Question

Find all the rational roots of the polynomial.
x^3-3x^2-2x+6

I kind of know how to do this but I'm still confused about how to do this kind of problem.

Answer 1

I know you use r/s

Answer 2

and
r=factors of constant
and
s=factors of leading coefficient

Answer 3

(x^2-2)(x-3)

Answer 4

x = 3
x = SQRT(2)
x = -SQRT(2)

Answer 5

f(3) = 27 - 27 - 6 + 6 = 0

so (x-3) is a factor

diiding by x-3 gives (x^2 - 2)

so rots are 3 , sqrt2 and - sqrt2

Answer 6

That is "trial and error". Better to look for factors of 6 such that you get -3x^2 and -2x.

Answer 7

i used factors of 6 : first i tried 2 than 3

Answer 8

If you do, then you get (x^2-2)(x-3) = 0. That shows 3 is a factor.

Answer 9

those arent the answers according to my hw site btw...

Answer 10

What are the answers according to your site?

Answer 11

It doesn't give me the answers... I have 100 tries. lol

Answer 12

It just said it's wrong

Answer 13

\[x^2(x-3)-2(x-3)=0\]
\[(x-3)(x^2-2)=0\]
trail factors not needed

Answer 14

Apparently, 3, sqrt(2) and -sqrt(2) are wrong answers. I can't figure out how they can be wrong.

Answer 15

trail factors?

Answer 16

"trial" I think

Answer 17

did you post the question right mario?

Answer 18

yes

Answer 19

and entering in:
\[3, \sqrt{2}, -\sqrt{2}\]
is not showing you are correct?
Does it say anything else like round answers to nearest tenths or whatever?