Question

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

Answer 1

did you find \(f(-x)\) ?

Answer 2

No

Answer 3

that is what you need to do

Answer 4

I'm confused on how the do it

Answer 5

you can do it by replacing \(x\) everywhere by \(-x\) and then see what you get, or more simply you can do it by changing the sign of each coefficient of terms of odd degree, because for example \((-x)^5=-x^5\) and leaving the terms of coefficients of even degree alone, becaue for example \((-x)^4=x^4\)

Answer 6

\[ f(x)=-2x^7 - 2x^6 + 7x^5 - 7x^4 + 4x^3 + 4x^2 \]
\[f(-x)=2x^7-2x^6-7x^5-7x^4-4x^3+4x^2\]
all terms of odd degree, change the sign, all terms of even degree, leave alone

Answer 7

So there are 4 negative?

Answer 8

now to find out the possible number of negative zeros, don't count the number of negative coefficients, count the "changes in sign" of the coefficients

Answer 9

2