Find all polynomial P(x) of degree 2 such that P(x2)=(P(x))2. still cant answer question...?

Answer 1

It's unusual to write 2x as x2, so it's not clear if you're referred to P(2x) = 2*P(x) or P(x^2) = (P(x))^2. In either case, let's assume P(x) = ax^2 + bx + c.

In the first case:
P(2x) = 4ax^2 + 2bx + c
2*P(x) = 2ax^2 + 2bx + 2c

This gives rise to a series of equations, related to each coefficient of a power of x.

x^2: 4a = 2a ==> a = 0
x: 2b = 2b ==> b is a real number
1: c = 2c ==> c = 0

This is a problem, because we found that a = 0, but then the polynomial is of degree 1, not 2. So there are no solutions in this case.

In the second case:
P(x^2) = ax^4 + bx^2 + c
(P(x))^2 = a^2*x^4 + 2abx^3 + (2ac + b^2)x^2 + 2bcx + c^2

This again gives rise to a series of equations.

x^4: a^2 = a ==> a = 0 or 1
x^3: 2ab = 0 ==> a = 0 or b = 0
x^2: 2ac + b^2 = b
x: 2bc = 0 ==> b = 0 or c = 0
1: c^2 = c ==> c = 0 or 1

As I mentioned, we cannot have a = 0. Therefore, a = 1 from the x^4 equation. And as a result, b = 0 due to the x^3 equation. The only remaining unknown is c. It can be equal to 0 or 1. We need to try each one, and see if the x^2 equation holds true.

c = 0 ==> 2ac + b^2 = 2*1*0 + 0^2 = 0 + 0 = 0 = b
c = 1 ==> 2ac + b^2 = 2*1*1 + 0^2 = 2 + 0 = 2 != b

So the only such polynomial is x^2.

Answer 2

You must mean P(2x) = 2P(x). If you mean P(x^2) = 2P(x)), it makes no sense because you would no longer have a polynomial of degree 2.

So take P(x) = ax^2 +bx+c
Then P(2x) = 4ax^2 + 2bx +c, and
2P(x) = 2ax^2+2bx+c
So 4ax^2 +2bx+c = 2ax^2 +2bx + c
2ax^2 = 0 <-- Answer