e^x(x-2)(-x+1)(x+1)²/(x+1)(x²+ x+1)>=0
factor out an x+1
e^x(x-2)(-x+1)(x+1)/(x²+ x+1)>=0
e^x is always positive, regardless of the value of x, so divide e^x out
(x-2)(-x+1)(x+1)/(x²+ x+1)>=0
since we're restricting our possible answers to positive integers, x^2 + x + 1 is always positive, regardless of the value of x, so multiply x^2 + x + 1 out
(x-2)(-x+1)(x+1)>=0
x - 2 is >=0 when x>=2
-x+1 is <=0 when x>=1
x+1 is >=0 when x>=1 (remember, x can't = -1 or 0)
x = 0 gives us - + +, a negative result, isn't >= 0
x = 1 gives us - 0 +, a zero result, is >= 0
x = 2 gives us 0 - +, a zero result, is >= 0
x = 3 gives us + - +, a negative result, isn't >= 0
Note: Any value for x above 2 won't change the sign of the original equation.
Answer: There are 2 positive integers (1 and 2) that satisfy the inequality e^x(x-2)(-x+1)(x+1)²/(x+1)(x²+ x+1)>=0.