What are the possible number of negative zeros of f(x) = 2x7 + 2x6 + 7x5 + 7x4 - 4x3 + 4x2 ?

Question

zale101 · Answer

Do you know Descartes' rule of signs?

zale101 · Answer

"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.

zale101 · Answer

use f(-x) to find  negative zeros

zale101 · Answer

2?

anonymous · Answer

Positive: 1; Negative: 2 or 0; Complex 2 or 0

zale101 · Answer

f(x)=x^2(2x^5+2x^4+7x^3+7x^2−4x+4)

anonymous · Answer

Is that right?

zale101 · Answer

How did u solve that?

anonymous · Answer

used f(-x), did it in my head

zale101 · Answer

x^2(2x^5 - 2x^4 + 7x^3 - 7x^2 - 4x + 4) flip the sign on the odd-numbered degrees

zale101 · Answer

So how many negative roots are there?

anonymous · Answer

3

anonymous · Answer

or one

zale101 · Answer

it's one, because between 7x^2 and the +4x the sign changes, once not three :)