Correct Answer: x(x+4)

Explanation:

First portion,

$x^{3}-16x$

=$x(x^{2}-16)$

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

=$x(x+4)(x-4)$

Second portion,

$2x^{3}+9x^{2} +4x$

=$x(2x^{2}+9x+4)$

=$x(2x^{2}+8x+1x+4)$

=x{2x(x+4)+1(x+4)}

=$x(x+4)(2x+1)$

Third portion

$2x^{3}+x^{2}-28x$

$= x (2x^{2} + x -28)$

$= x(2x^{2} + 8x -7x -28)$

= x{2x(x+4) – 7( x+4)}

=x(x+4)(2x-7)

Now, H.C.F = x(x+4)