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

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

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:

$$ a = 2 , b = -9 ~ \text{ and } ~ c = -18 $$

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

$$ a \cdot c = -36 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -36 $ and add to $ b = -9 $.

Step 4: All pairs of numbers with a product of $ -36 $ are:

PRODUCT = -36
-1     36	1     -36
-2     18	2     -18
-3     12	3     -12
-4     9	4     -9
-6     6	6     -6

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

PRODUCT = -36 and SUM = -9
-1     36	1     -36
-2     18	2     -18
-3     12	3     -12
-4     9	4     -9
-6     6	6     -6

Step 6: Replace middle term $ -9 x $ with $ 3x-12x $:

$$ 2x^{2}-9x-18 = 2x^{2}+3x-12x-18 $$

Step 7: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -6 $ out of the last two terms.

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

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