Question

What are all the possible rational zeros for `f(x)=x^3-13x^2+23x-11` and how do you find all zeros?

Answer 1

The "possible" rational zeros are: `+-1, +-11`

The actual zeros are: `1, 1, 11`

Explanation:

`f(x) = x^3-13x^2+23x-11`

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 `-11` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-11`

Note that the sum of the coefficients of `f(x)` is zero. That is:

`1-13+23-11 = 0`

Hence `f(1) = 0`, `x=1` is a zero and `(x-1)` a factor:

`x^3-13x^2+23x-11 = (x-1)(x^2-12x+11)`

The sum of the coefficients of the remaining quadratic is zero too:

`1-12+11 = 0`

So `x=1` is a zero again and `(x-1)` a factor:

`(x^2-12x+11) = (x-1)(x-11)`

So the zeros of `f(x)` are `x=1` with multiplicity `2` and `x=11`.