use descartes rule of signs to determine the possible number of positive real zeros and negative zeros. 
List all the terms of 
Term #1. 
. Therefore, term #1 has a NEGATIVE sign
Term #2. 
. Therefore, term #2 has a POSITIVE sign
Therefore in going from Term #1 to Term #2 there was a change in
sign, from negative to positive. So far that's 1 sign change.
Term #3. 
Therefore, term #2 has a NEGATIVE sign
Therefore in going from Term #2 to Term #3 there was another change
in sign, from positive to negative. So far that's 2 sign changes.
Term #4. 
. Therefore, term #2 has a POSITIVE sign
Therefore in going from Term #3 to Term #4 there was another change in
sign, from negative to positive. So far that's 3 sign change.
Term #5. 
. Therefore, term #5 has a NEGATIVE sign
Therefore in going from Term #4 to Term #5 there was another change
in sign, from positive to negative. So far that's 4 sign changes.
So the fact that there are 4 sign changes means that the maximum number of
positive zeros is 4. But since complex imaginary zeros come in pairs,
There may be only 2 positive zeros, (in case there happens to be one pair
of complex imaginary zeros). Also there may be 0 positive zeros, (in case
there happens to be two pairs of complex imaginary zeros).
So there are either 4 or 2 or 0 positive 0's.
Now the number negative zeros of will be the number of
positive zeros of 
So let's substitute 
for in to get
Simplify:
List all the terms of
Term #1. . Therefore Term #1 has a NEGATIVE sign.
Term #2. 
. Therefore Term #2 also has a NEGATIVE sign.
Therefore there is no sign change from Term #1 to Term #2. So far 0 sign changes.
Term #3. 
. Therefore Term #3 also has a NEGATIVE sign.
Therefore there is no sign change from Term #2 to Term #3. So far there
are still 0 sign changes.
Term #4. . Therefore Term #4 also has a NEGATIVE sign.
Therefore there is no sign change from Term #3 to Term #4. So far there
are still 0 sign changes.
Term #5. . Therefore Term #5 also has a NEGATIVE sign.
Therefore there is no sign change from Term #4 to Term #5. So far there
are still 0 sign changes.
So there are 0 positive zeros of and therefore there are
0 negative zeros of P(x).
Answer: P(x) has 4 or 2 or 0 POSITIVE zeros and 0 NEGATIVE zeros.
[By doing further study on P(x), which you will eventually learn to
do, we could discover that P(x) actually has 0 real zeros, and 2 pairs
of complex imaginary zeros.]
Edwin
