Answer
(a) (b) (c) see proves below.
(d) $P(x)=[x^5+6x^3+(-2x^1)]+[(-x^2)+5x^0]$
Work Step by Step
(a) Function $P(x)$ can be written as:
$P(x)=\sum a_{2n+1}x^{2n+1}$ where $n=0,1,2,...$ are integers.
We have $P(-x)=\sum a_{2n+1}(-x)^{2n+1}=-\sum a_{2n+1}x^{2n+1}=-P(x)$
which means that $P(x)$ is an odd function in this case.
(b) Similarly, in this case, Function P(x) can be written as:
$P(x)=\sum a_{2n}x^{2n}$ where $n=0,1,2,...$ are integers.
We have $P(-x)=\sum a_{2n}(-x)^{2n}=\sum a_{2n}x^{2n}=P(x)$
which means that $P(x)$ is an even function in this case.
(c) We write $P(x)$ as:
$P(x)=\sum (a_{2n+1}x^{2n+1}+ a_{2n}x^{2n})$ where $n=0,1,2,...$ are integers.
$P(-x)=\sum (a_{2n+1}(-x)^{2n+1}+ a_{2n}(-x)^{2n})=\sum (a_{2n}(x)^{2n}-a_{2n+1}(x)^{2n+1})$
and this does not equal to neither $P(x)$ nor $-P(x)$
(d) Express the function as the sum of an odd function and an even function.
$P(x)=[x^5+6x^3+(-2x^1)]+[(-x^2)+5x^0]$
