Step 1 :

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

$$ 3y^{4}-10y^{3}-8y^{2} = y^{2} ( 3y^{2}-10y-8 ) $$

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:

$$ a = 3 , b = -10 ~ \text{ and } ~ c = -8 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -8} $.

$$ a \cdot c = -24 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -24 $ and add to $ b = -10 $.

Step 5: All pairs of numbers with a product of $ -24 $ are:

PRODUCT = -24
-1     24	1     -24
-2     12	2     -12
-3     8	3     -8
-4     6	4     -6

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -10 }$

PRODUCT = -24 and SUM = -10
-1     24	1     -24
-2     12	2     -12
-3     8	3     -8
-4     6	4     -6

Step 7: Replace middle term $ -10 x $ with $ 2x-12x $:

$$ 3x^{2}-10x-8 = 3x^{2}+2x-12x-8 $$

Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -4 $ out of the last two terms.

$$ 3x^{2}+2x-12x-8 = x\left(3x+2\right) -4\left(3x+2\right) = \left(x-4\right) \left(3x+2\right) $$

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