algebra 2 problem

Question

What are the possible number of positive and negative real zeros for f(x)=3^5-3x^2+x-5

Answer 1

First of all, I assume you mean f(x) = 3x⁵ - 3x² + x - 5

 

Use Descartes's rule of signs.

 

First look at the number of times the sign changes from one coefficient to the next.  The first coefficient is 3 (positive), the next is -3 (negative).  That's a sign change.  The next is 1 (positive) -- another sign change.  And the last is -5 (negative).  There are a total of 3 sign changes, which means there are a maximum of 3 positive zeros.

 

Now plug in -1 for x and evaluate each term:

 

f(-1) = 3(-1)⁵ - 3(-1)² + (-1) - 5 = -1 -3 -1 -5.   There are NO sign changes -- all the terms are negative.  So the maximum number of NEGATIVE zeros is 0, which of course means there are no negative zeros.

 

So we have a maximum of 3 positive zeros and no negative zeros.  But it's a 5th degree equation, which means it has 5 zeros when you include complex numbers.  That means there have to be at least 2 complex zeros.

 

But the number of positive zeros we got (3) is the MAXIMUM, not the specific number.  So what are the other possibilities?

 

Since complex zeros always come in pairs (a + bi, and a - bi), there could be another 2 complex zeros.  That would mean that there would only be 1 positive zero.  But there can't be any more complex zeros, since that would add up to 6, which is too many.  So there are no other possibilities.

 

So we end up with either 1 or 3 positive zeros, and no negative zeros.