Question

Find all the real roots of the polynomial.
x^3+x^2-27x-27

Why isn't the answer -1?

Answer 1

it equals to zero so why is it wrong?

Answer 2

One of the roots is definitely -1.

x^3 + x^2 - 27x - 27

-1 is a root, so it's divisible by (x+1)

factoring:

(x+1)(x^2 - 27)

x^2 - 27 = 0

x= +-3sqrt3

Answer 3

thats wrong

Answer 4

use the rational roots test... r/s

Answer 5

when u do, u get +-1, +-3,+-9,+-27

Answer 6

and when u plug those into the function, the only one that equals to zero is -1

Answer 7

That will not give you all the roots.

http://www.wolframalpha.com/input/?i=x^3%2Bx^2-27x-27

wolframalpha seems to agree with me. So I don't think I'm wrong.

Answer 8

i put -+ \[\sqrt{3}\]

Answer 9

i mean -+ 3 sqrt{3}

Answer 10

and it said im wrong on my hw site

Answer 11

Then your hw site is wrong.

Answer 12

doubt it

Answer 13

I don't. Or it's user error. wolframalpha > your hw site

Answer 14

ok i just did a very similar problem the exact same way i did this one and got the right answer -_-

Answer 15

i have no idea what im doing wrong for this one

Answer 16

maybe you have the wrong input to the site.

Answer 17

The problem with using the rational roots test is that \(\sqrt{3}\) is not a rational root, it's an irrational one.

Answer 18

Try putting in sqrt{3} and -sqrt{3} separately on the site, perhaps?

Answer 19

thats not going to do it

Answer 20

my hw website gives u the option to use the square root sign already... i use webassign btw

Answer 21

by the way the answer was -1, sqrt{27}, and - sqrt{27}

Answer 22

sqrt27 = 3sqrt3

Answer 23

i know but it didnt accept it the way you had it so i tried every other way until it accepted it

Answer 24

Ouch. That's pretty retarded. I'm sorry to hear the system isn't smart enough to understand that :/