The value of $4 x^3-4 x^2-7 x+127$ when $x=\frac{4+5 \sqrt{-1}}{2}$ is

1. $1$
2. $2$
3. $3$
4. $4$

Solution

$\because x=\frac{4+5 \sqrt{-1}}{2} \Rightarrow 2 x=4+5 \sqrt{-1}$ ...(i) $(2 x)^2=(4+5 \sqrt{-1})^2$ $\Rightarrow 4 x^2=16-25+40 \sqrt{-1}$ $\Rightarrow 4 x^2=-9+40 \sqrt{-1}$ ...(ii) Equation (ii) $-8 \times$ (i) $4 x^2-16 x=-41 \Rightarrow 4 x^2-16 x+41=0$ ...(iii) Now, $4 x^3-4 x^2-7 x+127$ $=x \cdot 4 x^2-4 x^2-7 x+127$ $=x(16 x-41)-4 x^2-7 x+127 \quad\{$ from (iii) $\}$ $\begin{aligned} & =12 x^2-48 x+127 \\ & =3\left(4 x^2-16 x\right)+127\end{aligned}$
$=4$

Asked in: AP EAMCET 2023 (18 May Shift 1)

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