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$${n^2 - 9n = 0}$$ How does that turn into ${n(n - 9) = 0}$?

Can someone, please, explain the logic behind this?

This arose in a problem involving the number of diagonals in a particular polygon:

$$\begin{align} \frac{n(n - 3)}{2} = 3n &\quad\to\quad n^2 - 3n = 6n \\ &\quad\to\quad n^2 - 9n = 0 \\[4pt] &\quad\to\quad n(n - 9) = 0 \\[4pt] &\quad\to\quad n = 0, 9 \end{align}$$

The problem is easy to solve. I just don't get how $n^2−9n=0$ expands to $n(n−9)=0$.

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It is verified by the distributive law. It is carried through originally by remembering the distributive law.

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Choose one vertex, $n$ choices, choose another one from the remaining ones which is not adjacent or equal, $n-3$ this gives $n^2-3n$. Now since you can do it in any order, you divide by $2$ to account for double counting giving

$$\#d = {1\over 2}n(n-3).$$

Not sure how you got the $9$. Clearly that's negative for a square, but there are $2$ diagonals for a square. Also not sure why you have the $=0$, are you sure this is how the problem was posed?

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    $\begingroup$ @W.Zlacki well if that's all, just note that $n(n-9) = n\cdot n -n\cdot 9$ by the distributive property, so $n(n-9)=n^2-9n$ is the same way of saying this without so many multiplication symbols. $\endgroup$ – Adam Hughes May 29 '16 at 22:57
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    $\begingroup$ You actually end up double counting the diagonals, so the final formula is $\frac{n(n-3)}{2}$ $\endgroup$ – Hrhm May 29 '16 at 23:01
  • $\begingroup$ I felt his answer was more suited to what I was looking for. $\endgroup$ – William Zlacki Jun 7 '16 at 3:08
  • $\begingroup$ d=1n!!!!!!!!!!! $\endgroup$ – William Zlacki Dec 3 '17 at 14:23

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