Question

Solve the inequlity by the test point method: x^3-2x^2-x+2≤0.

Answer 1

first let's look at
\[f(x)=x^3-2x^2-x+2\]
how do we find when f is 0
well it is obvious that x=-1 gives us f=0
f(-1)=0
x=-1 => x+1 is a factor of f

We can use synthetic division to find the other zeros

-1 | 1 -2 -1 2
| -1 3 -2
---------------------
1 -3 2 | 0

so we have
\[f=(x+1)(x^2-3x+2)=(x+1)(x-1)(x-2)\]


so you need to choose test points around x=-1 and x=1 and x=2

----------|----------|-----------|--------
-1 1 2

Answer 2

do you have to use synthetic division?
and how do you know its -1, do you guess? or can you tell by looking at it?

Answer 3

I think myininaya guessed it by the general trial and error method, but there is actually an analytically approach for finding the rational roots of the any equation ... if you want I could tell you about it ? or is it advance ?

Answer 4

Go for it. Maybe that will help be better understand.

Answer 5

I appreciate your eagerness, it is actually known as Rational root theorem ( http://en.wikipedia.org/wiki/Rational_root_theorem)

Answer 6

yes i just notice a real root

Answer 7

that root being x=-1

Answer 8

it's actually the consequence of Gauss lemma ( http://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)) which is really very old ;)

Answer 9

but for this problem it's really easy to factor out so all this are probably overkill here.

Answer 10

it's really late here .. I have to go now, I nearly forgot I have classes tomorrow. Leave your queries here, if unanswered I would check those tomorrow :)

Thanks,

Answer 11

you couldn't notice x=-1 was a solution to f=0
then you could have used synthetic division in the beginning to get this
just go through the factors of 2
\[\pm1 ,\pm2\]

Answer 12

sometimes this doesn't always work out
and we get disappointed because we don't have any rational roots
but this problem we have all rational roots which is crazy cool