Answer
$-2,1,3,4$
Work Step by Step
List all possible rational zeros as $\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24$
Test small values first for solutions such as $\pm1,\pm2\pm3,\pm4$, we found three zeros $1,3,4$.
Use synthetic division as shown to reduce P into linear and quadratic form
$P(x)=(x-4)(x-3)(x-1)(x^2+x-2)$
The zeros are $-2,1,3,4$ with $x=1$ having a multiplicity of $2$.