Iffind the value of

Solution:

Given, $x=\frac{\sqrt{3}+1}{2}$, and given to find the value of $4 x^{3}+2 x^{2}-8 x+7$

$2 x=\sqrt{3}+1$

$2 x-1=\sqrt{3}$

Now, squaring on both the sides, we get, $(2 x-1)^{2}=3$

$4 x^{2}-4 x+1=3$

$4 x^{2}-4 x+1-3=0$

$4 x^{2}-4 x-2=0$

$2 x^{2}-2 x-1=0$

Now taking $4 x^{3}+2 x^{2}-8 x+7$

$2 x\left(2 x^{2}-2 x-1\right)+4 x^{2}+2 x+2 x^{2}-8 x+7$

$2 x\left(2 x^{2}-2 x-1\right)+6 x^{2}-6 x+7$

As, $2 x^{2}-2 x-1=0$

$\left.2 x(0)+3\left(2 x^{2}-2 x-1\right)\right)+7+3$

0 + 3(0) + 10

10