# How do I find the non-integer roots of the equation x^3=3x^2+6x+2 ?

Sep 5, 2016

The non-integer roots are:

$x = 2 \pm \sqrt{6}$

#### Explanation:

First subtract the right hand side of the equation from the left, to get it into standard form:

${x}^{3} - 3 {x}^{2} - 6 x - 2 = 0$

Note that if we reverse the signs of the coefficient on terms of odd degree then the sum is zero. That is:

$- 1 - 3 + 6 - 2 = 0$

Hence we can tell that $x = - 1$ is a root and $\left(x + 1\right)$ a factor:

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

We can find the zeros of the remaining quadratic by completing the square and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x - 2\right)$ and $b = \sqrt{6}$ as follows:

${x}^{2} - 4 x - 2 = {x}^{2} - 4 x + 4 - 6$

$\textcolor{w h i t e}{{x}^{2} - 4 x - 2} = {\left(x - 2\right)}^{2} - {\left(\sqrt{6}\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - 4 x - 2} = \left(\left(x - 2\right) - \sqrt{6}\right) \left(\left(x - 2\right) + \sqrt{6}\right)$

$\textcolor{w h i t e}{{x}^{2} - 4 x - 2} = \left(x - 2 - \sqrt{6}\right) \left(x - 2 + \sqrt{6}\right)$

Hence zeros:

$x = 2 \pm \sqrt{6}$