Factor polynomial $$ 20x^{2}+18x-18 $$

$$ 20x^{2}+18x-18 = 2( 5x-3 )( 2x+3 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 20x^{2}+18x-18 = 2 ( 10x^{2}+9x-9 ) $$

Step 2 :

Step 2: 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:

$$ a = 10 , b = 9 ~ \text{ and } ~ c = -9 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 10 }$ by the constant term $\color{blue}{c = -9} $.

$$ a \cdot c = -90 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -90 $ and add to $ b = 9 $.

Step 5: All pairs of numbers with a product of $ -90 $ are:

PRODUCT = -90
-1     90	1     -90
-2     45	2     -45
-3     30	3     -30
-5     18	5     -18
-6     15	6     -15
-9     10	9     -10

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 9 }$

PRODUCT = -90 and SUM = 9
-1     90	1     -90
-2     45	2     -45
-3     30	3     -30
-5     18	5     -18
-6     15	6     -15
-9     10	9     -10

Step 7: Replace middle term $ 9 x $ with $ 15x-6x $:

$$ 10x^{2}+9x-9 = 10x^{2}+15x-6x-9 $$

Step 8: Apply factoring by grouping. Factor $ 5x $ out of the first two terms and $ -3 $ out of the last two terms.

$$ 10x^{2}+15x-6x-9 = 5x\left(2x+3\right) -3\left(2x+3\right) = \left(5x-3\right) \left(2x+3\right) $$

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