Question

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

Answer 1

So the first given is $p(x)=x^2+ax+b=0$.

The second given is $p(p(p(x)))=0$. So we solve this one from the inner towards the outwards. So first, we evaluate p(p(x)). Since p(x) = 0, therefore, we will evaluate p(0).

$p(p(x))=p(0)=0^2+a(0)+b=b$.

Now, we evaluate $p(p(p(x)))=p(b)=b^2+ab+b.$

Since p(p(p(x)))=0, therefore, $b^2+ab+b=0$.

Since they have equal roots, therefore, if we compare $x^2+ax+x=0$ and $b^2+ab+b=0$, we will find that x and b have to be equal to each other. Therefore, $b=x.$

Substitute with that in p(x): $p(x)=x^2+ax+x$.

The requirement is p(0).p(1). So we will first calculate p(0): $p(0)=0^2+a(0)+0=0.$Then we calculate p(1): $p(1)=1+a(1)+1+a+2.$.

So p(0).p(1) = 0(a+2) = 0.