Factor polynomial $$ r^{2}+21r+108 $$

$$ r^{2}+21r+108 = ( r+9 )( r+12 ) $$

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:

$$ \color{blue}{ b = 21 } ~ \text{ and } ~ \color{red}{ c = 108 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 21 } $ and multiply to $ \color{red}{ 108 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 108 }$.

PRODUCT = 108
1     108	-1     -108
2     54	-2     -54
3     36	-3     -36
4     27	-4     -27
6     18	-6     -18
9     12	-9     -12

Step 3: Find out which pair sums up to $\color{blue}{ b = 21 }$

PRODUCT = 108 and SUM = 21
1     108	-1     -108
2     54	-2     -54
3     36	-3     -36
4     27	-4     -27
6     18	-6     -18
9     12	-9     -12

Step 4: Put 9 and 12 into placeholders to get factored form.

$$ \begin{aligned} r^{2}+21r+108 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ r^{2}+21r+108 & = (x + 9)(x + 12) \end{aligned} $$

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