How do I find all real and complex zeros of x^3+4x^2+5x?

Sep 25, 2014

First set the expression equal to 0.

${x}^{3} + 4 {x}^{2} + 5 x = 0$

Factor out an $x$

$x \left({x}^{2} + 4 x + 5\right) = 0$

$x = 0$, this is one of the roots

Factor the polynomial $\implies \left({x}^{2} + 4 x + 5\right) \implies$Use the quadratic formula

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = 1 , b = 4 , \mathmr{and} c = 5$

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

Simplify

$x = \frac{- 4 \pm \sqrt{16 - 20}}{2}$

$x = \frac{- 4 \pm \sqrt{- 4}}{2}$

$x = \frac{- 4 \pm 2 i}{2}$

$x = - 2 \pm i \implies$2 complex roots

This function has 3 roots. One of the roots is real and other 2 roots are complex numbers.

The roots are $0 , - 2 + i , \mathmr{and} - 2 - i .$