Question

PLEASE HELP! WILL FAN AND MEDAL!!!




What are the possible numbers of positive real, negative real, and complex zeros of f(x) = -7x^4 - 12x^3 + 9x^2 - 17x + 3?

Positive Real: 3 or 1
Negative Real: 1
Complex: 2 or 0

Positive Real: 3 or 1
Negative Real: 2 or 0
Complex: 1

Positive Real: 1
Negative Real: 3 or 1
Complex: 2 or 0

Positive Real: 4, 2 or 0
Negative Real: 1
Complex: 0 or 1 or 3

Answer 1

you need to use descartes rule for this...

Answer 2

@dtan5457 Yep I'm pretty sure the answer is B right???

Answer 3

did you just guess that cause i didn't even do it yet

Answer 4

@dtan5457 No I didn't guess! Can you check my answer?

Answer 5

No, it's not B, lol, do you want me to show my work? it's quite a bit.

Answer 6

@dtan5457 Yes please

Answer 7

Positive roots=

original equation, and locate sign changes

-7x^4 - 12x^3 + 9x^2 - 17x + 3

you can see it turns from negative to positive, to negative, and back to positive.

that is 3 sign changes.

however, descartes, states that for positive roots, you must subtract by 2, and that number will work as well

therefore you have : 3 or 1 positive roots

Answer 8

so you got the positive part right, now let me explain the negatives

Answer 9

-7x^4 - 12x^3 + 9x^2 - 17x + 3

negate all the x's

-7(-x)^4 - 12(-x)^3 + 9(-x)^2 - 17(-x) + 3

even exponent: the same term

odd exponent: change the x sign

the first one:-7(-x)^4, even, so it stays as =-7x^4

the 2nd one: -12(-x)^3

odd exponent therefore it changes to 12x^3

you do that for all the other ones

and you end up with

- 7x^4 + 12x^3 + 9x^2 + 17x + 3

1 sign change


1 negative root

Answer 10

Wow thank you so much! :)