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## Get ready for Algebra 2

### Unit 1: Lesson 4

Multiplying binomials

# Warmup: Multiplying binomials

In this article we are going to get some initial practice with multiplying binomials, to prepare you for the Multiply binomials intro exercise.
If you don't know or remember the distributive property clearly enough, we recommend that you check out this lesson.

## Example 1: Expanding $(x+2)(x+3)$left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 3, right parenthesis

There are two ways we can think about this operation. Both are equally valid; you can use whichever you feel more comfortable with.

### First method: Area model

We imagine a rectangle whose height is x, plus, 2 and width is x, plus, 3, and divide it into four sub-rectangles:
An area model for a rectangle that has a height of x plus two and a width of x plus three. The rectangle is broken into four rectangles to isolate each term in the height and the width. The top left rectangle has a height of x and a width of x. The top right rectangle has a height of x and width of three. The bottom left rectangle has a height of two and a width of x. The bottom right rectangle has a height of two and a width of three.
Now we find the area of each sub-rectangle by multiplying its width and height:
An area model for a rectangle that has a height of x plus two and a width of x plus three. The rectangle is broken into four rectangles to isolate each term in the height and the width. The top left rectangle has a height of x and a width of x. The area of the top left rectangle is x squared. The top right rectangle has a height of x and width of three. The area of the top right rectangle is three x. The bottom left rectangle has a height of two and a width of x. The area of the bottom left rectangle is two x. The bottom right rectangle has a height of two and a width of three. The area of the bottom right rectangle is six.
Now we know that this is the area of the entire rectangle, which is the expression we are looking for:
start color #11accd, x, squared, end color #11accd, plus, start color #ed5fa6, 3, x, end color #ed5fa6, plus, start color #74cf70, 2, x, end color #74cf70, plus, start color #ff9c39, 6, end color #ff9c39
We can combine the x-terms to get a standard trinomial:
x, squared, plus, 5, x, plus, 6

### Second method: The distributive property

We can apply the distributive property twice to expand the expression:
\begin{aligned} &\phantom{=}\blueD{(x+2)}(x+3) \\\\ &=\blueD{(x+2)}x+\blueD{(x+2)}3 \\\\ &=\blueD x\cdot x+\blueD 2\cdot x+\blueD x\cdot 3+\blueD 2\cdot 3 \\\\ &=x^2+2x+3x+6 \\\\ &=x^2+5x+6 \end{aligned}
In any way, we reached the same result! left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 3, right parenthesis expanded is x, squared, plus, 5, x, plus, 6.

Problem 1.1
Expand and combine like terms.
left parenthesis, x, plus, 3, right parenthesis, left parenthesis, x, plus, 4, right parenthesis, equals

## Example 2: Expanding $(x-4)(x+7)$left parenthesis, x, minus, 4, right parenthesis, left parenthesis, x, plus, 7, right parenthesis

Why do we have another example? Well, multiplying binomials becomes a little more tricky when subtraction is involved. Let's see how it's done.

### First method: Area model

As always, we draw a rectangle. However, don't forget to put a minus sign on the 4.
An area model for a rectangle that has a height of x minus four and a width of x plus seven. The rectangle is broken into four rectangles to isolate each term in the height and the width. The top left rectangle has a height of x and a width of x. The top right rectangle has a height of x and width of seven. The bottom left rectangle has a height of negative four and a width of x. The bottom right rectangle has a height of negative four and a width of seven.
Now we find the area of each sub-rectangle, keeping in mind that the height of the bottom-left rectangle is minus, 4, not 4.
This doesn't make a lot of sense when thinking about actual rectangles and areas, but it works out with the algebra.
An area model for a rectangle that has a height of x minus four and a width of x plus seven. The rectangle is broken into four rectangles to isolate each term in the height and the width. The top left rectangle has a height of x and a width of x. The area of the top left rectangle is x squared. The top right rectangle has a height of x and width of seven. The area of the top right rectangle is seven x. The bottom left rectangle has a height of negative four and a width of x. The area of the bottom left rectangle is negative four x. The bottom right rectangle has a height of negative four and a width of seven. The area of the bottom right rectangle is negative twenty-eight.
Now we add the areas of all the sub-rectangles:
\begin{aligned} &\phantom{=}\blueD{x^2}+\maroonC{7x}+(\greenC{-4x})+(\goldC{-28}) \\\\ &=x^2+3x-28 \end{aligned}

### Second method: The distributive property

We can apply the distributive property twice, making sure to remember that minus sign!
\begin{aligned} &\phantom{=}\blueD{(x-4)}(x+7) \\\\ &=\blueD{(x-4)}x+\blueD{(x-4)}7 \\\\ &=\blueD x\cdot x+(\blueD{-4})\cdot x+\blueD x\cdot 7+(\blueD{-4})\cdot 7 \\\\ &=x^2-4x+7x-28 \\\\ &=x^2+3x-28 \end{aligned}