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