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

Dec 25, 2016

The "possible" rational zeros are: $\pm 1 , \pm 2 , \pm 4$

The actual zeros are: $4$, $- 1 + \sqrt{2}$, $- 1 - \sqrt{2}$

#### Explanation:

$f \left(x\right) = {x}^{3} - 6 {x}^{2} + 7 x + 4$

By the rational roots 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:

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

Trying each in turn, we eventually find:

$f \left(4\right) = 64 - 96 + 28 + 4 = 0$

So $x = 4$ is a zero and $\left(x - 4\right)$ a factor:

${x}^{3} - 6 {x}^{2} + 7 x + 4 = \left(x - 4\right) \left({x}^{2} - 2 x - 1\right)$

The remaining quadratic can be factored by completing the square:

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

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

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

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

Hence the other two zeros are:

$x = - 1 \pm \sqrt{2}$