Question

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

Answer 1

not sure i understand the question?

Answer 2

is the que is 2x^7

Answer 3

\[\Large f(x) = x^2\left( 2x^5+ 2x^4 + 7x^3 + 7x^2 - 4x + 4 \right)\] ... i suppose you could have upto three negative zeros

Answer 4

Do you know Descartes' Rule of Signs?

Answer 5

Picking up on PaxP's work, zero is a double root but is not negative.

Looking at g(x) = 2x^5 + 2x^4 + 7 x^3 + 7x^2 - 4x +4, use Descartes' Rule of Signs to do the following:

In g(x), replace x with -x to find g(-x).

g(-x) = 2* (-) ^5 + 2*(-)^4 + 7* (-)^3 + 7*(-)^2 - 4*(-) + 4

g(-x) = (-) + (+) + (-) + (+) + (+) + (+)

We are interested in the number of sign changes.

There are 3 which indicates 3 possible negative real roots or less that by 2.

So, there are 3 negative possible real roots or 1 negative real root.

As it turns out, the original function has one negative real root.

Note that my sign change work above is sloppy and not very mathematical.

Once you read about the theorem, the work will make more sense.

Please read the info at the link given here:

http://www.purplemath.com/modules/drofsign.htm