Step 1 :

After factoring out $ 2y^{2} $ we have:

$$ 6y^{4}+2y^{3}+20y^{2} = 2y^{2} ( 3y^{2}+y+10 ) $$

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:

$$ a = 3 , b = 1 ~ \text{ and } ~ c = 10 $$

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

$$ a \cdot c = 30 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 30 $ and add to $ b = 1 $.

Step 5: All pairs of numbers with a product of $ 30 $ are:

PRODUCT = 30
1     30	-1     -30
2     15	-2     -15
3     10	-3     -10
5     6	-5     -6

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

Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.

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