Factor polynomial $$ n^{2}-18n+72 $$

$$ n^{2}-18n+72 = ( n-6 )( n-12 ) $$

Step 1: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = -18 } ~ \text{ and } ~ \color{red}{ c = 72 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -18 } $ and multiply to $ \color{red}{ 72 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 72 }$.

PRODUCT = 72
1     72	-1     -72
2     36	-2     -36
3     24	-3     -24
4     18	-4     -18
6     12	-6     -12
8     9	-8     -9

Step 3: Find out which pair sums up to $\color{blue}{ b = -18 }$

PRODUCT = 72 and SUM = -18
1     72	-1     -72
2     36	-2     -36
3     24	-3     -24
4     18	-4     -18
6     12	-6     -12
8     9	-8     -9

Step 4: Put -6 and -12 into placeholders to get factored form.

$$ \begin{aligned} n^{2}-18n+72 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ n^{2}-18n+72 & = (x -6)(x -12) \end{aligned} $$

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