Question

What are all the possible rational zeros for `f(x)=3x^3-7x^2+29x-9` and how do you find all zeros?

Answer 1

The "possible" rational zeros are:

`+-1/3, +-1, +-3, +-9`

The actual zeros are:

`1/3` and `1+-2sqrt(2)i`

Explanation:

`f(x) = 3x^3-7x^2+29x-9`

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 fo the constant term `-9` and `q` a divisor of the coefficient `3` of the leading term.

That means that the only possible rational zeros are:

`+-1/3, +-1, +-3, +-9`

Note also that the signs of the coefficients of `f(x)` are in the pattern: `+ - + -`. By Descartes's rule of signs, since this has `3` changes of sign, `f(x)` must have `3` or `1` positive real zero. Note that `f(-x) = -3x^3-7x^2-29x-9` has all negative coefficients, so `f(x)` has no negative real zeros.

So the only possible rational real zeros are:

`1/3, 1, 3, 9`

Trying each in turn we immediately find:

`f(1/3) = 3(1/27)-7(1/9)+29(1/3)-9 = (1-7+87-81)/9 = 0`

So `x=1/3` is a zero and `(3x-1)` a factor:

`3x^3-7x^2+29x-9 = (3x-1)(x^2-2x+9)`

We can factor the remaining quadratic by completing the square, but it does require some Complex coefficients:

`x^2-2x+9 = x^2-2x+1+8`

`color(white)(x^2-9x+9) = (x-1)^2-(2sqrt(2)i)^2`

`color(white)(x^2-9x+9) = ((x-1)-2sqrt(2)i)((x-1)+2sqrt(2)i)`

`color(white)(x^2-9x+9) = (x-1-2sqrt(2)i)(x-1+2sqrt(2)i)`

Hence the other two zeros are:

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