Random Polynomial Question (1 Viewer)

[xwrathbringerx]

Hi

If P(x) is monic of degree 7, if P(x) is odd, prove P(x) = ax + bx^3 + Cx^5 + x^7.
I think we need a reasoning where the eqn. has infintely many roots so it's the zero polynomial or something?

Could someone please help?

Thanx a lot!<!-- google_ad_section_end -->

 

[blaze.bluetane]

Well if it is odd, then f(-x)=-f(x)

and the roots must be A, -A, B, -B, C, -C, D
and D must be 0 since it is odd and has an odd power
Therefore sum of the roots (co-efficient of x⁶) = 0

Derivative of an odd function is even and vice versa

Therefore, second derivative (x⁵)
has roots X, -X, Y, -Y, Z

Therefore sum of the roots of the second derivative (co-efficient of x⁴) = 0

Fourth derivative (x³)
has roots x, -x, y and y=0 because it is an odd function

Therefore sum of the roots of the fourth derivative (co-efficient of x²) = 0

and again for the sixth derivative


Does that work??

 

[Timothy.Siu]

well,

P(x)=x^7+ax^6+bx^5.....+k
for odd, P(-x)=-P(x)
-x^7+ax^6-bx^5+....+k=-x^7-ax^6-bx^5.....-k
equating coefficients....
-a=a a=0 similarly for the rest....this way is pretty dodgey lol.
but then basically all the coefficients of even degree x will be 0 and you'll get the required answer.

 

[Trebla]