Let f(x) be a quadratic polynomial with leading coefficient 1 such that f(0) = p, $\mathrm{p}\ne 0$ and $\mathrm{f}\left(1\right)=\frac{1}{3}$. If the equation f(x) = 0 and fofofof(x) = 0 have a common real root, then f(–3) is equal to.......................

$$\begin{gathered} Let f\left(x\right)=x^2+bx+c \\ f\left(0\right)=p\Rightarrow c=p\therefore f\left(x\right)=x^2+bx+p \\ and f\left(1\right)=\frac{1}{3}\Rightarrow 1+b+p=\frac{1}{3}\Rightarrow \boxed{b+p=\frac{-2}{3}} \\ Let \alpha be the common root of f\left(x\right)=0 and fofofof\left(x\right)=0 \\ then f\left(\alpha \right)=0 and fofofof\left(\alpha \right)=0\Rightarrow fofof\left(0\right)=0 \\ \Rightarrow fof\left(p\right)=0\therefore f\left(p^2+bp+p\right)=0 \\ \Rightarrow p^2+bp+p=\alpha \Rightarrow p\left(p+b+1\right)=\alpha \\ \Rightarrow \frac{p}{3}=\alpha . \\ Let \beta be the other root of f\left(x\right)=0. \\ \mathrm{Product} \mathrm{of} \mathrm{roots} of f\left(x\right)=\mathrm{p} \\ \Rightarrow \frac{\mathrm{p}}{3}\left(\mathrm{\beta }\right)=\mathrm{p}\therefore \mathrm{\beta }=3 \\ \Rightarrow \left(\mathrm{\alpha },\mathrm{\beta }\right)=\left(\frac{\mathrm{p}}{3},3\right). \\ \sin ce \beta is a root \\ \mathrm{f}\left(\mathrm{\beta }\right)=0\Rightarrow \boxed{9+3\mathrm{b}+\mathrm{p}=0} \\ \Rightarrow \mathrm{b}=\frac{-25}{6},\mathrm{p}=\frac{7}{2} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-\frac{25}{6}\mathrm{x}+\frac{7}{2} \\ \therefore \mathrm{f}\left(-3\right)=9+\frac{25}{2}+\frac{7}{2}=25 \end{gathered}$$