Question

http://gyazo.com/683b16a18861f1ff1b00d46ece8a9c08 is my question and did I solve it right?

Answer 1

Also I can't find it with the Descartes’s rule of sign. Do I need to revers the zero? which changes all my signs first?

Answer 2

what was your original question?
you need to find the zeroes of the cubic polynomial obtained ?

Answer 3

My factors don't seem to work for the p/q

Answer 4

•The equation for the volume of the object written in terms of x

Answer 5

I know I need to find 3 possitive and no zeros and three complex but now im totally lost

Answer 6

i mean no negative zeros

Answer 7

this is the graph that I got

Answer 8

I have to f
ind the solutions to this equation algebraically using the Fundamental Theorem of Algebra, the Rational Root Theorem, Descartes' Rule of Signs, and the Factor Theorem.

Answer 9

Descarte's rule of signs says it has UP TO 3 positive roots. However, only one of them is real and it is irrational. Are you expecting rational roots?

Answer 10

I'm wondering if I'm going to have irrational because I'm questioning my whole problem. I'm wondering if my -48 in my equation should have been a positive 48 because I can't get a zero remainder http://gyazo.com/683b16a18861f1ff1b00d46ece8a9c08

Answer 11

If 192 is supposed to be a volume, then x represents one of the sides, and it should not be negative.

Answer 12

It is true that if 192 is negative (volume?), then there is one rational real solution.

Answer 13

I re-wrote my problem and I got -x^3+7x^2-6x+48=0 and it zero's out using V=1wh Length: 8
Width: 2
Height: 12
Volume= 192
I believe I messed up my original equation right from the beginning

Answer 14

You need to check further. This equation still gives x=-2 as the real root.
x^3+7*x^2+6*x-48=0
will give x=+2 as the real root.
So perhaps you could check the equation again.

Answer 15

Alright thank you!!