Of course by the size condition on $p$, all the constants $24, 5, 2011$ are invertible modulo $p$ so we really only need to prove that the polynomial $X^4-2X^2+9$ factors modulo $p$. Suppose this was not the case. Since $X^4-2X^2+9=(X^2-1)^2+8$ we find that $-8$ and hence $-2$ is not a quadratic residue modulo $p$. Moreover we have no solution to $(X^2+aX+b)(X^2-aX+b)=X^4-2X^2+9$ in $\mathbb{Z}_p[X]$ and hence no solution to $a^2-2b \equiv 2\mod p, b^2 \equiv 9\mod p$ and hence no solution to $a^2 \equiv 2\pm 6 \mod p$ whence neither $-4$ nor $8$ and hence neither $-1$ nor $2$ nor $-2$ is a quadratic residue which is of course impossible since the product of two non-residues is a residue.