Find the polynomial P(x) with real coefficients such that P(2)=12 and P(x^2)=x^2(x^2+1)P(x) for each x in RR?

3 Answers
Oct 28, 2016

P(x) = x^4-x^2

Explanation:

First, observe that the degree of P(x^2) will be twice the degree of P(x), and that the degree of x^2(x^2+1)P(x) will be 4 greater than the degree of P(x). Taken together, we know that P(x) is a polynomial of degree 4.

Further observations:

P(0^2) = 0(1)P(0) = 0

Thus x is a factor of P(x).

P(1) = P(1^2) = 1(2)(P(1))

=> P(1) = 2P(1)

=> P(1) = 0

Thus (x-1) is a factor of P(x)

0 = P(1) = P((-1)^2) = 1(2)P(-1)

=> P(-1) = 0

Thus (x+1) is a factor of P(x)


With those, we can write P(x) as, for some a, c in RR,

P(x) = cx(x-1)(x+1)(x-a)=cx(x^2-1)(x-a)

Plugging in P(2) = 12, we get

c(2)(3)(2-a) = 12
=> 12c - 6ac = 12
=> 2c - ac = 2

Plugging in P(4) = P(2^2), we get

P(4) = 4(5)(12) = 240
=>c(4)(15)(4-a) = 240
=>240c-60ac = 240
=>4c - ac = 4

Solving the system {(2c-ac=2),(4c-ac=4):}, we arrive at

{(a = 0),(c = 1):}

:. P(x) = x^2(x^2-1) = x^4-x^2

Oct 28, 2016

P(x) = x^4-x^2

Explanation:

For x ne 0 we have P(x)=(P(x^2))/(x^2(x^2+1)) so

P(x)=P(-x) an even function. Also

deg(P(x^2))=2deg(P(x))=4+deg(P(x)) so

deg(P(x))=4

If deg(P(x))=4 and P(x) is even then

P(x)=ax^4+bx^2+c but

P(0)=0 so c=0. Now

P(2)=2^4a+2^2b=12 and
P((sqrt(2))^2)=2(2+1)(2^2a+2b)=12

solving for a,b we get a = 1 and b=-1 so

P(x)=x^4-x^2

Oct 28, 2016

x^2(x^2-1)

Explanation:

x^2(x^2+1)=(P(x^2))/(P(x))~=_(x->oo)x^(2n)/(x^n)=x^n\ => n=4

{(P(x)=P(-x)=(P(x^2))/(x^2(x^2+1))), (P(0)=0):}\ \ => \ P(x)=ax^4+bx^2

P(x^2)=ax^8+bx^4=x^2(x^2+1)(ax^4+bx^2)=(b+a)x^6+...

=> b=-a

P(x)=a(x^4-x^2)

P(2)=a(16-4)=12 \ \ => a= 1

P(x)=x^2(x^2-1)