Lesson Box Method

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This lesson will go over the box method. This is a visual way to multiply polynomials.

Let's say we wanted to multiply (x+2) with (x+5)
The FOIL method says:
(x+2)(x+5)
x*x + x*5 + 2*x + 2*5
x^2+5x+2x+10
x^2+7x+10

Therefore, (x+2)(x+5) = x^2+7x+10

Alternatively, we can use the distributive rule.
Let y = x+2
So,
(x+2)(x+5)
y(x+5)
yx+y*5
xy+5y
x(y)+5(y)
x(x+2)+5(x+2)
x^2+2x+5x+10
x^2+7x+10
As you can see, the y term allows us to fairly easily distribute. Then we replace y with x+2 to do two more sets of distributions.

Those examples above are motivation for this lesson. The box method is a visual way to multiply the various terms, and keep things organized.
Once again, we want to multiply (x+2) and (x+5)
We'll set up a two by two table like this

	x	2
x
5

The terms of x+2 and x+5 are placed along the top row and left most column.

Each inner cell is the result of multiplying the corresponding row and column headers
Example: the upper left corner we have x times x = x*x = x^2
Another example: The bottom right corner will have 2*5 = 10
This is what the filled out table looks like

	x	2
x	x^2	2x
5	5x	10

the four terms inside are x^2, 2x, 5x, 10
They add to x^2+2x+5x+10 = x^2+7x+10

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Another problem where the box method is useful.

Let's say we are tasked with finding (2x+3y)(7x-10y)

This is the set up of the box method table needed.

	2x	3y
7x
-10y

and this is the filled in table

	2x	3y
7x	14x^2	21xy
-10y	-20xy	-30y^2

Example: 14x^2 is in the upper left corner because 2x times 7x = 14x^2

Therefore,
(2x+3y)(7x-10y) = 14x^2+21xy-20xy-30y^2 = 14x^2+xy-30y^2

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One last example: Expand (a+b)^3

Recall that x^3 = x*x^2
So (a+b)^3 = (a+b)(a+b)^2

Table for (a+b)^2

	a	b
a	a^2	ab
b	ab	b^2

Showing that (a+b)^2 = a^2+ab+ab+b^2 = a^2+2ab+b^2

This means (a+b)(a+b)^2 is the same as (a+b)(a^2+2ab+b^2)
Now we use the box method again to multiply (a+b) with (a^2+2ab+b^2)

The filled out table looks like this

	a^2	2ab	b^2
a	a^3	2a^2b	ab^2
b	a^2b	2ab^2	b^3

Example: b^3 is in the bottom right corner because b*b^2 = b^3

Which leads to,
(a+b)(a^2+2ab+b^2) = a^3+2a^2b+ab^2+a^2b+2ab^2+b^3 = a^3+3a^2b+3ab^2+b^3

Ultimately,
(a+b)^3 = a^3+3a^2b+3ab^2+b^3

You can use the FOIL rule to confirm this, or the binomial theorem.

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