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If ${(1 + x)}^{1 + x^{2} + 2 x} = 16$ , then x =

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answer is B.

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Detailed Solution

x = 0 \Rightarrow {(1 + x)}^{1 + x^{2} + 2 x} = 16 \Rightarrow {(1 + 0)}^{1 + 0^{2} + 2 (0)} = 16 \Rightarrow 1 \neq 16

If $x = 1 \Rightarrow {(1 + x)}^{1 + x^{2} + 2 x} = 16 \Rightarrow {(1 + 1)}^{1 + 1^{2} + 2 (1)} = 16 \Rightarrow {(2)}^{4} = 16$

If $x = - 1 \Rightarrow {(1 + x)}^{1 + x^{2} + 2 x} = 16 \Rightarrow {(1 - 1)}^{1 + {(- 1)}^{2} + 2 (- 1)} = 16 \Rightarrow 0 \neq 16$

If $x = - 2 \Rightarrow {(1 + x)}^{1 + x^{2} + 2 x} = 16 \Rightarrow {(1 - 2)}^{1 + {(- 2)}^{2} + 2 (- 2)} = 16 \Rightarrow - 1 \neq 16$

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