Factor polynomial $$ 6x^{2}-19x+14 $$

$$ 6x^{2}-19x+14 = ( x-2 )( 6x-7 ) $$

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:

$$ a = 6 , b = -19 ~ \text{ and } ~ c = 14 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 6 }$ by the constant term $\color{blue}{c = 14} $.

$$ a \cdot c = 84 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 84 $ and add to $ b = -19 $.

Step 4: All pairs of numbers with a product of $ 84 $ are:

PRODUCT = 84
1     84	-1     -84
2     42	-2     -42
3     28	-3     -28
4     21	-4     -21
6     14	-6     -14
7     12	-7     -12

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -19 }$

PRODUCT = 84 and SUM = -19
1     84	-1     -84
2     42	-2     -42
3     28	-3     -28
4     21	-4     -21
6     14	-6     -14
7     12	-7     -12

Step 6: Replace middle term $ -19 x $ with $ -7x-12x $:

$$ 6x^{2}-19x+14 = 6x^{2}-7x-12x+14 $$

Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -2 $ out of the last two terms.

$$ 6x^{2}-7x-12x+14 = x\left(6x-7\right) -2\left(6x-7\right) = \left(x-2\right) \left(6x-7\right) $$

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