For a brain-twist

we will first find an equation which has roots $x = \frac{\sqrt{3}\pm1}{2}$

We get the equation $2x^2-2x-1$ which has $\frac{\sqrt{3}\pm1}{2}$ as it's zeroes. $4x^2+2x^2-8x+7$ $= 2x(2x^2-2x-1) + 3(2x^2-2x-1)+10$ which gives us $\boxed{10}$ as value at $x=\frac{\sqrt{3}+1}{2}$

Seriously, both answers that I saw here seem to overcomplicate the problem.

This is ridiculously simple algebra.

Just plug x into that polynomial and simplify it.

$4\left(\frac{\sqrt{3}+1}{2}\right)^3+2\left(\frac{\sqrt{3}+1}{2}\right)^2-8\left(\frac{\sqrt{3}+1}{2}\right)+7$

$4\left(\frac{6\sqrt{3}+10}{8}\right)+2\left(\frac{2\sqrt{3}+4}{4}\right)-8\left(\frac{\sqrt{3}+1}{2}\right)+7$

$4\left(\frac{3\sqrt{3}+5}{4}\right)+2\left(\frac{\sqrt{3}+2}{2}\right)-4\left(\sqrt{3}+1\right)+7$

$3\sqrt{3}+5+\sqrt{3}+2-4\sqrt{3}-4+7$

$5+2-4+7$

$10$

$x^2=x+\frac{1}{2}$ $EXP=4(x+\frac{1}{2})x+2 \times (x+\frac{1}{2})-8x+7=4x^2-4x+8=4(x+\frac{1}{2})-4x+8=\boxed{10}$

We just need to put the value of $x$ in the equation.

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

The given equation is: $4x^{3}+2x^{2}-8x+7$

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

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

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

$= 4\times\frac{6\sqrt{3}+10}{8}+2\times\frac{2(2+\sqrt{3})}{4}-4\sqrt{3}-4+7$

$= 4\times\frac{2(3\sqrt{3}+5)}{8}+2\times\frac{2(2+\sqrt{3})}{4}-4\sqrt{3}-4+7$

$= 3\sqrt{3}+5+2+\sqrt{3}-4\sqrt{3}-4+7$

$= 5+2+7-4+3\sqrt{3}+\sqrt{3}-4\sqrt{3}$

$= 10+4\sqrt{3}-4\sqrt{3} = 10$

Thus, the answer is: $4x^{3}+2x^{2}-8x+7=\boxed{10}$

My solution may not be as simple as it could possibly be but I just found it 'most clean' (in my opinion). We just need some trigonometric identities:

1. Pythagorean, $\boxed{\sin^{2}\theta + \cos^{2}\theta = 1}$ , and

2. sine of double-angle, $\boxed{\sin(2\theta) = 2\sin\theta\cos\theta}$ .

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

which $\frac{\sqrt{3} + 1}{2} = \frac{\sqrt{3}}{2} + \frac{1}{2} = \sin \frac{\pi}{3} + \cos \frac{\pi}{3}$

If let $\boxed{m = \sin \frac{\pi}{3} , n = \cos \frac{\pi}{3}}$ Then $\boxed{x = m + n}$ .

Now let's temporarily forget about $x$ and do some tricks to $m$ and $n$ .

By the above referred identities, what we have in hands will be:

1. $\boxed{m^{2} + n^{2} = 1}$ , and

2. (a bit more tricky) From $\sin(2\theta) = 2\sin\theta\cos\theta$ . As we have $\frac{\pi}{3}$ as our main angle, luckily that $2(\frac{\pi}{3}) = 2\pi - \frac{\pi}{3}$ , that results into $\sin(2(\frac{\pi}{3})) = \sin\frac{\pi}{3}$ . So $\boxed{m = 2mn}$ .

Let's write power of $x$ in terms of $m$ and $n$ .

1. $\boxed{x = m + n}$

2. $x^{2} = (m + n)^{2} = m^{2} + 2mn + n^{2}$ . Then, $\boxed{x^{2} = m + 1}$

3. $x^{3} = x^{2}x = (m + 1)(m + n)$ . Then, $\boxed{x^{3} = m^{2} + m + mn + n}$

Now let's go to the problem:

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

Get rid of $x$ ,

$= 4(m^{2} + m + mn + n) + 2(m + 1) - 8(m + n) + 7$

$= 4m^{2} + 4m + 4mn + 4n + 2m + 2 - 8m - 8n + 7$

$= 4m^{2} - 2m + 4mn - 4n + 9$

$= 4m^{2} - 2m + 2m - 4n + 9$ (from $m = 2mn$ )

$= 4m^{2} - 4n + 9$

Finalize $m$ and $n$ ,

$= 4(\sin\frac{\pi}{3})^{2} - 4(\cos\frac{\pi}{3}) + 9$

$= 4(\frac{\sqrt{3}}{2})^{2} - 4(\frac{1}{2}) + 9$

$= 3 - 2 + 9$

$= 10$

I answered this in less than 1 minutes using my old calculator.

put the value of x everywhere in the question,use algebraic identities,you will get the answer

2 x = sqrt(3) + 1

=> (2 x - 1)^2 = 3

=> 4 x^2 = 4 x + 2

Therefore x (4 x^2 + 2 x - 8) + 7

= x (6 x - 6) + 7

= 6 x (x - 1) + 7

= 6 [sqrt(3) + 1][sqrt(3) - 1]/ 4 + 7

= (3/2)(3 - 1) + 7

= 3 + 7

=10

$x[(2x+1)^2 -9-2x]+7$

$x[(\sqrt{3}+2)^2 -9 -\sqrt{3} -1] +7$

$x(7+4\sqrt{3}-9-\sqrt{3}-1)+7$

$\frac{\sqrt{3}+1}{2} [3(\sqrt{3}-1)+7$

$2 \times \frac{3}{2} +7$

$=10$

This can look simpler with substitution. Let $y=\sqrt { 3 } +1$ . Then then the value you want is half of: ${ y }^{ 3 }+{ y }^{ 2 }-8y+14$ . You still have to crank out the algebra for a cube and a square but it's a bit easier without fractions and still comes to 10.