Step 1 :

To factor $ 4y^{3}-7y^{2}-16y+28 $ we can use factoring by grouping:

Group $ \color{blue}{ 4x^{3} }$ with $ \color{blue}{ -7x^{2} }$ and $ \color{red}{ -16x }$ with $ \color{red}{ 28 }$ then factor each group.

$$ \begin{aligned} 4y^{3}-7y^{2}-16y+28 = ( \color{blue}{ 4x^{3}-7x^{2} } ) + ( \color{red}{ -16x+28 }) &= \\ &= \color{blue}{ x^{2}( 4x-7 )} + \color{red}{ -4( 4x-7 ) } = \\ &= (x^{2}-4)(4x-7) \end{aligned} $$

Step 2 :

Rewrite $ y^{2}-4 $ as:

$$ y^{2}-4 = (y)^2 - (2)^2 $$

Now we can apply the difference of squares formula.

$$ I^2 - II^2 = (I - II)(I + II) $$

After putting $ I = y $ and $ II = 2 $ , we have:

$$ y^{2}-4 = (y)^2 - (2)^2 = ( y-2 ) ( y+2 ) $$

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