We are given the polynomial $p(x) = 2x^4 + ax^3 + bx - 1$, and the fact that when it is divided by $x^2 - x + 1$, the remainder is $3x + 2$. We need to find the values of $a$ and $b$.

Step 1: Division Structure

The division of $p(x)$ by $x^2 - x + 1$ follows the structure:

$p(x) = (x^2 - x + 1)q(x) + (3x + 2)$

Where $q(x)$ is the quotient, and $3x + 2$ is the remainder.

Step 2: Express the Polynomial

Write out $p(x) = (x^2 - x + 1)q(x) + (3x + 2)$. Here, we want to match the degrees of the polynomial on both sides. Since the degree of $p(x)$ is 4, the degree of $q(x)$ should be 2, so assume:

$q(x) = px^2 + qx + r$

Thus, we have:

$p(x) = (x^2 - x + 1)(px^2 + qx + r) + (3x + 2)$

Step 3: Expand the Product

Now, expand $(x^2 - x + 1)(px^2 + qx + r)$:

$(x^2 - x + 1)(px^2 + qx + r) = x^2(px^2 + qx + r) - x(px^2 + qx + r) + (px^2 + qx + r)$

Expand each term:

1. $x^2(px^2 + qx + r) = px^4 + qx^3 + rx^2$
2. $-x(px^2 + qx + r) = -px^3 - qx^2 - rx$
3. $(px^2 + qx + r) = px^2 + qx + r$

Now combine the terms:

$p(x) = px^4 + (q - p)x^3 + (r - q + p)x^2 + (-r + q)x + r$

Step 4: Match Coefficients

Next, match the coefficients of this expansion with the terms of $p(x) = 2x^4 + ax^3 + bx - 1$.

From the $x^4$ term: $p = 2$

From the $x^3$ term: $q - p = a \quad \Rightarrow \quad q - 2 = a \quad \Rightarrow \quad q = a + 2$

From the $x^2$ term: $r - q + p = 0 \quad \Rightarrow \quad r - (a + 2) + 2 = 0 \quad \Rightarrow \quad r = a$

From the $x^1$ term: $-q + r = b \quad \Rightarrow \quad -(a + 2) + a = b \quad \Rightarrow \quad b = -2$

From the constant term: $r = -1 \quad \Rightarrow \quad a = -1$

Step 5: Final Values

Thus, we have:

$a = -1 \quad \text{and} \quad b = -2$

Conclusion:

The values of $a$ and $b$ are $\boxed{a = -1}$ and $\boxed{b = -2}$.

---

Would you like a more detailed explanation of any part of the solution?

Here are some related questions to explore:

1. What is the quotient $q(x)$ in this division?
2. How do we verify the remainder using synthetic or polynomial division?
3. What happens when we change the divisor to a higher-degree polynomial?
4. How would the solution differ if the remainder was a constant?
5. Can this method be extended to non-polynomial divisors?

Tip: Always check your result by substituting the values of $a$ and $b$ back into the original polynomial to verify consistency with the remainder.