No, the numbers you state would be potential rational roots.
You still have to check to see if they actually work
f(1/2) = 2(1/2)^4 - 5(1/2)^3 + 8(1/2)^2 + 4(1/2) + 7 = 10.5 ≠ 0
f(1) = 2 - 5 + 8 + 4 + 7 ≠ 0
none of them work. A quick look at "Wolfram" shows that there are no real roots at all.
For negative zeros, substitute -x for x, but do it carefully, using the exact same coefficients as originally:
(-x)^4 - (-x)^3 + 2(-x)^2 + (-x) - 5
x^4 + x^3 + 2x^2 - x - 5
There's only one change of sign there, so at most one negative root.
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