Alg2.1.18

Find the value of $4x^3+2x^2-8x+7 if x=\frac{\sqrt{3}+1}{2}\,$

Substituting the x value in the given expression

$4x^3+2x^2-8x+7=4(\frac{\sqrt{3}+1}{2})^{3}+2(\frac{\sqrt{3}+1}{2})^{2}-8(\frac{\sqrt{3}+1}{2})+7\,$

Expanding the cubes and squares in the equation,it looks like,

$4(\frac{3\sqrt{3}+1+9+3\sqrt{3}}{8})+2(\frac{3+1+2\sqrt{3}}{4})-8(\frac{\sqrt{3}+1}{2})+7\,$

$(\frac{6\sqrt{3}+10}{2})+(\frac{4+2\sqrt{3}}{2})-4\sqrt{3}-1+7\,$

Taking out the common terms and cancelling them, we get

$3\sqrt{3}+5+2+\sqrt{3}-4\sqrt{3}+6\,$

$4\sqrt{3}+5+2+\sqrt{3}-4\sqrt{3}+6\,$

Simplifying we get

$4x^3+2x^2-8x+7=13\,$