# How does this make sense?

Just today I encountered a problem at brilliant.org:

As we saw in the last problem, when multiplying sums and differences, each term in each sum or difference must be multiplied by every term in the other. Geometrically, this can be thought of as finding the areas of all of the sections of a rectangle and summing them to find the total area. The area of the square below is the sum of the areas of the blue (9), yellow ($$x^2$$), and green ($$3x$$) rectangles:

It then shows me something like this:

\begin{align} &(3 + x)^2\\ =& (3 + x)(3+x)\\ =& 9 + 3x + 3x + x^2\\ =& 9 + 6x + x^2 \end{align}

My question is:

1. Why is the $$3x + 3x$$ (line 3) there? How did it appear? It's possible to visualize it thanks to the rectangle in the picture above. But what if I had only $$(3 + x)^2$$ (the equation) and not the rectangle? How do I know why the $$3x + 3x$$ can be found in $$(3 + x)^2$$?

2. As far as I can tell, $$(3 + x)(3 + x)$$ can be written as $$9(x^2)$$. Is that correct?

3. ~~(irrelevent) Can this question be tagged "abstract-algebra"?~~

• Regarding your third question, no this is not an abstract algebra question. You can read the tag description here. Dec 12, 2018 at 2:18
• @André3000 okay. I was about to change the tag when you changed it. Thanks. :) Dec 12, 2018 at 2:19

1. It's called the distributive property. $$(3 + x)(3 + x) = 3(3 + x) + x(3 + x) = 9 + 3x + 3x + x^2$$

Referring to such a rectangle can be helpful to visualize where all of the terms in this last expression come from, but it's usually not used once you know how to multiply.

1. No. Try this with $$x = 5$$.

2. It can be tagged thus, but it shouldn't. Abstract algebra comes way after this (pre-algebra).

• actually. I just took a closer look. It might be annoying (after saying "got it"), but why is $3(3 + x) + x (3 + x)$ neccessary? Dec 12, 2018 at 2:09
• $(3 + x)^2$ does not equal $9x^2$. It's necessary because anything else you do will lead you to wrong conclusions! Dec 12, 2018 at 2:12
• oh. I just visualized 3(3+x)+x(3+x) with the help of the rectangle (again). But I guess it's more or less my problem now. Thanks! (btw, your answer to no 2. should be $x = 2$) Dec 12, 2018 at 2:17
• @JeffWhite: $x=5$ is one of the counterexamples showing that $(3+x)(3+x)\neq 9x^2$.
– user587192
Dec 12, 2018 at 2:19
• @user587192 $x$ seems to be more suitable being 2 according to the picture. but yes, you're right. Dec 12, 2018 at 2:21

What is multiplication?

Forget for a moment repeated addition, and areas and scaling factors...

Multiplication is a operation that distribute over addition!

It has a few basic rules:

$$a\cdot 0 = 0\cdot a = 0\\ a\cdot 1 = 1\cdot a = a\\ a(b+1) = ab + a$$

That is the definition!

Now remembering that $$b + 1 = \underbrace {1+1+1+\cdots}_{b \text { times}} + 1$$

$$a(b+1) = a(\underbrace {1+1+1+\cdots}_{b+1 \text{ times}}) = (\underbrace {a+a+a+\cdots}_{b+1 \text{ times}})$$

It gets to the same place as repeated addition.

We might also write an example of the distributive property as

$$a(b+c) = ab + ac$$

and multipication distributes right as well as left

$$(b+c)a = ba + ca$$

And if we have two sets of brackets we can choose the order to distribute

$$(x+3)(x-2) = x(x-2) + 3(x-2) = x\cdot x - x\cdot 2 + 3\cdot x - 3\cdot 2$$

or in the example above

$$(x+3)(x+3) = x(x+3) + 3(x+3) = x\cdot x + x\cdot 3 + 3\cdot x + 3\cdot 3$$

Now we can also use multiplication to represent areas.

That is the nice picture you show.

You have a square with side each of length $$x + 3$$

and it creates 2 squares, and 2 rectangles.

One square has area $$x^2$$, one has area $$3^2$$ and the rectangles each have area $$3x$$ (and there are 2 of them.)

The total area is $$x^2 + 2\cdot 3x + 3^2$$

regarding assertion 2)

"As far as I can tell $$(x+3)(x+3) = 9x^2$$"

You can always plug numbers. let $$x = 1$$

$$(1+3)(1+3) = 16 \ne 9(1^2)$$

$$(1+3)(1+3) = 1^2 + 3 + 3 + 3^2 = 16$$

$$(3x)(3x) = 9x^2$$

3) This is not abstract algebra. Abstract algebra discusses groups, rings, fields, and vectors spaces. It doesn't matter if you don't know what those are, if you are not using those terms, it probably isn't abstract algebra.

Hope this helps.

• Thank you for the answer. It did really help. Thank you once again! Dec 12, 2018 at 6:10