Suppose `x^(2)-x +1` is factor of `2x^(6) - x^(5) + ax^(4) + x^(3)+bx^(2)-4x-3`. Find `a-4b`.

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that \( x^2 - x + 1 \) is a factor of the polynomial \( 2x^6 - x^5 + ax^4 + x^3 + bx^2 - 4x - 3 \).

### Step-by-Step Solution:

1. **Identify the Roots of the Factor**:
The roots of the quadratic \( x^2 - x + 1 = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2} \]
Let \( \omega = \frac{1 + i\sqrt{3}}{2} \) and \( \omega^2 = \frac{1 - i\sqrt{3}}{2} \).

2. **Substitute the Roots into the Polynomial**:
Since \( x^2 - x + 1 \) is a factor, substituting \( x = \omega \) into the polynomial should yield 0:
\[ P(\omega) = 2\omega^6 - \omega^5 + a\omega^4 + \omega^3 + b\omega^2 - 4\omega - 3 = 0 \]

3. **Calculate Powers of \( \omega \)**:
Using the properties of \( \omega \):
- \( \omega^3 = 1 \)
- \( \omega^4 = \omega \)
- \( \omega^5 = \omega^2 \)
- \( \omega^6 = 1 \)

Substitute these values into \( P(\omega) \):
\[ P(\omega) = 2(1) - \omega^2 + a\omega + 1 + b\omega^2 - 4\omega - 3 \]
Simplifying this gives:
\[ P(\omega) = 2 - \omega^2 + a\omega + 1 + b\omega^2 - 4\omega - 3 = 0 \]
\[ P(\omega) = (b - 1)\omega^2 + (a - 4)\omega + 0 = 0 \]

4. **Set Coefficients to Zero**:
Since \( P(\omega) = 0 \), we can equate the coefficients of \( \omega^2 \) and \( \omega \) to zero:
\[ b - 1 = 0 \quad \Rightarrow \quad b = 1 \]
\[ a - 4 = 0 \quad \Rightarrow \quad a = 4 \]

5. **Calculate \( a - 4b \)**:
Now that we have \( a \) and \( b \):
\[ a - 4b = 4 - 4(1) = 4 - 4 = 0 \]

### Final Answer:
The value of \( a - 4b \) is \( 0 \).