Question

How do you find all the real and complex roots of `3x^4+8x^3+6x^2+3x-2`?

Answer 1

`x = 1/3`, `x = -2`, `x = -1/2+-sqrt(3)/2i`

Explanation:

`f(x) = 3x^4+8x^3+6x^2+3x-2`

By the rational root theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of hte constant term `-2` and `q` a divisor of the coefficient `3` of the leading term.

So the only possible rational zeros are:

`+-1/3`, `+-2/3`, `+-1`, `+-2`

We find:

`f(1/3) = 1/27+8/27+2/3+1-2 = 0`

`f(-2) = 48-64+24-6-2 = 0`

So `x = 1/3` and `x=-2` are zeros and `(3x-1)` and `(x+2)` are factors:

`3x^4+8x^3+6x^2+3x-2`

`= (3x-1)(x^3+3x^2+3x+2)`

`= (3x-1)(x+2)(x^2+x+1)`

The zeros of `x^2+x+1` are the Complex cube roots of `1`, since `(x-1)(x^2+x+1) = x^3-1`

We can find them using the quadratic formula or whatever your preferred method is to get:

`x = -1/2+-sqrt(3)/2i`