# Thread: Random Polynomial Question

1. ## Random Polynomial Question

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!

2. No, it's not a zero-polynomial.

A monic 7th-degree polynomial would look like this:
$\displaystyle P(x) = x^7 + a_6 x^6 + a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0$.
(You should write polynomials in standard form, ie. with exponents in descending order. I'm changing the coefficients to $\displaystyle a_0, a_1,... a_6$.)

If P(x) is odd, then P(-x) = -P(x):
$\displaystyle P(-x)$
$\displaystyle = ({\color{red}-x})^7 + a_6 ({\color{red}-x})^6 + a_5 ({\color{red}-x})^5 + a_4 ({\color{red}-x})^4 + a_3 ({\color{red}-x})^3 + a_2 ({\color{red}-x})^2 + a_1 ({\color{red}-x}) +a_0$
$\displaystyle = {\color{red}-}x^7 + a_6 x^6 \;{\color{red}-}\; a_5 x^5 + a_4 x^4 \;{\color{red}-}\; a_3 x^3 + a_2 x^2 \;{\color{red}-}\; a_1 x +a_0$

$\displaystyle -P(x)$
$\displaystyle = {\color{red}-}(x^7 + a_6 x^6 + a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x +a_0)$
$\displaystyle = \;{\color{red}-}\;x^7 \;{\color{red}-}\; a_6 x^6 \;{\color{red}-}\; a_5 x^5 \;{\color{red}-}\; a_4 x^4 \;{\color{red}-}\; a_3 x^3 \;{\color{red}-}\; a_2 x^2 \;{\color{red}-}\; a_1 x \;{\color{red}-}\; a_0)$

In order for P(-x) to equal -P(x),
$\displaystyle a_6 = a_4 = a_2 = a_0 = 0$,
so
$\displaystyle P(x) = x^7 + a_5 x^5 + a_3 x^3 + a_1 x$.

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