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

Mar 10, 2018

$f \left(x\right)$ has zeros $x = 2$ and $x = 1 \pm i$

Explanation:

Given:

$f \left(x\right) = {x}^{3} - 4 {x}^{2} + 6 x - 4$

Rational root theorem

By the rational root theorem any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ 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:

$\pm 1 , \pm 2 , \pm 4$

Descartes' Rule of Signs

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

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

So the only possible rational zeros are the positive ones:

$1 , 2 , 4$

We find:

$f \left(2\right) = 8 - 16 + 12 - 4 = 0$

So $x = 2$ is a zero and $\left(x - 2\right)$ a factor:

${x}^{3} - 4 {x}^{2} + 6 x - 4 = \left(x - 2\right) \left({x}^{2} - 2 x + 2\right)$

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

${x}^{2} - 2 x + 2 = {x}^{2} - 2 x + 1 + 1$

$\textcolor{w h i t e}{{x}^{2} - 2 x + 2} = {\left(x - 1\right)}^{2} - {i}^{2}$

$\textcolor{w h i t e}{{x}^{2} - 2 x + 2} = \left(\left(x - 1\right) - i\right) \left(\left(x - 1\right) + i\right)$

$\textcolor{w h i t e}{{x}^{2} - 2 x + 2} = \left(x - 1 - i\right) \left(x - 1 + i\right)$

So the remaining zeros are:

$x = 1 \pm i$