Factor polynomial $$ 12n^{2}-17n-5 $$

$$ 12n^{2}-17n-5 = ( 3n-5 )( 4n+1 ) $$

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:

$$ a = 12 , b = -17 ~ \text{ and } ~ c = -5 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 12 }$ by the constant term $\color{blue}{c = -5} $.

$$ a \cdot c = -60 $$

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 $ 3x-20x $:

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

Step 7: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 12x^{2}+3x-20x-5 = 3x\left(4x+1\right) -5\left(4x+1\right) = \left(3x-5\right) \left(4x+1\right) $$

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