Question

what are the zeroes of the equation x^3-x^2-3x-3? I know becs. only one sign change there is 1 negative real root which is -1, but are the other 2 roots +- isq rt. 3 or just +-sq rt. 3

Answer 1

-1 is not a root.

Answer 2

If there is one sign change in f(x) then that means there is 1 positive real root. The other 2 roots are either negative or complex.

Answer 3

oh-I had it backwards. so the real root is +1 but what are the other 2 roots?

Answer 4

+1 is not a root either.

Answer 5

my 4 choices for answers all have either - or + 1 as a zero of equation

Answer 6

Are you sure you wrote the right equation?

Answer 7

this is how it originally appeared and I set it = to 0 x^3-x^2=3x+3

Answer 8

\[x^3-x^2-3x-3\]
Has no rational roots.
\[x^3+x^2-3x-3\]
However is factorable . . .

Answer 9

I couldn't factor it out becs. my negative and positive didn't come out right

Answer 10

Yeah. Possible typo in the problem statement?

Answer 11

that's what I am thinking

Answer 12

Here is what the graph of x^3-x^2-3x-3 looks like

Answer 13

and that's even an option as an answer

Answer 14

not even an option

Answer 15

Ugh, that drawing came out terrible . . . :"<
But you get the idea.

If your choices are -1 or +1 then the equation must have been
x^3+x^2-3x-3 which factors to (x^2-3)(x+1)
or
x^3-x^2+3x-3 which factors to (x^2+3)(x-1)

Answer 16

If it's a typo, then you're just going to have to guess at what it was supposed to be.

Answer 17

do both equations have i sq rt 3 as a zero?

Answer 18

x^3+x^2-3x-3 has ±√3
x^3-x^2+3x-3 has ±i√3

Answer 19

thanks I will just take a guess

Answer 20

You should complain to the proper authorities.

Answer 21

Tha would my teacher