Question

What are all the possible rational zeros for `f(x)=x^3-4x^2-3x+14` and how do you find all zeros?

Answer 1

`f(x)` has zeros `2` and `1+-2sqrt(2)`

Explanation:

`f(x) = x^3-4x^2-3x+14`

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

That means that the only possible rational zeros are:

`+-1, +-2, +-7, +-14`

Trying each in turn, we find:

`f(2) = 8-4(4)-3(2)+14 = 8-16-6+14 = 0`

So `x=2` is a zero and `(x-2)` a factor:

`x^3-4x^2-3x+14 = (x-2)(x^2-2x-7)`

We can factor the remaining quadratic by completing the square:

`x^2-2x-7`

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

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

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

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

Hence zeros:

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