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[azureus88]

Prove that if P(x)=ax^4 + bx^3 +cx^2 + dx + e is even, then b=d=0.

I know its obvious from just looking at it but how would you prove it?

 

[gcchick]

Not sure how to prove it, but all even functions have even powers, eg. squares or quartics. b and d are coefficients of odd powers.

 

[Azreil]

P(x)=ax^4 + bx^3 +cx^2 + dx + e
For even function P(x)=P(-x)
P(-x)=a(-x)^4 + b(-x)^3 + c(-x)^2 +d(-x) + e

ax^4 + bx^3 +cx^2 + dx + e = ax^4 - bx^3 + cx^2 - dx + e
2bx^3 + 2dx = 0
x(b^2 + d) = 0
b^2 + d = 0
Therefore for P(x) to be even, b=d=0.

 

[azureus88]

you know for the last step how you got, bx^2 + d = 0, how does that lead to b=d=0.

 

[u-borat]

b^2=-d, which is only true if they are complex numbers(not happening in 3unit) or if they are both zero.