Let $f(x)$ be a quadratic polynomial with leading coefficient 1 such that $f(0)=p, p \neq 0$, and $f(1)=\frac{1}{3}$. If the equations $f(x)=0$ and $f \circ f \circ f \circ f(x)=0$ have a common real root, then $f(-3)$ is equal to ________________.

<p>Let $f(x) = (x - \alpha )(x - \beta )$</p> <p>It is given that $f(0) = p \Rightarrow \alpha \beta = p$</p> <p>and $f(1) = {1 \over 3} \Rightarrow (1 - \alpha )(1 - \beta ) = {1 \over 3}$</p> <p>Now, let us assume that, $\alpha$ is the common root of $f(x) = 0$ and $fofofof(x) = 0$</p> <p>$fofofof(x) = 0$</p> <p>$\Rightarrow fofof(0) = 0$</p> <p>$\Rightarrow fof(p) = 0$</p> <p>So, $f(p)$ is either $\alpha$ or $\beta$.</p> <p>$(p - \alpha )(p - \beta ) = \alpha$</p> <p>$$(\alpha \beta - \alpha )(\alpha \beta - \beta ) = \alpha \Rightarrow (\beta - 1)(\alpha - 1)\beta = 1$$ ($\because$ $\alpha \ne 0$)</p> <p>So, $\beta = 3$</p> <p>$(1 - \alpha )(1 - 3) = {1 \over 3}$</p> <p>$\alpha = {7 \over 6}$</p> <p>$f(x) = \left( {x - {7 \over 6}} \right)(x - 3)$</p> <p>$f( - 3) = \left( { - 3 - {7 \over 6}} \right)(3 - 3) = 25$</p>