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## Get ready for Algebra 2

### Course: 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 $\left(x+2\right)\left(x+3\right)$‍

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 $x+2$ and width is $x+3$, and divide it into four sub-rectangles:
Now we find the area of each sub-rectangle by multiplying its width and height:
Now we know that this is the area of the entire rectangle, which is the expression we are looking for:
${x}^{2}+3x+2x+6$
We can combine the $x$-terms to get a standard trinomial:
${x}^{2}+5x+6$

### Second method: The distributive property

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

Problem 1.1
Expand and combine like terms.
$\left(x+3\right)\left(x+4\right)=$

## Example 2: Expanding $\left(x-4\right)\left(x+7\right)$‍

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 $4$.
Now we find the area of each sub-rectangle, keeping in mind that the height of the bottom-left rectangle is $-4$, not $4$.
This doesn't make a lot of sense when thinking about actual rectangles and areas, but it works out with the algebra.
Now we add the areas of all the sub-rectangles:
$\begin{array}{rl}& \phantom{=}{x}^{2}+7x+\left(-4x\right)+\left(-28\right)\\ \\ & ={x}^{2}+3x-28\end{array}$

### Second method: The distributive property

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

$\left(x-2\right)\left(x+5\right)=$