How we can find the non-integer roots of the equation&nbsp;x3=3x2+6x+2?

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How we can find the non-integer roots of the equation&amp;nbsp;x3=3x2+6x+2?

Michaela Boyle · Accepted Answer

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 means:
$- 1 - 3 + 6 - 2 = 0$
Thus we can tell that $x = - 1$ is a root and $(x + 1)$ a factor:
$x^{2} - 3 x^{2} - 6 x - 2 = (x + 1) (x^{2} - 4 x - 2)$
By completing the square and applying the identity derived from the difference of squares, we can determine the zeros of the remaining quadratic:
$a^{2} - b^{2} = (a - b) (a + b)$
with
$a = (x - 2)$
and
$b = \sqrt{6}$
$x^{2} - 4 x - 2 = x^{2} - 4 x + 4 - 6$
$= {(x - 2)}^{2} - {(\sqrt{6})}^{2}$
$= ((x - 2) - \sqrt{6}) ((x - 2) + \sqrt{6})$
$= (x - 2 - \sqrt{6}) (x - 2 + \sqrt{6})$
Thus zeros:
$x = 2 \pm \sqrt{6}$

How do I find the non-integer roots of the equation

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