Question 45475: Hello, I need help with roots and zeros. I don't understand how to find the number of positive real zeros, negative real zeros, or imaginary zeros for a function. I know that you need to count the number of changes in sign for the coefficients but I don't understand how to interpret that.
Ex.
g(x)=x^4+x^3+2x^2-3x-1
Thank you.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Descartes' Rule of Signs
Given a polynomial such as:
x^4 + 7x^3 - 4x^2 - x - 7
Is it possible to say anything about how many positive real roots it has, just by looking at it?
Here's a striking theorem due to Descartes in 1637, often known as "Descartes' rule of signs": The number of positive real roots of a polynomial is bounded by the number of changes of sign in its coefficients. Gauss later showed that the number of positive real roots, counted with multiplicity, is of the same parity as the number of changes of sign.
Thus for the polynomial above, there is at most one positive root, and therefore exactly one.
In fact, an easy corollary of Descartes' rule is that the number of negative real roots of a polynomial f(x) is determined by the number of changes of sign in the coefficients of f(-x). So in the example above, the number of negative real roots must be either 1 or 3.
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Your problem:
g(x)=x^4+x^3+2x^2-3x-1
#of changes of signs = 1
So, #of positive real zeroes =1
g(-x)=x^4-x^3+2x^2+3x-1
Its # of changes of signs = 3
So, # of negative real zeroes = 1 or 3
Cheers,
Stan H.
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