Factor polynomial $$ 3x^{2}-20x+12 $$

$$ 3x^{2}-20x+12 = ( x-6 )( 3x-2 ) $$

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:

$$ a = 3 , b = -20 ~ \text{ and } ~ c = 12 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = 12} $.

$$ a \cdot c = 36 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 36 $ and add to $ b = -20 $.

Step 4: All pairs of numbers with a product of $ 36 $ are:

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

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -20 }$

PRODUCT = 36 and SUM = -20
1     36	-1     -36
2     18	-2     -18
3     12	-3     -12
4     9	-4     -9
6     6	-6     -6

Step 6: Replace middle term $ -20 x $ with $ -2x-18x $:

$$ 3x^{2}-20x+12 = 3x^{2}-2x-18x+12 $$

Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -6 $ out of the last two terms.

$$ 3x^{2}-2x-18x+12 = x\left(3x-2\right) -6\left(3x-2\right) = \left(x-6\right) \left(3x-2\right) $$

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