We are given the polynomial $P(x) = x^3 - px + q$. When $P(x)$ is divided by $x^2 - 3x + 2$, the remainder is $4x - 1$. We need to find the constants $p$ and $q$.

1. Problem Description

We are given the polynomial $P(x) = x^3 - px + q$. When $P(x)$ is divided by $x^2 - 3x + 2$, the remainder is $4x - 1$. We need to find the constants $p$ and $q$.

2. Solution Steps

First, we can factor the divisor $x^2 - 3x + 2$ as $(x-1)(x-2)$.

By the division algorithm, we can write $P(x) = (x^2 - 3x + 2)Q(x) + R(x)$, where $Q(x)$ is the quotient and $R(x)$ is the remainder.

In this case, we have $P(x) = (x-1)(x-2)Q(x) + (4x-1)$.

Since $x-1$ and $x-2$ are factors of $(x-1)(x-2)$, we can find the roots of the divisor, which are $x=1$ and $x=2$.

When $x = 1$, we have $P(1) = (1)^3 - p(1) + q = 1 - p + q$.

Also, $P(1) = (1-1)(1-2)Q(1) + (4(1) - 1) = 0 + 4 - 1 = 3$.

Thus, we have the equation $1 - p + q = 3$, which simplifies to $-p + q = 2$.

When $x = 2$, we have $P(2) = (2)^3 - p(2) + q = 8 - 2p + q$.

Also, $P(2) = (2-1)(2-2)Q(2) + (4(2) - 1) = 0 + 8 - 1 = 7$.

Thus, we have the equation $8 - 2p + q = 7$, which simplifies to $-2p + q = -1$.

Now we have a system of two linear equations with two variables:

$-p + q = 2$

$-2p + q = -1$

Subtracting the second equation from the first, we get $(-p + q) - (-2p + q) = 2 - (-1)$, which simplifies to $p = 3$.

Substituting $p = 3$ into the first equation, we have $-3 + q = 2$, so $q = 5$.

3. Final Answer

$p = 3$ and $q = 5$