According one polynomial determine the values of p and q if(X-1) and (X+2) are factors of 2x³+px²-x+q

The Factor Theorem states that if $(x-a)$ is a factor of a polynomial $P(x)$, then $P(a) = 0$.
Given $P(x) = 2x^3 + px^2 - x + q$ and $(x-1)$ is a factor, we set $x=1$:

$$P(1) = 2(1)^3 + p(1)^2 - (1) + q = 0$$ $$2 + p - 1 + q = 0$$ $$p + q + 1 = 0$$

This gives our first equation:

$$p + q = -1 \quad (*)$$

Given $(x+2)$ is a factor, we set $x=-2$:

$$P(-2) = 2(-2)^3 + p(-2)^2 - (-2) + q = 0$$ $$2(-8) + p(4) + 2 + q = 0$$ $$-16 + 4p + 2 + q = 0$$ $$4p + q - 14 = 0$$

This gives our second equation:

$$4p + q = 14 \quad (**)$$

We have the system:

1. $p + q = -1$
2. $4p + q = 14$

Subtract equation (1) from equation (2):

$$(4p + q) - (p + q) = 14 - (-1)$$ $$4p + q - p - q = 14 + 1$$ $$3p = 15$$ $$p = \frac{15}{3}$$ $$p = 5$$

Using $p + q = -1$:

$$5 + q = -1$$ $$q = -1 - 5$$ $$q = -6$$

The values are $p = \boxed{\text{5}}$ and $q = \boxed{\text{-6}}$.