The value of 4x^3 - 4x^2 - 7x + 127 when x = | vedclass

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Question

(D) Given $x = \frac{4 + 5i}{2}$ $\Rightarrow 2x = 4 + 5i$ $\Rightarrow 2x - 4 = 5i$. Squaring both sides: $(2x - 4)^2 = (5i)^2$ $\Rightarrow 4x^2 - 1

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$4$

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(D) Given $x = \frac{4 + 5i}{2}$ $\Rightarrow 2x = 4 + 5i$ $\Rightarrow 2x - 4 = 5i$. Squaring both sides: $(2x - 4)^2 = (5i)^2$ $\Rightarrow 4x^2 - 16x + 16 = -25$ $\Rightarrow 4x^2 - 16x + 41 = 0$. Now,we divide the polynomial $4x^3 - 4x^2 - 7x + 127$ by $4x^2 - 16x + 41$: $4x^3 - 4x^2 - 7x + 127 = x(4x^2 - 16x + 41) + 12x^2 - 48x + 127$ $= x(0) + 3(4x^2 - 16x) + 127$ $= 3(-41) + 127$ $= -123 + 127 = 4$.

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

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