Using the Rational Zero Test, we know that all the possible rational solutions are factors of 4:

±2, ±4, ±1

After checking the graph, we see the only plausible rational factor is 1.

x³+3x²-8x+4=(x-1)(x²+4x-4)

Using the Quadratic Formula, we can then factor x²+4x-4=0 where a=1, b=4, c=-4

Good thing we checked the factorization otherwise we would not have caught the -2+2√2 solution, since it is so close so x=1 (also a factor).

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So all the real zeros of the function are: x=1, -2-2√2, -2+2√2.