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$4\left (x+y \right)=4x+4y $ because $4\left (x+y \right) =\left (x+y \right) +\left (x+y \right) +\left (x+y \right) +\left (x+y \right)$ , but why is $\left (x+y \right) \left (x+y \right) =xx+xy+yx+yy$?

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  • $\begingroup$ It is also the distributive law or rule. $\endgroup$ – Dr. Sonnhard Graubner Dec 14 '18 at 16:19
  • $\begingroup$ This is distribution done twice. $\endgroup$ – Randall Dec 14 '18 at 16:20
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Because:

$$\left (x+y \right) \left (x+y \right) = \underbrace{(x+y)+(x+y)+...(x+y)}_{x + y\text{ times}}$$

$$=\underbrace{(x+y)+(x+y)+...(x+y)}_{x\text{ times}}+\underbrace{(x+y)+(x+y)+...(x+y)}_{y\text{ times}}$$

$$=\underbrace{x}_{x\text{ times}}+\underbrace{y}_{x\text{ times}}+\underbrace{x}_{y\text{ times}}+\underbrace{y}_{y\text{ times}}$$

$$=xx+xy+yx+yy$$

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  • $\begingroup$ If $x+y$ isn't an integer, what does "$x+y$ times" mean then? The same for $x$ and $y$ separately. $\endgroup$ – StackTD Dec 14 '18 at 16:30
  • $\begingroup$ @StackTD Excellent question! Obviously one needs to do something else in that case ... but I figured this would be in the spirit of the OP $\endgroup$ – Bram28 Dec 14 '18 at 16:32
  • $\begingroup$ I'm not sure what OP means, but my comment is intended to make OP think about that too because even in his first example, $x$ and $y$ need not be integers (and the argument still holds, thanks to the 4!). $\endgroup$ – StackTD Dec 14 '18 at 16:33
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$4\left (x+y \right)=4x+4y $ because $4\left (x+y \right) =\left (x+y \right) +\left (x+y \right) +\left (x+y \right) +\left (x+y \right)$

This works because $4$ is a (positive) integer; but for non-integer factors you can't use the "repeated addition"-argument.

but why is $\left (x+y \right) \left (x+y \right) =xx+xy+yx+yy$?

The distributive law works even if $x$ and/or $y$ are not integers.

Keep one of the factors together in a first step, and apply distributivity twice: $$\begin{align} (\color{blue}{x}+\color{red}{y})(x+y) & =\color{blue}{x}(x+y)+\color{red}{y}(x+y) \\ & =\color{blue}{x}x+\color{blue}{x}y+\color{red}{y}x+\color{red}{y}y \end{align}$$

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Do it in steps.

You accept that $M(x+y) = Mx + My$

So replace $M$ with $(x+y)$ and you get:

$(x+y)(x+y) = M(x+y) =$

$Mx + My = $

$(x + y)x + (x+y)y$

Now distribute a second time: Replace $x$ with $A$ and $y$ with $B$ to get:

$(x+y)(x+y) = M(x+y) =$

$Mx + My = $

$(x + y)x + (x+y)y=$

$(x+y)A + (x+y)B=$

$xA + yA + xB + yB =$

$xx + yx + xy + yy =$

$x^2 + 2xy + y^2$.

Of you you don't have to, and you shouldn't, do all that replacement. You should do it directly.

$(x+y)(x+y)=$ we treat one of the $(x+y)$ as a single thing and distribute across the other $x + y$.

$(x+y)(x+y) = (x+y)x + (x+y)y=$.

Now we have to sums to distribute: $(x+y)x = xx + yx$ and $(x + y)y = xy + yy$. So puting them together:

$(x+y)(x+y) = (x+y)x + (x+y)y=xx + yx + xy + yy=$.

And then some clean up:

$=x^2 + 2xy + y^2$

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