Question

How do you find all the zeros of `f(x)=2x^3-x^2-3x-1`?

Answer 1

Zeros: `-1/2` and `1/2+-sqrt(5)/2`

Explanation:

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

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 the constant term `-1` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros of `f(x)` are:

`+-1/2, +-1`

We find:

`f(-1/2) = 2(-1/8)-(1/4)-3(-1/2)-1 = -1/4-1/4+3/2-1 = 0`

So `x=-1/2` is a zero and `(2x+1)` a factor:

`2x^3-x^2-3x-1`

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

`= (2x+1)((x-1/2)^2-5/4)`

`= (2x+1)((x-1/2)^2-(sqrt(5)/2)^2)`

`= (2x+1)(x-1/2-sqrt(5)/2)(x-1/2+sqrt(5)/2)`

Hence the other two zeros are:

`x = 1/2+-sqrt(5)/2`