Factor polynomial $$ 18j^{2}+9j-27 $$

$$ 18j^{2}+9j-27 = 9( j-1 )( 2j+3 ) $$

Step 1 :

After factoring out $ 9 $ we have:

$$ 18j^{2}+9j-27 = 9 ( 2j^{2}+j-3 ) $$

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:

$$ a = 2 , b = 1 ~ \text{ and } ~ c = -3 $$

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

$$ a \cdot c = -6 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -6 $ and add to $ b = 1 $.

Step 5: All pairs of numbers with a product of $ -6 $ are:

PRODUCT = -6
-1     6	1     -6
-2     3	2     -3

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 1 }$

PRODUCT = -6 and SUM = 1
-1     6	1     -6
-2     3	2     -3

Step 7: Replace middle term $ 1 x $ with $ 3x-2x $:

$$ 2x^{2}+x-3 = 2x^{2}+3x-2x-3 $$

Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -1 $ out of the last two terms.

$$ 2x^{2}+3x-2x-3 = x\left(2x+3\right) -1\left(2x+3\right) = \left(x-1\right) \left(2x+3\right) $$

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