Factor polynomial $$ 36y^{2}+114y-20 $$

$$ 36y^{2}+114y-20 = 2( 6y-1 )( 3y+10 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 36y^{2}+114y-20 = 2 ( 18y^{2}+57y-10 ) $$

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:

$$ a = 18 , b = 57 ~ \text{ and } ~ c = -10 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 18 }$ by the constant term $\color{blue}{c = -10} $.

$$ a \cdot c = -180 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -180 $ and add to $ b = 57 $.

Step 5: All pairs of numbers with a product of $ -180 $ are:

PRODUCT = -180
-1     180	1     -180
-2     90	2     -90
-3     60	3     -60
-4     45	4     -45
-5     36	5     -36
-6     30	6     -30
-9     20	9     -20
-10     18	10     -18
-12     15	12     -15

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 57 }$

PRODUCT = -180 and SUM = 57
-1     180	1     -180
-2     90	2     -90
-3     60	3     -60
-4     45	4     -45
-5     36	5     -36
-6     30	6     -30
-9     20	9     -20
-10     18	10     -18
-12     15	12     -15

Step 7: Replace middle term $ 57 x $ with $ 60x-3x $:

$$ 18x^{2}+57x-10 = 18x^{2}+60x-3x-10 $$

Step 8: Apply factoring by grouping. Factor $ 6x $ out of the first two terms and $ -1 $ out of the last two terms.

$$ 18x^{2}+60x-3x-10 = 6x\left(3x+10\right) -1\left(3x+10\right) = \left(6x-1\right) \left(3x+10\right) $$

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