Question

How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given `f(x)=x^3-2x^2-8x`?

Answer 1

The zeros of `f(x)` are `0`, `-2` and `4`

Explanation:

Given:

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

Note that the signs of the coefficients of `f(x)` are in the pattern `+ - -`. With one change of sign, Descartes' Rule of Signs tells us that `f(x)` has exactly one positive real zero.

The signs of the coefficients of `f(-x)` are in the pattern `- - +`. With one change of sign, Descartes' Rule of Signs tells us that `f(x)` has exactly one negative real zero.

In order to usefully apply the rational zeros theorem we need a non-zero constant term, but `f(x)` has none, or in other words it is zero. That's because `x=0` is a zero of `f(x)`. Let's separate `x` out as a factor first:

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

Now applying the rational zeros theorem to `x^2-2x-8`, any rational zeros of this quadratic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-8` and `q` a divisor of the coefficient `1` of the leading term.

That means that its only possible rational zeros are:

`+-1, +-2, +-4, +-8`

Trying these, we find:

`(color(blue)(-2))^2-2(color(blue)(-2))-8 = 4+4-8 = 0`

`(color(blue)(4))^2-2(color(blue)(4))-8 = 16-8-8 = 0`

So the zeros of `f(x)` are `0`, `-2` and `4`