Let f(x) be a quadratic polynomial with leading coefficient 1 such...

Let $f\left(x\right) = \left(x - \alpha\right)\left(x - \beta\right)$

It is given that $f\left(0\right) = p ⇒ \alpha\beta = p$

and $f\left(1\right) = \dfrac{1}{3} ⇒ \left(1 - \alpha\right)\left(1 - \beta\right) = \dfrac{1}{3}$

Now let us assume that $\alpha$ is the common root of $f\left(x\right) = 0$ and $fofofof\left(x\right) = 0$

$fofofof\left(x\right) = 0$

$⇒ fofofof\left(\alpha\right) = 0$

$⇒ fofof\left(0\right) = 0$

$⇒ fof\left(p\right) = 0$

So, $f\left(p\right)$ is either $\alpha$ or $\beta$.

Now assuming \left(p - \alpha\right)\left(p - \beta\right) = \alpha$</p><p>$⇒ \left(\alpha\beta - \alpha\right)\left(\alpha\beta - \beta\right) = \alpha ⇒ \left(\beta - 1\right)\left(\alpha - 1\right)\beta = 1$</p><p>$⇒ \dfrac{\beta}{3} = 1$\left(\text{as}\left(1 - \alpha\right)\left(1 - \beta\right) = \dfrac{1}{3}\right)$

So, $\beta = 3$

Now finding $\alpha$ by putting the value of $\beta\text{in}\left(1 - \alpha\right)\left(1 - \beta\right) = \dfrac{1}{3}$,

$⇒ \left(1 - \alpha\right)\left(1 - 3\right) = \dfrac{1}{3}$

$⇒ \alpha = \dfrac{7}{6}$

So, $f\left(x\right) = \left(x - \dfrac{7}{6}\right)\left(x - 3\right)$

So, $f\left(- 3\right) = \left(- 3 - \dfrac{7}{6}\right)\left(- 3 - 3\right) = 25$