# 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 $x(x+2)=x^2+2x$? – Git Gud Nov 3 '14 at 16:49
• Sorry I should have included I just want to generalize this concept, instead of trying to attempt understand it explicitly. – Bright Nov 3 '14 at 16:53
• In that case Vladimir's answer below should be useful. – Git Gud Nov 3 '14 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. – Dave L. Renfro Nov 3 '14 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. – Bright Nov 3 '14 at 16:57
• You're welcome. – Vladimir Vargas Nov 3 '14 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). – user26486 Nov 3 '14 at 18:38
• Do you have a counter example? – Vladimir Vargas Nov 3 '14 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 $\pi$. – user26486 Nov 9 '14 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 $k(a+b)(c+d)=(ka+kb)(c+d)$:
$(x^2+2x)(x-3)$