Why is $x(x+2)(x-3)$ not $x^2+2x(x-3)$?

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How would you explain the principle why $x(x+2)(x-3)$ is not $x^2+2x(x-3)$ but $(x^2+2x)(x-3)$? This may involve the fundamentals of eliminating parenthesis.

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This is the distributive property for multiplication: $(a+b)(c+d)=a(c+d)+b(c+d)$. In your case:

$$k(a+b)(c+d)=(ka+kb)(c+d)=ka(c+d)+kb(c+d).$$

This comes from:

$$k(a+b) = \overbrace{(a+b)+(a+b)+\dotsc+(a+b)}^{k\text{-times}} = a+b+a+b+\dotsc+a+b=\overbrace{a+a+\dotsc+a}^{k\text{-times}}+\overbrace{b+b+\dotsc+b}^{k\text{-times}}=ka+kb.$$

Generalizing a bit more:

$$(k+p)(a+b) = \overbrace{(a+b)+(a+b)+\dotsc+(a+b)}^{(k+p)\text{-times}} = a+b+a+b+\dotsc+a+b=\overbrace{a+a+\dotsc+a}^{(k+p)\text{-times}}+\overbrace{b+b+\dotsc+b}^{(k+p)\text{-times}}=(k+p)a+(k+p)b.$$

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Here's my attempt to explain it in a very super simple way based on what you already know. Make sure you make yourself familiar with Vladimir's answer as well.

Consider...

$2*3*4=24$, or in other words, $6*4=24$ most of people understand it very well that the both are the same.

Now consider $3+3*4=15$ you know can't replace $2*3$, with $3+3$ in that equation. It breaks the equation if you do that.

Consider now $(3+3)*4=24$, hey this works.

Conclusion, the simplified multiplication of form $a+a+...+a$ within an equation has to be written within parenthesis. In order to preserve the consistency within the equation.

Answering why the parenthesis

How would you explain the principle why $x(x+2)(x−3)$ is not $x^2+2x(x-3)$ but $(x^2+2x)(x-3)$? (it's about the lack of parenthesis)

$x(x+2)(x−3)=\overbrace{((x+2)+(x+2)+...+(x+2))}^{x\text{-times (notice parenthesis)}}(x−3)$

or the same can be noted, as Vladimir explained $k(a+b)(c+d)=(ka+kb)(c+d)$:

$(x^2+2x)(x-3)$