Step 1 :
To factor $ 48x^{3}+112x^{2}-3x-7 $ we can use factoring by grouping:
Group $ \color{blue}{ 48x^{3} }$ with $ \color{blue}{ 112x^{2} }$ and $ \color{red}{ -3x }$ with $ \color{red}{ -7 }$ then factor each group.
$$ \begin{aligned} 48x^{3}+112x^{2}-3x-7 = ( \color{blue}{ 48x^{3}+112x^{2} } ) + ( \color{red}{ -3x-7 }) &= \\ &= \color{blue}{ 16x^{2}( 3x+7 )} + \color{red}{ -1( 3x+7 ) } = \\ &= (16x^{2}-1)(3x+7) \end{aligned} $$Step 2 :
Rewrite $ 16x^{2}-1 $ as:
$$ 16x^{2}-1 = (4x)^2 - (1)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 4x $ and $ II = 1 $ , we have:
$$ 16x^{2}-1 = (4x)^2 - (1)^2 = ( 4x-1 ) ( 4x+1 ) $$