SOLUTION: Consider polynomial {{{g(x) = x^n + 4x^(n-1)}}}; what is its maximum number of zeroes? @ The maximum number of zeroes = n, regardless of the value of n. a. 0 b. 1 c. Cannot be d

We apply Descartes' rule of signs, which says:

If f(x) = a polynomial in descending order, then:

1. The maximum number of positive zeros of polynomial f(x),
is the number of sign changes in f(x), going left to right.

2. The maximum number of negative zeros of polynomial f(x),
is the number of sign changes in f(-x), going left to right.



has only two terms, and NO sign changes. Therefore it has NO
POSITIVE zeros.

Now let's test it for NEGATIVE zeros.

We form  by replacing  by .

  

Now since we do not know whether n is even or odd, we have to
consider both cases:

If  is even, then  is odd, so  and


So in this case



That has 1 sign change, so g(x) would have a maximum of 1 negative zero.

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If  is odd, then  is even, so  and


So in this case



That also has 1 sign change, so g(x) would have a maximum of 1 
negative zero regardless of whether n is even or odd.

---------

Now we know that it has no positive zeros, a maximium of
1 negative zero.

However we must also check to see if 0, which is neither
positive nor negative, is a zero.  So we find 

 



Therefore 0 is also a zero, regardless of n.

So the maximium number of zeros is 2. 

Since 2 is not listed as a choice, then the only correct
choice is d "None of the above".

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Comment:
It's easy to see the correct choice cannot be 0 or 1 
because when 




This has two zeros 0 and -4.

And we showed above that c cannot be the answer,
for we were able to determine that the maximum number
of zeros is 2. 

Edwin