Answer
$-2,1,\frac{3}{4},\frac{2\pm\sqrt 2}{2}$
Work Step by Step
1. List possible rational zeros: $\pm1,\pm1/2,\pm1/4,,\pm1/8,\pm2,\pm3,\pm3/2,\pm3/4,\pm3/8,\pm6$
2. Use synthetic division or remainder theorem to test with some easy values $\pm1,\pm2,\pm3,\pm6$ and
we can find two zeros $x=-2, 1$
3. Use Descartes' Rule, we know that there should be 2 positive rational zeros.
4. Test positive fractions like $1/2,1/4,1/8,3/2,3/4,3/8$ we can find $x=3/4$ is a zero.
5. Use synthetic division to reduce P down to linear and quadratic terms
$P(x)=(x+2)(x-1)(4x-3)(2x^2-4x+1)$
6. Find all the zeros as $-2,1,\frac{3}{4},\frac{2\pm\sqrt 2}{2}$
