SOLUTION: Possible number of imaginary zeros in g(X)=x^4+3x^3+7x^2-6x-13

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Question 1010745: Possible number of imaginary zeros in g(X)=x^4+3x^3+7x^2-6x-13
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
The analysis you are using is called Descartes' rule of signs.
When you have determined the number of sign changes for f%28x%29, then you know the (maximum) possible number of positive zeros for your function.
You then have to evaluate f%28-x%29 to get the (maximum) possible number of negative zeros.
Positive and negative possible zeros always decrease by two's. You can't go below none.
In your problem:
g%28x%29=x%5E4%2B3x%5E3%2B7x%5E2-6x-13+
there are 1 sign changes for %28+%2B+x%29, which means there is 1 positive real zero
now, rewrite the given polynomial by substituting -x for x:
g%28x%29=%28-x%29%5E4%2B3%28-x%29%5E3%2B7%28-x%29%5E2-6%28-x%29-13+
g%28x%29=x%5E4+-3x%5E3%2B7x%5E2%2B6x-13+
there are 2 sign changes for %28+-x%29, which means there are a maximum of two negative real zeros
since we have 4th degree function,means there are 4 zeros in all, and we know that complex zeros always come in pairs, it means there will be 1 negative zero and 2 imaginary or complex zeros
so, in all will be:
1 positive real zero
1 negative real zero
2 complex zeros