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

p(x)=$x^{2}+a x+b$

Also, p(x)=0

And p(p(p(x)))=0

→ p(p(0))=0

→p(b)=0

→$b^2+ab+b=0$

→b(b+a+1)=0

→Either ,b=0  ∧  b+a+1=0

Now, p(0)=0+a×0+b=b

p(1)=1×1+a×1+b=1+a+b

Now, p(0).p(1)=b(1+a+b)

If b=0, then p(0).p(1)=0

If , b+a+1=0, then also p(0).p(1)=0

Answer 2

P(x) = x² + ax + b such that P(x) = 0 and P(P(P(x))) = 0 have equal root

so, P(P(x)) = (x² + ax + b)² + a(x² + ax + b) + b

P(P(P(x))) = {(x² + ax + b)² + a(x² + ax + b) + b}² + a{(x² + ax + b)² + a(x² + ax + b) + b} + b

here you see such a complicated terms.

Let's do something different.

if we assume , a = 0 and b ≤ 0 e.g., b = -1 {assume}

so, p(x) = x² - 1

then, P(P(x)) = (x² - 1)² - 1 = x⁴ - 2x² + 1 - 1

= x⁴ - 2x²

P(P(P(x))) = (x² - 1)⁴ - 2(x² - 1)²

= (x² - 1)²[x⁴ - 2x² + 1 - 2]

= (x² - 1)²(x⁴ - 2x² - 1)

here it is clear that P(x) and P(P(P(x))) have no equal root. because of P(x) have only two roots -1, 1 but P(P(P(x)) have many roots as you can see .

so, condition will satisfy when we assume b = 0

then, P(x) = x² and P(P(P(x))) = x^8

and both have equal root e.g., x = 0

so, P(0) = 0 and P(1) = 1

so, P(0).P(1) = 0.1 = 0