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Use Descartes' rule of signs to describe the roots of the polynomial: f(x)=3x^5-4x^3+3x^2+2x-1
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Choices? Or no multiple choice?
No, I have to supply the answer, I kind of remember how to do it... But I'm a little fuzzy.
For the positive roots. 3x^5-4x^3+3x^2+2x-1 Find how many sign changes goes from a negative to a positive, back to a negative. so 2 or 0 root change
For the negative roots. make every x negative. 3(-x^5)-4(-x^3)+3(-x)^2+2(-x)-1 If you simplify that out you should get -3x^5+4x^3+3x^2-2x-1 again, 2 or 0 roots
So...\[\le2\]? For both?
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No, a root can't be 1 so it has to be 2 or 0
for both
Ok. c: Thank yhu n.n
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