Factor polynomial $$ 3x^{2}-19x-14 $$

$$ 3x^{2}-19x-14 = ( x-7 )( 3x+2 ) $$

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:

$$ a = 3 , b = -19 ~ \text{ and } ~ c = -14 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -14} $.

$$ a \cdot c = -42 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -42 $ and add to $ b = -19 $.

Step 4: All pairs of numbers with a product of $ -42 $ are:

PRODUCT = -42
-1     42	1     -42
-2     21	2     -21
-3     14	3     -14
-6     7	6     -7

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

PRODUCT = -42 and SUM = -19
-1     42	1     -42
-2     21	2     -21
-3     14	3     -14
-6     7	6     -7

Step 6: Replace middle term $ -19 x $ with $ 2x-21x $:

$$ 3x^{2}-19x-14 = 3x^{2}+2x-21x-14 $$

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

$$ 3x^{2}+2x-21x-14 = x\left(3x+2\right) -7\left(3x+2\right) = \left(x-7\right) \left(3x+2\right) $$

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