Question 1181839
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None of those are prime.  Three of them can be easily factored by grouping.<br>
3x^3 + 3x^2 - 2x - 2 = (3x^2)(x+1)-2(x+1) : (x+1) is a factor<br>
3x^3 - 2x^2 + 3x - 4 : x=1 makes the expression equal to 0, so (x-1) is a factor<br>
4x^3 + 2x^2 + 6x + 3 = (2x^2)(2x+1)+3(2x+1) : (2x+1) is a factor<br>
4x^3 + 4x^2 - 3x - 3 =4x^2(x+1)-3(x+1) : (x+1) is a factor<br>