What are the possible number of negative zeros of f(x) = 2x^7 – 2x^6 + 7x^5 – 7x^4 – 4x^3 + 4x^2 ?
i got 1 not sure help?

Question

What are the possible number of negative zeros of f(x) = 2x^7 – 2x^6 + 7x^5 – 7x^4 – 4x^3 + 4x^2 ?
i got 1 not sure help?

campbell_st · Answer

take x^2 as a factor 

$$P(x) = x^2(2x^5 - 2x^4 + 7x^3 - 7x^2 - 4x + 4)$$

x - 1 is a factor 

since P(1) = 1(2 - 2 + 7 - 7 - 4 + 4)
        P(1) = 0

which gives 

$$P(x) = x^2(x -1)(2x^4 + 7x^2 - 4) $$

$$(2x^4 + 7x^2 - 4) = (2x^2 - 1)(x^2 +4) $$

so $$P(x) = x^2(x+1)(2x^2 -1)(x^2 + 4) $$

anonymous · Answer

1

 2 or 0

 4, 2, or 0

 7, 5, 3, or 1

campbell_st · Answer

so there are 7 possible zeros

anonymous · Answer

7,5,3, or 1

anonymous · Answer

is that the answer

campbell_st · Answer

no I just read the negative bit 

$$P(x) = x^2 (x -1)(2x^2 -1)(x^2 +4) $$

then there are 2

anonymous · Answer

so its 2 or 0

anonymous · Answer

I was going to tell you that (x+1) in your previous post should be (x-1)

anonymous · Answer

? can you help with this question please

anonymous · Answer

Actually there is only one that is < 0
$$
x=-\sqrt{\frac 1 2}
$$

anonymous · Answer

im really confused everyone is giving me a different answer

campbell_st · Answer

what about the complex from x^2 + 4

anonymous · Answer

These are not negative numbers. Negative numbers are only real numbers.

anonymous · Answer

x^2 + 4 > 0 and never zero

anonymous · Answer

for x real.

anonymous · Answer

A  1

B   2 or 0

C  4, 2, or 0

D   7, 5, 3, or 1                 ????????????????????????

anonymous · Answer

A is your answer. You can take it to the bank.

anonymous · Answer

thats what i got first

anonymous · Answer

You are right. There is only one negative real root.

anonymous · Answer

thanks

anonymous · Answer

yw