Question

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

Answer 1

Let $f\left(x\right)=(x-\alpha )(x-\beta )$
It is given that $f\left(0\right)=p\Rightarrow \alpha \beta =p$
and $f\left(1\right)=\frac{1}{3}\Rightarrow (1-\alpha )(1-\beta )=\frac{1}{3}...\left(1\right)$

Now, let us assume that $\alpha$ is the common root of $f\left(x\right)=0$ add $fofofof\left(x\right)=0$
$fofofof\left(x\right)=0$
$\Rightarrow fofof\left(0\right)=0$
$\Rightarrow fof\left(p\right)=0$
So, $f\left(p\right)$ is either $\alpha$ or $\beta$.
$(p-\alpha )(p-\beta )=\alpha$
$(\alpha \beta -\alpha )(\alpha \beta -\beta )=\alpha \Rightarrow (\beta -1)(\alpha -1)\beta =1$ $(\because \alpha \ne 0)$
So, $\beta =3$... from eq.(1)
$(1-\alpha )(1-3)=\frac{1}{3}$
$\alpha =\frac{7}{6}$
$f\left(x\right)=(x-\frac{7}{6})(x-3)=25$
$f(-3)=(-3-\frac{7}{6})(-3-3)=25$