Question

What are all the possible rational zeros for `f(x)=x^3+x^2-8x-6` and how do you find all zeros?

Answer 1

Possible rational zeros: `+-1, +-2, +-3, +-6`

Actual zeros: `-3` and `1+-sqrt(3)`

Explanation:

Given:

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

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant 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`

Trying each in turn, we find:

`f(-3) = -27+9+24-6 = 0`

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

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

We can factor the remaining quadratic by completing the square:

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

`color(white)(x^2-2x-2) = (x-1)^2-(sqrt(3))^2`

`color(white)(x^2-2x-2) = ((x-1)-sqrt(3))((x-1)+sqrt(3))`

`color(white)(x^2-2x-2) = (x-1-sqrt(3))(x-1+sqrt(3))`

Hence the other two zeros of our cubic function are:

`x = 1+-sqrt(3)`