If the polynomial  $ƒ\left(x\right)=x^{2025}+x^{2023}-x^{2021}+x^5+1$ is divided by $x^3-x,$  then remainder is  $g\left(x\right).$ The value of $\sqrt{g\left(4\right)}+\sqrt{g\left(12\right)}+\sqrt{g\left(24\right)}$  __________.

$ƒ\left(x\right)=(x^3-x)q\left(x\right)+g\left(x\right)$                                here   $g\left(x\right)=a{x}^2+bx+c$

$x=0\Rightarrow 1=c$

$x=1\Rightarrow a+b+c=1$

$x=-1\Rightarrow a-b+c=-1$

By solving, we get  $a=0,b=2,c=1$
 $\therefore g\left(x\right)=2x+1$
 $\therefore g\left(12\right)=25.$