you can't just add 24 and expect the zeros to remain the same.

What you can do is scale the entire thing

$A(x-2)(x-1)(x+2)$

Now find A so that the above expression is 24 when x=0.

$A(x-2)(x-1)(x+2)|_{x=0}=24$

$A(-2)(-1)(2)=24$

$4A=24$

$A=6$

so

$6(x-2)(x-1)(x+2)$ has the properties you want.