10.a)Solve the inequality x^3-5x^2+2x+8<0 by

Question

i) using intervals
ii) considering all cases

I did this but my teacher said that when i solved this: x^3-5x^2+2x+8<0
(x+1)(x^2-6x+8)
(x+1)(x-4)(x-2)<0

She told me to show all the steps, but i don't know how to show more. so can you please help me to show all the works and rest of the works i did too but she said " the same answer has to happen for each method. i don't really understand that. can you help me please and thank you....

Steve explained this, but still i didn't got the steps. can someone do it for me step by step so i can understand the steps of it please.

Answer 1

Ok, so all the algebra work has been done
x^3-5x^2+2x+8<0
(x+1)(x-4)(x-2)<0

Quickest way:
I would look at the graph of
y = x^3-5x^2+2x+8
= (x+1)(x-4)(x-2)
I know that this is standard cubic function opening up in quadrant I and having x-intercepts of -1, 2, and 4
So a very rough sketch shows me that this curve is below the x-axis ( < 0 ) for x < - 1 or 2 < x < 4

looking at all cases , the critical values of -2,2, and 4 split the line into 4 parts :

if x < -1, try some value, e.g. x = -5
(-)(-)(+) < 0 , so yes
x < -1

if x is between -2 and 2 , e.g. x = 0
(+)(-)(-2) > 0 , so NO

if x is between 2 and 4 , e.g. x = 3
(+)(-)(+) < 0 , so YES
x is between 2 and 4

if x > 4 , e.g. x = 5
(+)(+)(+) > 0 , NO

then x < -2 OR 2 < x < 4

Answer 2

Reiny help me with using intervals. u forgot to help me with that. its urgent please