Power of 4?

$3x^{4}-x^{3}-(24x^{2}+x-3)=0$

$\implies x^{3}(3x-1)-((3x-1)(8x+3))=0$

$\implies (3x-1)(x^3-8x-3)=0$

$\implies (3x-1)((x-3)(x^2+3x+1))=0$

$\implies x=1/3,3,\underbrace{\dfrac{-3\pm\sqrt 5}2}_\text{both negative}$

By rearranging the equation into $3x^{4}+3-x^{3}-x-24x^{2}=0$ and divide every term by $x^{2}$ , the equation is equivalent to
$3(x^{2}+\frac{1}{x^{2}})-(x+\frac{1}{x})-24=0$ ,
therefore by completing the square, we can get
$3(x+\frac{1}{x})^2-(x+\frac{1}{x})-30=0$ .
Let $x+\frac{1}{x} = y$ , then by solving the equation
$3y^{2}-y-30=0$
$y=\frac{10}{3}$ or $y=-3$
$x+\frac{1}{x}=\frac{10}{3}$ or $x+\frac{1}{x}=-3$
Rejecting the other three possible values, $x=\boxed{3}$

<A quick but not so complete solution>

Denote the given function as $f(x)$ . Note that $f(0)=3>0$ and $f(1)=-20<0$ , so there exists a real root between $(0,1)$ . Again note that $f(2)<0$ and $f(4)>0$ , so there exists a root between $(2,4)$ . Since we are searching for a positive integer root, plugging in $x=3$ we get $f(3)=0$ .