Let f x be a quadratic polynomial with leading coefficient 1 such that f 0 = p , p ≠ 0 , and f 1 = 1 3…

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

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

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

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

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

$\Rightarrow fofofof \left(\alpha \right)=0$

$\Rightarrow fofof \left(0\right)=0$

$\Rightarrow fof\left(p\right)=0$

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

Now assuming $\left(p-\alpha \right)\left(p-\beta \right)=\alpha$

$\Rightarrow \left(\alpha \beta -\alpha \right)\left(\alpha \beta -\beta \right)=\alpha \Rightarrow \left(\beta -1\right)\left(\alpha -1\right)\beta =1$

$\Rightarrow \frac{\beta }{3}=1$ $\left(\text{as }\left(1-\alpha \right)\left(1-\beta \right)=\frac{1}{3}\right)$

So, $\beta =3$

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

$\Rightarrow \left(1-\alpha \right)\left(1-3\right)=\frac{1}{3}$

$\Rightarrow \alpha =\frac{7}{6}$

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

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