Factor polynomial $$ 3q^{2}-11q+6 $$

$$ 3q^{2}-11q+6 = ( q-3 )( 3q-2 ) $$

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:

$$ a = 3 , b = -11 ~ \text{ and } ~ c = 6 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = 6} $.

$$ a \cdot c = 18 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 18 $ and add to $ b = -11 $.

Step 4: All pairs of numbers with a product of $ 18 $ are:

PRODUCT = 18
1     18	-1     -18
2     9	-2     -9
3     6	-3     -6

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -11 }$

PRODUCT = 18 and SUM = -11
1     18	-1     -18
2     9	-2     -9
3     6	-3     -6

Step 6: Replace middle term $ -11 x $ with $ -2x-9x $:

$$ 3x^{2}-11x+6 = 3x^{2}-2x-9x+6 $$

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

$$ 3x^{2}-2x-9x+6 = x\left(3x-2\right) -3\left(3x-2\right) = \left(x-3\right) \left(3x-2\right) $$

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