Step 1 :
After factoring out $ x $ we have:
$$ 3x^{4}+5x^{3}-12x^{2}-20x = x ( 3x^{3}+5x^{2}-12x-20 ) $$Step 2 :
To factor $ 3x^{3}+5x^{2}-12x-20 $ we can use factoring by grouping:
Group $ \color{blue}{ 3x^{3} }$ with $ \color{blue}{ 5x^{2} }$ and $ \color{red}{ -12x }$ with $ \color{red}{ -20 }$ then factor each group.
$$ \begin{aligned} 3x^{3}+5x^{2}-12x-20 = ( \color{blue}{ 3x^{3}+5x^{2} } ) + ( \color{red}{ -12x-20 }) &= \\ &= \color{blue}{ x^{2}( 3x+5 )} + \color{red}{ -4( 3x+5 ) } = \\ &= (x^{2}-4)(3x+5) \end{aligned} $$Step 3 :
Rewrite $ x^{2}-4 $ as:
$$ x^{2}-4 = (x)^2 - (2)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = x $ and $ II = 2 $ , we have:
$$ x^{2}-4 = (x)^2 - (2)^2 = ( x-2 ) ( x+2 ) $$