Question

Help me please. Algebra 2/Trig

Answer 1

Find the solutions for the equation.
\[3x^3-x^2-9x+3=0\]

Answer 2

I've used Descartes Rule and the Rational Root Theorem already.

Answer 3

I know that there are 2 or 0 possible positive roots and 1 possible negative roots and that the possible roots are +/- 1, +/-3, and +/-\[\frac{ 1 }{ 3 }\]

Answer 4

3x^3-x^2-9x+3 = 0
(3x-1)(x^2 - 3) = 0

So

3x-1 = 0
x = 1/3
or

x^2 - 3 = 0
x^2 = 3
x = sqrt(3)

Answer 5

I tried using synthetic division but the problem seems to be a bit weird.

Answer 6

So the solutions are 1/3 and sqrt(3)?

Answer 7

There is one negative roots though...

Answer 8

yes

Answer 9

*root

Answer 10

yes

Answer 11

sqrt (3) is 1.73 or -1.73

Answer 12

the solusions are 1/3, 1.73, and -1.73

Answer 13

Ah, that makes sense.

Answer 14

\[x^2 - 3 = 0 \rightarrow x = \pm \sqrt{3}\]

Answer 15

Thank you.

Answer 16

or the solusions are the solusions are 1/3, sqrt(3), and -sqrt(3)

Answer 17

you are welcome :)