Step 1 :

After factoring out $ 3 $ we have:

$$ 12n^{2}+60n+45 = 3 ( 4n^{2}+20n+15 ) $$

Step 2 :

Step 2: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 4 , b = 20 ~ \text{ and } ~ c = 15 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = 15} $.

$$ a \cdot c = 60 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 60 $ and add to $ b = 20 $.

Step 5: All pairs of numbers with a product of $ 60 $ are:

PRODUCT = 60
1     60	-1     -60
2     30	-2     -30
3     20	-3     -20
4     15	-4     -15
5     12	-5     -12
6     10	-6     -10

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 20 }$

Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ 20 }$, we conclude the polynomial cannot be factored.

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