# How do you factor x^3 - 2x + 1?

It is

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

Now for the trinomial ${x}^{2} + x - 1$ the roots are

${x}_{1 , 2} = \frac{- {1}^{2} \left(\pm\right) \sqrt{{1}^{2} + 4}}{2} \implies {x}_{1} = \frac{- 1 + \sqrt{5}}{2} \mathmr{and} {x}_{2} = \frac{1 + \sqrt{5}}{2}$

So finally the factorization is

${x}^{3} - 2 x + 1 = - \frac{1}{4} \left(x - 1\right) \cdot \left(- 2 x + \sqrt{5} - 1\right) \cdot \left(2 x + \sqrt{5} + 1\right)$