Question

If $p\left(x\right)=x^2+ax+b$ such that $p\left(x\right)=0$ and $p\left(p\right(p\left(x\right)\left)\right)=0$ have equal root then find the integral value of $p\left(0\right).p\left(1\right)$

Answer 1

$P\left(x\right)=x^2+ax+b$ such that $P\left(x\right)=0\&P\left(P\right(P\left(x\right)\left)\right)=0$ have equal roots.

$P\left(P\right(x\left)\right)=(x^2+ax+b{)}^2+a(x^2+ax+b)+b$

$P\left(P\right(P\left(x\right)\left)\right)=\left\{\right(x^2+ax+b{)}^2+a(x^2+ax+b)+b{\}}^2+a(x^2+ax+b{)}^2+a(x^2+ax+b)+b$

Let $a=0$ & $b\le 0$ eq ; $b=-1$

So $P\left(x\right)=x^2-1$ ; $P\left(P\right(x\left)\right)=(x^2-1{)}^2-1$

$=x^4-2x^2+1-1=x^4-2x^2$

$P\left(P\right(P\left(x\right)\left)\right)=(x^2-1{)}^4-2(x^2-1{)}^2=(x^2-1{)}^2[x^4-2x^2+1-2]$

$=(x^2-1{)}^2(x^4-2x^2-1)$

Now it is clean $P\left(x\right)$ & $P\left(P\right(P\left(x\right)\left)\right)$ have no equal root because of $P\left(x\right)$ have only two roots $-1,1$ but $P\left(P\right(P\left(x\right)\left)\right)$ have many roots as you see.

So, condition will satisfy when $b=0$

then $P\left(x\right)=x^2$ & $P\left(P\right(P\left(x\right)\left)\right)=x^8$ and both have equal roots eq : $x=0$

So $P\left(0\right)=0$ & $P\left(1\right)=1$ now $P\left(0\right).P\left(1\right)=0.1=0$