If x = 2 + sqrt{3},then what is the value of | vedclass

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(C) Given $x = 2 + \sqrt{3}$,so $x - 2 = \sqrt{3}$.Squaring both sides,we get $(x - 2)^2 = 3$,which simplifies to $x^2 - 4x + 4 = 3$,or $x^2 - 4x + 1

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$-9$

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(C) Given $x = 2 + \sqrt{3}$,so $x - 2 = \sqrt{3}$.Squaring both sides,we get $(x - 2)^2 = 3$,which simplifies to $x^2 - 4x + 4 = 3$,or $x^2 - 4x + 1 = 0$.Now,we divide the polynomial $x^3 - 7x^2 + 13x - 12$ by $(x^2 - 4x + 1)$ or rewrite it:$x^3 - 7x^2 + 13x - 12 = x(x^2 - 4x + 1) - 3x^2 + 12x - 12$$= x(x^2 - 4x + 1) - 3(x^2 - 4x + 1) - 9$Since $x^2 - 4x + 1 = 0$,the expression becomes $x(0) - 3(0) - 9 = -9$.

If $x = 2 + \sqrt{3}$,then what is the value of $x^3 - 7x^2 + 13x - 12$?

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