You posted no instructions at all!!!!!

So I'll just have to guess what they might have been:

My guesses:
 
Use DesCartes rule of signs to determine the number of positive
and negative zeros of P(x). 

Find all real zeros of P(x).

If neither of these were the instructions, post again and
this time give the instructions as to what you are supposed to do
with P(x)

P(x) = x5 + 3x3 + 2x - 6

x5 and 3x3 are both positive, so there is no sign change going
from x5 to 3x3.

3x3 and 2x are both positive, so there is no sign change going
from 3x3 to 2x.

2x is positive and -6 is negative, so that is 1 sign change
going from 2x to -6.

So there is exactly 1 positive zero of P(x).

Now we form P(-x)

P(-x) = (-x)5 + 3(-x)3 + 2(-x) - 6

P(-x) = -x5 - 3x3 - 2x - 6

-x5 and -3x3 are both negative, so there is no sign change going
from -x5 to -3x3.

-3x3 and -2x are both positive, so there is no sign change going
from -3x3 to -2x.

-2x and -6 are both negative, so there is no sign change
going from -2x to -6.

So there are no negative zeros of P(x).

There is only one positive zero.

Since the leading coefficient is 1, the possible rational 
zeros are the ± divisors of the absolute value of the last
term, |-6| or 6.  So we try the smallest divisor of 6, which
is 1. 

For synthetic division, we must insert placeholders for the
powers of x which are missing in P(x)

P(x) = x5 + 0x4 + 3x3 + 0x2 + 2x - 6


1|1  0  3  0  2 -6
 |   1  1  4  4  6
  1  1  4  4  6  0

Yes that is a zero since the last number on the bottom is 0.

So 1 is the only real zero of P(x).

Edwin
AnlytcPhil@aol.com