`x = (sqrt3+1)/2` , find `4x^3 +2x^2 -8x +7`

Options

A. 10
B. 4
C. 8
D. 6

Answer

To solve the expression \(4x^3 + 2x^2 - 8x + 7\) given that \(x = \frac{\sqrt{3}+1}{2}\), we will substitute the value of \(x\) into the expression and simplify step by step.

### Step 1: Substitute \(x\) into the expression
We start with the expression:
\[ 4x^3 + 2x^2 - 8x + 7 \]
Substituting \(x = \frac{\sqrt{3}+1}{2}\):
\[ 4\left(\frac{\sqrt{3}+1}{2}\right)^3 + 2\left(\frac{\sqrt{3}+1}{2}\right)^2 - 8\left(\frac{\sqrt{3}+1}{2}\right) + 7 \]

### Step 2: Calculate \(x^2\) and \(x^3\)
First, we calculate \(x^2\):
\[ x^2 = \left(\frac{\sqrt{3}+1}{2}\right)^2 = \frac{(\sqrt{3}+1)^2}{4} = \frac{3 + 2\sqrt{3} + 1}{4} = \frac{4 + 2\sqrt{3}}{4} = 1 + \frac{\sqrt{3}}{2} \]

Next, we calculate \(x^3\):
\[ x^3 = \left(\frac{\sqrt{3}+1}{2}\right)^3 = \frac{(\sqrt{3}+1)^3}{8} = \frac{3\sqrt{3} + 3 + 1 + \sqrt{3}}{8} = \frac{4 + 4\sqrt{3}}{8} = \frac{1 + \sqrt{3}}{2} \]

### Step 3: Substitute \(x^2\) and \(x^3\) back into the expression
Now we can substitute \(x^2\) and \(x^3\) back into the expression:
\[ 4\left(\frac{1+\sqrt{3}}{2}\right) + 2\left(1 + \frac{\sqrt{3}}{2}\right) - 8\left(\frac{\sqrt{3}+1}{2}\right) + 7 \]

### Step 4: Simplify each term
Calculating each term:
1. \(4x^3 = 4 \cdot \frac{1+\sqrt{3}}{2} = 2(1+\sqrt{3}) = 2 + 2\sqrt{3}\)
2. \(2x^2 = 2(1 + \frac{\sqrt{3}}{2}) = 2 + \sqrt{3}\)
3. \(-8x = -8\left(\frac{\sqrt{3}+1}{2}\right) = -4(\sqrt{3}+1) = -4\sqrt{3} - 4\)
4. The constant \(7\) remains as is.

### Step 5: Combine all terms
Now we combine all the terms:
\[ (2 + 2\sqrt{3}) + (2 + \sqrt{3}) + (-4\sqrt{3} - 4) + 7 \]
Combining like terms:
- Constant terms: \(2 + 2 - 4 + 7 = 7\)
- Terms with \(\sqrt{3}\): \(2\sqrt{3} + \sqrt{3} - 4\sqrt{3} = -1\sqrt{3}\)

Thus, the final result is:
\[ 7 - \sqrt{3} \]

### Final Answer:
\[ \boxed{7 - \sqrt{3}} \]