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

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.

• So people can better help you, do you understand why $x(x+2)=x^2+2x$? Nov 3, 2014 at 16:49
• Sorry I should have included I just want to generalize this concept, instead of trying to attempt understand it explicitly. Nov 3, 2014 at 16:53
• In that case Vladimir's answer below should be useful. Nov 3, 2014 at 16:54
• Not a proof, but consider looking at what happens when $x$ is some specified number. For example, suppose $x=5.$ Then $x(x+2)(x-3)$ is the product of $5$ and $7$ and $2,$ which equals $70.$ However, $x^2 + 2x(x-3)$ is the sum of $5^2 = 25$ and $(2)(5)(2) = 20,$ which equals $45.$ Finally, $(x^2 + 2x)(x-3)$ is the product of $5^2 + (2)(5) = 25 + 10 = 35$ and $5-3 = 2,$ which equals $70.$ In particular, note that in the original version, it's $35$ that gets multiplied by $2,$ which the $(x^2+2x)(x-3)$ version maintains. Nov 3, 2014 at 17:02

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:

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

• This is exactly the answer to my question. Many thanks. Nov 3, 2014 at 16:57
• You're welcome. Nov 3, 2014 at 16:59
• The explanation doesn't work when $k$ and $p$ are not natural numbers. The better explanation would be using rectangle areas, like this picture shows (except, of course, you can use real numbers for the areas instead). Nov 3, 2014 at 18:38
• Do you have a counter example? Nov 3, 2014 at 18:43
• Use @mathh when replying so that I get a notification. And you can't add something $\pi$ times. But you can find the area of a rectangle with a side $\pi$. Nov 9, 2014 at 16:56

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.

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 $k(a+b)(c+d)=(ka+kb)(c+d)$:
$(x^2+2x)(x-3)$