Factor polynomial $$ 3n^{2}-17n+20 $$

$$ 3n^{2}-17n+20 = ( n-4 )( 3n-5 ) $$

Step 1: 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:

$$ a = 3 , b = -17 ~ \text{ and } ~ c = 20 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = 20} $.

$$ a \cdot c = 60 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 60 $ and add to $ b = -17 $.

Step 4: 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 5: Find out which factor pair sums up to $\color{blue}{ b = -17 }$

PRODUCT = 60 and SUM = -17
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: Replace middle term $ -17 x $ with $ -5x-12x $:

$$ 3x^{2}-17x+20 = 3x^{2}-5x-12x+20 $$

Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -4 $ out of the last two terms.

$$ 3x^{2}-5x-12x+20 = x\left(3x-5\right) -4\left(3x-5\right) = \left(x-4\right) \left(3x-5\right) $$

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