Question

What are all the zeroes of `f(x)= x^3 -4x^2 +6x - 4`?

Answer 1

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

Explanation:

Given:

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

Rational root theorem

By the rational root 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`

Descartes' Rule of Signs

The pattern of signs of the coefficients of `f(x)` is `+ - + -`. With `3` changes of signs, Descartes' Rule of Signs tells us that `f(x)` has `3` or `1` positive real zeros.

The pattern of signs of the coefficients of `f(-x)` is `- - - -`. With no changes of signs, Descartes' Rule of Signs tells us that `f(x)` has no negative real zeros.

So the only possible rational zeros are the positive ones:

`1, 2, 4`

We find:

`f(2) = 8-16+12-4 = 0`

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

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

The zeros of the remaining quadratic are non-real complex, but we can find them by completing the square:

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

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

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

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

So the remaining zeros are:

`x = 1+-i`