Solve the equations:
$x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 5 = 0$ , one root being $\sqrt{- 1}$ .

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Solution

$p (x) = x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 5 = 0$

If $i$ is one of the roots then $- i$ is also a root

$\Rightarrow {x - (i)} {x - (- i)}$ is a factor of $p (x)$

$\Rightarrow {(x)}^{2} - {(i)}^{2}$ is a factor of $p (x)$

$\Rightarrow x^{2} + 1$ is a factor off $p (x)$

Dividing $p (x)$ by $\Rightarrow x^{2} + 1$
$\Rightarrow p (x) = (x^{2} + 4 x + 5) (x^{2} + 1) \Rightarrow (x^{2} + 4 x + 5) (x^{2} + 1) = 0 \Rightarrow x^{2} + 4 x + 5 = 0 \Rightarrow x = \frac{- 4 \pm \sqrt{16 - 4 (1) (5)}}{2} \Rightarrow x = \frac{- 4 \pm \sqrt{- 4}}{2} \Rightarrow x = - 2 \pm i$

So the other two roots are $- 2 + i$ and $- 2 - i$

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Solve the equation $x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 5 = 0$ , given one root is $\sqrt{(- 1)}$ . Find the value of other three roots.

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x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1 = 0

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