Question

What are the possible number of negative zeros of f(x) = –2x^7 + 2x^6 + 7x^5 + 7x^4 + 4x^3 + 4x^2 ? When I changed the f(x) to f(-x) I only get one sign change, so one negative zero. One is not an answer... what am I doing wrong?

Answer 1

you know Descartes' rules of sign ?

Answer 2

Yes, that's what I tried. But I only came to one negative zero, so I guess I'm using the rule wrong...

Answer 3

when i change it to f(-x) i am getting
f(-x)=2x^7+2x^6-7x^5+7x^4-4x^3+4x^2
now find it !!

Answer 4

Oh. Would 4,2, or 0 be correct?

Answer 5

I think zero is the correct answer.

Answer 6

there are 4 sign changes !!

Answer 7

I think sami is right, now I see my mistake in using the rule. I was only reversing the signs, not substituting in -x for all the x's.

Answer 8

\[
4 x^2 + 4 x^3 + 7 x^4 + 7 x^5 + 2 x^6 -
2 x^7 == -x^2 (-4 - 4 x - 7 x^2 - 7 x^3 - 2 x^4 + 2 x^5)
\]

Answer 9

It is enough to study
\[
4 - 4 x - 7 x^2 - 7 x^3 - 2 x^4 + 2 x^5
\]

Answer 10

Use Descarte's rule. Use f(-x) instead of f(x), you will get to an equation that totals 4 sign changes.

Answer 11

@eliassaab Sir you are right :)

Answer 12

see the graph of the 5th degree polynomial.

Answer 13

@JustBoss there are zero negative roots.

Answer 14

@JustBoss if you are not convinced, give me two negative numbers a and b such that
f(a) and f(b) have different signs.

Answer 15

Sorry, had to step away from my computer for a minute. I see where you get your answer now, thank you.