Step 1 :

After factoring out $ 5y $ we have:

$$ 15y^{3}-70y^{2}-245y = 5y ( 3y^{2}-14y-49 ) $$

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:

$$ a = 3 , b = -14 ~ \text{ and } ~ c = -49 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -49} $.

$$ a \cdot c = -147 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -147 $ and add to $ b = -14 $.

Step 5: All pairs of numbers with a product of $ -147 $ are:

PRODUCT = -147
-1     147	1     -147
-3     49	3     -49
-7     21	7     -21

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -14 }$

PRODUCT = -147 and SUM = -14
-1     147	1     -147
-3     49	3     -49
-7     21	7     -21

Step 7: Replace middle term $ -14 x $ with $ 7x-21x $:

$$ 3x^{2}-14x-49 = 3x^{2}+7x-21x-49 $$

Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -7 $ out of the last two terms.

$$ 3x^{2}+7x-21x-49 = x\left(3x+7\right) -7\left(3x+7\right) = \left(x-7\right) \left(3x+7\right) $$

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