Question

What are all the possible rational zeros for `f(x)=3x^3-11x^2+5x+3` and how do you find all zeros?

Answer 1

`f(x)` has zeros `1`, `-1/3` and `3`

Explanation:

`f(x) = 3x^3-11x^2+5x+3`

By the rational roots theorem, any rational zero of `f(x)` is expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `3` 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`

Note first that there's a shortcut in this particular example, in that the sum of the coefficients is `0`. That is `3-11+5+3 = 0`

Hence we can deduce that `f(1) = 0`, `x=1` is a zero of `f(x)` and `(x-1)` is a factor:

`3x^3-11x^2+5x+3 = (x-1)(3x^2-8x-3)`

We could just try the remaining rational possibilities that we listed, but we can factor the remaining quadratic (`3x^2-8x-3`) easily enough using an AC method.

Find a pair of factors of `AC = 3*3 = 9` which differ by `B=8`.

The pair `9, 1` works in that `9*1 = 9` and `9-1 = 8`

Use this pair to split the middle term and factor by grouping:

`3x^2-8x-3 = 3x^2-9x+x-3`

`color(white)(3x^2-8x-3) = (3x^2-9x)+(x-3)`

`color(white)(3x^2-8x-3) = 3x(x-3)+1(x-3)`

`color(white)(3x^2-8x-3) = (3x+1)(x-3)`

Hence the other two zeros of `f(x)` are `x=-1/3` and `x=3`