Question

What are all the possible rational zeros for `f(x)=4x^3-9x^2+6x-1` and how do you find all zeros?

Answer 1

`f(x)` has "possible" zeros:

`+-1/4`, `+-1/2`, `+-1`

and actual zeros:

`1/4, 1, 1`

Explanation:

Given:

`f(x) = 4x^3-9x^2+6x-1`

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 `-1` and `q` a divisor of the coefficient `4` of the leading term.

That means that the only possible rational zeros are:

`+-1/4, +-1/2, +-1`

Here's a non-standard way of finding the zeros:

I notice that the coefficients are all fairly small and the signs alternate.

Taking them in reverse order and stringing them together we get:

`1694 = 2*7*11*11 = 14*11*11`

Hence we find:

`t^3+6t^2+9t+4 = (t+4)(t+1)(t+1)`

Then putting `t = -1/x` we find:

`4x^3-9x^2+6x-1 = (4x-1)(x-1)(x-1)`

So `f(x)` has zeros `1/4, 1, 1`