Question

How do you find all the zeros of `P(x) = 3x^3 – 6x^2 – 3x – 18`?

Answer 1

`P(x)` has zeros: `3` and `-1/2+-sqrt(7)/2i`

Explanation:

`P(x) = 3x^3-6x^2-3x-18 = 3(x^3-2x^2-x-6)`

By the rational root theorem, the only possible rational zeros of `x^3-2x^2-x-6` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constnat term `-6` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-3, +-6`

Substituting `x=3` we find:

`x^3-2x^2-x-6=(3)^3-2(3)^2-(3)-6=27-18-3-6 = 0`

So `x=3` is a zero and `(x-3)` a factor:

`x^3-2x^2-x-6`

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

`=(x-3)((x+1/2)^2+7/4)`

`=(x-3)((x+1/2)^2-(sqrt(7)/2i)^2)`

`=(x-3)(x+1/2-sqrt(7)/2i)(x+1/2+sqrt(7)/2i)`

So the other two zeros are:

`x = -1/2+-sqrt(7)/2i`