# What are the roots of x^3-8x-3=0?

Mar 24, 2017

Three roots of ${x}^{3} - 8 x - 3 = 0$ are $3$, $\frac{- 3 + \sqrt{5}}{2}$ and $\frac{- 3 - \sqrt{5}}{2}$

#### Explanation:

in an equation of type ${x}^{n} + {a}_{1} {x}^{n - 1} + {a}_{2} {x}^{n - 2} + \ldots \ldots \ldots + {a}_{n}$, the roots of the equation are factors of ${a}_{n}$.

Here, we have the equation as ${x}^{3} - 8 x - 3 = 0$ and hence roots are factors of $3$ i.e. ${+}_{1}$ and $\pm 3$.

It is apparent that $x = 3$ is a root as it satisfies the equation

${3}^{3} - 8 \times 3 - 3 = 27 - 24 - 3 = 0$ and hence

$x - 3$ is a factor of ${x}^{3} - 8 x - 3$ and dividing latter by former

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

As such we have $\left(x - 3\right) \left({x}^{2} + 3 x + 1\right) = 0$

And as ${x}^{2} + 3 x + 1$ cannot be factorized as rational factors, using quadratic formula roots are

$\frac{- 3 \pm \sqrt{{3}^{2} - 4 \times 1 \times 1}}{2} = \frac{- 3 \pm \sqrt{5}}{2}$

and hence three roots of ${x}^{3} - 8 x - 3 = 0$ are $3$, $\frac{- 3 + \sqrt{5}}{2}$ and $\frac{- 3 - \sqrt{5}}{2}$