Multiplying Binomials at a Glance

This is the last type of multiplication that we're going to look at in this unit. The good news is that there's nothing new to learn here. All we're really doing is applying the distributive property twice.

The most important part of multiplying two binomials is to make sure that we multiply each term in the first factor by each term in the second. There are several operations that need to happen, and there are different ways to track each operation to make sure we get 'em all. We're going to look at three methods. They all do the same math and get the same answer, but in different orders, so we can pick one and stick with it.

Method 1: Distributing the first factor to both terms in the second factor

Let's say we want to multiply these two binomials together:

(x – 3)(2x + 1)

First we'll make things easier by changing the subtraction symbol to adding a negative number.

(x + -3)(2x + 1)

We need to distribute (x + -3) to both terms in the second binomial, to both 2x and 1. If we apply the distributive property once, it looks like this:

distribution arrows (x + -3)(2x +1)

We're multiplying the entire first binomial by 2x and by 1.

(x + -3)(2x) + (x + -3)(1)

Now we can run through the distributive property again on each term, like so:

(x)(2x) + (-3)(2x) + (x)(1) + (-3)(1) =
2x2 + (-6x) + x + (-3) =
2x2 – 6x + x – 3

Hey, look at that. We've got two terms with x's in them, so we can combine like terms.

2x2 – 6x + x – 3 =
2x2 – 5x – 3

Whew, there's our answer. Let's see if we get that same answer using the other methods.

Method 2: Distributing the second factor to both terms in the first factor

We're starting with the exact same problem.

(x – 3)(2x + 1) =
(x + -3)(2x + 1)

In this method, we first distribute x to the entire binomial (2x + 1), keeping the addition sign between them. Then we distribute the -3 to the entire binomial (2x + 1).

(x)(2x + 1) + (-3)(2x + 1)

Now we run through the distributive property a second time: distribute the x to both 2x and 1, and distribute -3 to both 2x and 1.

(x)(2x) + (x)(1) + (-3)(2x) + (-3)(1) =
2x2 + x + (-6x) + (-3) =
2x2 + x – 6x – 3

And once again, we combing those x terms to get our final answer.

2x2 + x – 6x – 3 =
2x2 – 5x – 3

Nice! We got the same answer as we did using the first method.

But wait, there's more. We've got another trick up our sleeves.

Method 3: Box Method

Create a two-by-two table. Place one factor on top, and the other on the side. It doesn't matter which goes where, since multiplication is commutative. Be sure to keep the subtraction and addition signs with the correct terms.

box method

Now, multiply each factor in the rows with each factor in the columns and write the products in the boxes.

box method 2

Add them all together.

2x2 + x – 6x – 3

Combine like terms.

2x2 – 5x – 3

And we're done! Yup: same answer again. Use whichever method you like best.

Example 1

Multiply (5y + 3x)(8y – 1).


Example 2

Multiply (x-5)(x+5).


Example 3

Multiply (7x^2 + 3)(7x^2 + 3).


Exercise 1

Multiply (2x - 8)(9x +4).


Exercise 2

Multiply (6a + b)(6a – b).


Exercise 3

A rectangular lawn 12 ft by 15 ft is going to be increased by a uniform amount (x) on each side. What will the new area be?


Exercise 4

Multiply -3y(x+6y)(3x - 4y).