Finding the roots help

Solve for the roots:
3x^4 + 2x^3 - 7x^2 + 4x - 2=0
I cannot not find the roots only 1 is the only root I found.
I appreciate for help. thanks

skip the synthetic and use regular long division.
start with x+1

.........3x^3.. -x^2 - 6x. -2
x +1 | 3x^4 + 2x^3 - 7x^2 + 4x - 2
-------3x^4 + 3x^3
-------------.. - x^3
bring down next
--------------- - x^3 - 7x^2
.................... - x^3 - x^2
----------------------- -6x^2 + 4x
............................ -6x^2 + 6x
-----------------------------...-2x -2


(x+1)(3x^3 -x^2 - 6x - 2)

do it again with x+1

_____3x^2__ -4x __ -2
x + 1 | 3x^3 - x^2 - 6x - 2
-------- 3x^3 + 3x^2
.................... -4x^2 - 6x
---------------- -4x^2 - 4x
.............................. -2x - 2
------------------------ -2x - 2

(x+1)(x+1)(3x^2 - 4x - 2)

Now that it's a quadratic, factor the last part with usual factoring process.

However, using a part of the quadratic formula we know that
sqrt ( -4^2 - 4(-4)(-2) ) = sqrt ( 16 - 32) = sqrt (-16) = 4i.
So the quadratic is solved in the complex number set and not in real numbers. So stop there.

(x+1)(x+1)(3x^2 - 4x - 2) is as far as we can go at this level of algebra.

If you know 1 is a root then use synthetic devision to reduce the 4th degree polynomial to a 3rd degree. Use further synthetic devision as needed.