How do you find all solutions to x^3+1=0?

1 Answer
Aug 16, 2016

$x = - 1 \mathmr{and} \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

Explanation:

Using synthetic division and the fact that $x = - 1$ is obviously a solution we find that we can expand this to:

$\left(x + 1\right) \left({x}^{2} - x + 1\right) = 0$

In order to have LHS=RHS need one of the brackets to be equal to zero, ie

$\left(x + 1\right) = 0 \text{ } \textcolor{b l u e}{1}$

$\left({x}^{2} - x + 1\right) = 0 \text{ } \textcolor{b l u e}{2}$

From $1$ we note that $x = - 1$ is a solution. We shall solve $2$ using the quadratic formula:

${x}^{2} - x + 1 = 0$

$x = \frac{1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(1\right) \left(1\right)}}{2} = \frac{1 \pm \sqrt{- 3}}{2} = \frac{1 \pm \sqrt{3} i}{2}$