If p(x) = x^{2} - 2sqrt{2}x + 1,then p(2sqrt{ | vedclass

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(B) Given the polynomial $p(x) = x^{2} - 2\sqrt{2}x + 1$.To find $p(2\sqrt{2})$,we substitute $x = 2\sqrt{2}$ into the polynomial expression.$p(2\sqrt

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$1$

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(B) Given the polynomial $p(x) = x^{2} - 2\sqrt{2}x + 1$.To find $p(2\sqrt{2})$,we substitute $x = 2\sqrt{2}$ into the polynomial expression.$p(2\sqrt{2}) = (2\sqrt{2})^{2} - 2\sqrt{2}(2\sqrt{2}) + 1$Calculating the terms:$(2\sqrt{2})^{2} = 4 	imes 2 = 8$$2\sqrt{2}(2\sqrt{2}) = 4 	imes 2 = 8$Substituting these back into the expression:$p(2\sqrt{2}) = 8 - 8 + 1$$p(2\sqrt{2}) = 1$Therefore,the correct option is $(b)$.

If $p(x) = x^{2} - 2\sqrt{2}x + 1$,then $p(2\sqrt{2})$ is equal to

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