# What are the roots of the equation x^3 +4x^2-4x- 16=0?

Dec 18, 2016

The roots are:

$x = 2$, $x = - 2$ and $x = - 4$

#### Explanation:

The difference of squares identity can be written:

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

We use this with $a = x$ and $b = 2$ later.

Given:

${x}^{3} + 4 {x}^{2} - 4 x - 16 = 0$

Note that the ratio between the first and second terms is the same as the ratio between the third and fourth terms, so this cubic factors by grouping:

$0 = {x}^{3} + 4 {x}^{2} - 4 x - 16$

$\textcolor{w h i t e}{0} = \left({x}^{3} + 4 {x}^{2}\right) - \left(4 x + 16\right)$

$\textcolor{w h i t e}{0} = {x}^{2} \left(x + 4\right) - 4 \left(x + 4\right)$

$\textcolor{w h i t e}{0} = \left({x}^{2} - 4\right) \left(x + 4\right)$

$\textcolor{w h i t e}{0} = \left({x}^{2} - {2}^{2}\right) \left(x + 4\right)$

$\textcolor{w h i t e}{0} = \left(x - 2\right) \left(x + 2\right) \left(x + 4\right)$

Hence the roots are:

$x = 2$, $x = - 2$ and $x = - 4$

Dec 18, 2016

$x \in \left\{2 , - 4 , - 2\right\}$

#### Explanation:

${x}^{3} + 4 {x}^{2} - 4 x - 16 = 0$

$\Rightarrow {x}^{3} + 4 {x}^{2} = 4 x + 16$

$\rightarrow x \cdot {x}^{2} + 4 \cdot {x}^{2} = x \cdot {2}^{2} + 4 \cdot {2}^{2}$

$\rightarrow$(at least one possible solution is) $x = 2$
(i.e. $\left(x - 2\right)$ is a factor)

Dividing ${x}^{3} + 4 {x}^{2} - 4 x - 16$ by $\left(x - 2\right)$ (using either synthetic or long division) gives:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 6 x + 8$
which factors using standard operations as:
$\textcolor{w h i t e}{\text{XXX}} \left(x + 4\right) \left(x + 2\right)$

Therefore
${x}^{3} + 4 {x}^{2} - 4 x - 16 = 0$
$\Rightarrow \left(x - 2\right) \left(x + 4\right) \left(x + 2\right) = 0$

$\Rightarrow x \in \left\{2 , - 4 , - 2\right\}$