.** The acronym **FOIL** (First-Outside-Inside-Last) is derived from the **process** used to expand the two **binomials.**

The **concept of the FOIL method** has been described in **Figure** 2 for the following example:

(a+b)(c+d) = ac + ad + bc + bd

i.e. the **product** of binomials (a+b)(c+d) is the **sum** obtained by:

- Multiplying the
**First**terms. - Multiplying the
**Outer**terms. - Multiplying the
**Inner**terms. - Multiplying the
**Last**terms.

## Background Concepts

### Binomials

A **binomial** is an expression **having two numbers or variables**, sometimes called two terms, that are separated by a sign of addition or **subtraction. **

We know that a variable is a letter that represents **something.** Let’s take a look at this **example of a binomial, i.e., x + 10,** as shown in Figure 3.

Note that this is a **binomial** because this polynomial contains two terms that are **separated** by a plus sign where x is a variable and 10 is a constant. **Another example of a binomial is ‘x + 1,’** which also contains two terms that are separated by an addition sign. Now, if we need to multiply these two **binomials,** one easy way to **multiply** them is to use the **FOIL** method!

**Multiplying** three binomials like the above **pairs** is an extension of the case for standard FOIL. One needs to pair off two binomials at a time to perform multiplication.

**Distributivity**

In binary **operations** involved in mathematics, the **distributive** property generalizes the **distributive** law, which is represented by the **equality**

**a × (b + c) = (a × b) + (a × c**)

This **equality** is always true in elementary **algebra**. The variable or **number** that is outside the **brackets** is been multiplied by every element present **inside** the brackets, and later the answer of individual products are summed up**.**

## Proof of FOIL Method

To start with, let’s consider that the following **standalone** equation is given:

(a+b)(c+d) = ac+bd+bc+bd

and we want to multiply the binomials **appearing** on the left-hand side. The target is to **prove** that the left-hand expression of the **relation** can be **rearranged** to become equal to the right **side.** One of the first steps of this **proof** is to note that the above goal can be **rephrased.** It is similar to showing that the **right-hand** expression is similar to the **left-side** expression**.**

ac+bd+bc+bd = (a+b)(c+d)

The above **relation** is known as **flipped** expansion of the binomial equation.

Given the **left-hand** side of the equation, we can utilize multiple ways to rearrange the **variables.**

## Algebraic Context of FOIL Method

Before we **discuss** the details of algebraic **context,** let’s take a small detour to further **investigate** the hypotheses and properties of this **verification.** For now, we’ll consider our binomial **variables** to be the real numbers.

It **might** be obvious that we plan on utilizing certain properties. This query points out **another hypothesis** **because** our variables or numbers in binomial permits us to apply **additive** or multiplicative **axioms.** Regarding algebraic context, the concept of **‘field’** needs to be clarified. A ‘field’ represents a set that satisfies the following identities:

**Associative property of addition:**a + (b + c) = (a + b) + c**Associative property of multiplication:**a ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c**Commutative property of addition:**a + b = b + a**Commutative property of multiplication:**a ⋅ b = b ⋅ a**Additive property:**there exists a ‘0’ so that for any ‘a’ in the field, a + 0 = a**Multiplication property:**there exist a ‘1’ as an element so that for any ‘a’ in the field, a ⋅ 1 = a**Additive property of inverses:**for every ‘a’, there exists a ‘−a’, which is called the additive inverse so that a + (−a) = 0.**Inverse for Multiplication:**for every a ≠ 0, there exists a ‘1/a’, so that a ⋅ 1/a = 1.**Distributive property of multiplication and addition:**a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c).

All these** properties** seem very evident to us — we have understood these axioms from a young age. However, there lies the **assumption** that we are always working in a field. This is not always true.

Now, how does this **concept** of field apply to our problem? Well, we **presumed** that the numbers or variables considered binomial are real numbers. **Real numbers** are verified to be a field, therefore permitting us to take **benefit** of the above properties in our algebraic **manipulations.**

As **mentioned** in the above section, in multiplication, **distributivity,** and for addition, **commutativity** plays an **important** part in the process of **proof.** In fact, as far as the **binomial** elements belong to a set that satisfies these two **pre-requisite** properties, the **FOIL** relation will hold.

## Solved Example

### Example

Evaluate:(4x + 5)(3x + 2)

### Solution

As **FOIL** Formula says:

(4x + 5)(3x + 2) = (4x * 3x + 4x * 2) + (5 * 3x + 5 * 2)

(4x + 5)(3x + 2) = (12x^{2} + 8x + 15x + 10)

(4x + 5)(3x + 2) = (12x^{2} + 23x+ 10)

*All images/mathematical drawings were created with GeoGebra.*