# How do you find all the roots of f(x) = x^4 + 2x^3 + x^2 - 2x - 2?

May 23, 2016

This quartic polynomial has zeros $\pm 1$ and $- 1 \pm i$.

#### Explanation:

First note that the sum of the coefficients is $0$. That is, $1 + 2 + 1 - 2 - 2 = 0$. Hence we can deduce that $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

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

If you reverse the signs of the coefficients of the terms of odd degree in the remaining cubic $\left({x}^{3} + 3 {x}^{2} + 4 x + 2\right)$ you get: $- 1 + 3 - 4 + 2 = 0$. Hence $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

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

The remaining quadratic factor has negative discriminant, but you can factor it by completing the square with Complex coefficients:

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

Hence zeros $x = - 1 \pm i$