Step 1 :

After factoring out $ 3 $ we have:

$$ 9x^{2}-21x+12 = 3 ( 3x^{2}-7x+4 ) $$

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:

$$ a = 3 , b = -7 ~ \text{ and } ~ c = 4 $$

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

$$ a \cdot c = 12 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 12 $ and add to $ b = -7 $.

Step 5: All pairs of numbers with a product of $ 12 $ are:

PRODUCT = 12
1     12	-1     -12
2     6	-2     -6
3     4	-3     -4

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -7 }$

PRODUCT = 12 and SUM = -7
1     12	-1     -12
2     6	-2     -6
3     4	-3     -4

Step 7: Replace middle term $ -7 x $ with $ -3x-4x $:

$$ 3x^{2}-7x+4 = 3x^{2}-3x-4x+4 $$

Step 8: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -4 $ out of the last two terms.

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

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