# How do you solve x^ { 5} - 4x ^ { 4} - 3x ^ { 3} - 12x ^ { 2} - 4x + 16= 0?

How do you solve x^ { 5} - 4x ^ { 4} - 3x ^ { 3} - 12x ^ { 2} - 4x + 16= 0?

Answer 1

$\left(1\right) : \text{The Solution Set } \left[\subset \mathbb{R}\right] = \left\{\pm 2 , 4\right\}$.

$\left(2\right) : \text{The Solution Set } \left[\subset \mathbb{C}\right] = \left\{\pm 2 , 4 , \pm i\right\}$.

#### Explanation:

$\underline{{x}^{5} - 4 {x}^{4}} - \underline{3 {x}^{3} - 12 {x}^{2}} - \underline{4 x + 16} = 0$.

$\therefore {x}^{4} \left(x - 4\right) - 3 {x}^{2} \left(x - 4\right) - 4 \left(x - 4\right) = 0$.

$\therefore \left(x - 4\right) \left({x}^{4} - 3 {x}^{2} - 4\right) = 0$.

$\therefore \left(x - 4\right) \left\{\underline{{x}^{4} - 4 {x}^{2}} + \underline{{x}^{2} - 4}\right\} = 0$.

$\therefore \left(x - 4\right) \left\{\left({x}^{2} - 4\right) \left({x}^{2} + 1\right)\right\} = 0$.

$\therefore \left(x - 4\right) \left(x - 2\right) \left(x + 2\right) \left({x}^{2} + 1\right) = 0$.

Therefore, $\text{the Solution Set } \left[\subset \mathbb{R}\right] = \left\{\pm 2 , 4\right\}$.

Further, in $\mathbb{C} , {x}^{2} + 1 = 0 \Rightarrow {x}^{2} = - 1 \Rightarrow x = \pm i$.

So, $\text{the Solution Set } \left[\subset \mathbb{C}\right] = \left\{\pm 2 , 4 , \pm i\right\}$.

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