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
sente
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#

Cesareo R.
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#

Alberto P.
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)#

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