How many solutions exist in reals

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Let $f(x)= x^3+3x^2+6x+2009$ and $$g(x)=\dfrac{1}{x-f(1)}+\dfrac{2}{x-f(2)}+\dfrac{3}{(x-f(3)}.$$ The number of real solutions of $g(x)=0$ is

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Consider $$g(x)=\frac{\alpha}{x-a}+\frac{\beta}{x-b}+\frac{\gamma}{x-c}$$ Reduce to same denominator, expand and simplify to get $$g(x)=\frac{(a b \gamma +a \beta c+\alpha b c)-x (a \beta +a \gamma +b \alpha +b \gamma + c \alpha +c \beta)+x^2 (\alpha +\beta +\gamma )}{(x-a)(x-b)(x-c)}$$

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