Evaluate the following: x^4-4x^3+4x^2+8x+44, when x=3+2i

Question

Evaluate the following:
$\text{x}^4-4\text{x}^3+4\text{x}^2+8\text{x}+44,$ when $\text{x}=3+2\text{i}$

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Answer

We have,
$\text{x}=3+2\text{i}$
$\Rightarrow\text{x}-3=2\text{i}$
$\Rightarrow(\text{x}-3)^2=(2\text{i})^2$
$\Rightarrow\text{x}^2+3^2-2\times3\times\text{x}=-4$
$\Rightarrow\text{x}^2+9-6\text{x}+4=0$
$\Rightarrow\text{x}^2-6\text{x}+13=0 \ ...\text{(i)}$
Now,
$\text{x}^4-4\text{x}^3+4\text{x}^2+8\text{x}+44$
$=\text{x}^2(\text{x}^2-6\text{x}+13)+6\text{x}^2-13\text{x}^2-4\text{x}^3+4\text{x}^2+8\text{x}+44$ (adding and subtracting $6x^3$ and $13x^2$)
$=\text{x}^2\times0+2\text{x}^3-9\text{x}^2+8\text{x}^2+44$ (using (i))
$2\text{x}(\text{x}^2-6\text{x}+13)+12\text{x}^2-26\text{x}-9\text{x}^2+8\text{x}+44$ (adding and subtracting $12x^3 $and $26x^2$)
$2\text{x}\times0+3\text{x}^2-18\text{x}+44$ (using (i))
$=3(\text{x}^2-6\text{x}+13)+5$
$=3\times0+5$ (using (i))
$=5$

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