Factor polynomial $$ 14y^{2}-9y-18 $$

$$ 14y^{2}-9y-18 = ( 2y-3 )( 7y+6 ) $$

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:

$$ a = 14 , b = -9 ~ \text{ and } ~ c = -18 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 14 }$ by the constant term $\color{blue}{c = -18} $.

$$ a \cdot c = -252 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -252 $ and add to $ b = -9 $.

Step 4: All pairs of numbers with a product of $ -252 $ are:

PRODUCT = -252
-1     252	1     -252
-2     126	2     -126
-3     84	3     -84
-4     63	4     -63
-6     42	6     -42
-7     36	7     -36
-9     28	9     -28
-12     21	12     -21
-14     18	14     -18

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -9 }$

PRODUCT = -252 and SUM = -9
-1     252	1     -252
-2     126	2     -126
-3     84	3     -84
-4     63	4     -63
-6     42	6     -42
-7     36	7     -36
-9     28	9     -28
-12     21	12     -21
-14     18	14     -18

Step 6: Replace middle term $ -9 x $ with $ 12x-21x $:

$$ 14x^{2}-9x-18 = 14x^{2}+12x-21x-18 $$

Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -3 $ out of the last two terms.

$$ 14x^{2}+12x-21x-18 = 2x\left(7x+6\right) -3\left(7x+6\right) = \left(2x-3\right) \left(7x+6\right) $$

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