Roots of polynomial $-4x^3+3x^2-2x+1=0$

I've calculated the derivative of the function $f(x)=-4x^3+3x^2-2x+1$ and came up with $f'(x)=-12x^2+6x-2$ which is always negative and since $\lim_{x\to\infty} f(x)=-\infty$ and $\lim_{x\to-\infty} f(x)=\infty$ I assumed that $f(x)=0$ has only one real solution and two other complex ones. But I'm confused since the polynomial has integer coefficients, shouldn't there be a way to determine the real solution?

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No one says there isn't a way to determine the real solution... –  ᴊ ᴀ s ᴏ ɴ Mar 17 '13 at 8:05
wolframalpha.com/input/… It's ugly, but it's there. –  Arthur Mar 17 '13 at 8:12

You are right about the polynomial $f$ having only one real root. Since $f'$ is always negative $f$ is strictly decreasing ...