# How do you find the zeros of f(x)=x^3+3x^2+6x+4?

Jan 17, 2017

$f \left(x\right)$ has zeros $- 1$ and $- 1 \pm \sqrt{3} i$

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} + 3 {x}^{2} + 6 x + 4$

Note that $- 1 + 3 - 6 + 4 = 0$, so $f \left(- 1\right) = 0$, $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{3} + 3 {x}^{2} + 6 x + 4 = \left(x + 1\right) \left({x}^{2} + 2 x + 4\right)$

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x + 1\right)$ and $b = \sqrt{3} i$ as follows:

${x}^{2} + 2 x + 4 = {x}^{2} + 2 x + 1 + 3$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 4} = {\left(x + 1\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 4} = \left(\left(x + 1\right) - \sqrt{3} i\right) \left(\left(x + 1\right) + \sqrt{3} i\right)$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 4} = \left(x + 1 - \sqrt{3} i\right) \left(x + 1 + \sqrt{3} i\right)$

Hence the other two zeros are:

$x = - 1 \pm \sqrt{3} i$