x^4+4x^3+6x^2+4x+9, when x=-1+i2 - Vidyadip

Question

$\text{x}^4+4\text{x}^3+6\text{x}^2+4\text{x}+9,$ when $\text{x}=-1+\text{i}\sqrt{2}$

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Answer

We have
$\text{x}=-1+\text{i}\sqrt{2}$
$\Rightarrow\text{x}+1+\text{i}\sqrt{2}$
$\Rightarrow(\text{x}+1)^2=(\text{i}\sqrt{2})^2$ (squaring both sides)
$\Rightarrow\text{x}^2+1+2\text{x}=-2$
$\Rightarrow\text{x}^2+2\text{x}+3=0 \ ...(\text{i})$
Now,
$\text{x}^4+4\text{x}^3+6\text{x}^2+4\text{x}+9$
$=\text{x}^2(\text{x}^2+2\text{x}+3)+2\text{x}^3+3\text{x}^2+4\text{x}+9$
$=\text{x}^2\times0+2\text{x}(\text{x}^2+2\text{x}+3)-\text{x}^2-2\text{x}+9$ (using(i))
$=2\text{x}\times0-({\text{x}^2+2\text{x}+3})+3+9$ (using(i) and adding and subtracting 3)
$=-0+3+9$ (using(i))
$=12$

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