Question

What are all the possible rational zeros for `f(x)=x^3-6x^2+7x+4` and how do you find all zeros?

Answer 1

The "possible" rational zeros are: `+-1, +-2, +-4`

The actual zeros are: `4`, `-1+sqrt(2)`, `-1-sqrt(2)`

Explanation:

`f(x) = x^3-6x^2+7x+4`

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

That means that the only possible rational zeros are:

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

Trying each in turn, we eventually find:

`f(4) = 64-96+28+4 = 0`

So `x=4` is a zero and `(x-4)` a factor:

`x^3-6x^2+7x+4 = (x-4)(x^2-2x-1)`

The remaining quadratic can be factored by completing the square:

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

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

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

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

Hence the other two zeros are:

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