Find the values of p and q If (x2 + x - 2) is a factor of f(x) = x3 + px2 + 2x + q.

Factorize the quadratic expression $x^2 + x - 2$:

$$x^2 + x - 2 = 0$$ $$(x+2)(x-1) = 0$$

The roots are $x = -2$ and $x = 1$.

Since $(x+2)$ and $(x-1)$ are factors of $f(x) = x^3 + px^2 + 2x + q$, we know that $f(-2) = 0$ and $f(1) = 0$.

For $x=1$:

$$f(1) = (1)^3 + p(1)^2 + 2(1) + q = 0$$ $$1 + p + 2 + q = 0$$ $$p + q + 3 = 0$$ $$p + q = -3 \quad \text{(Equation 1)}$$

For $x=-2$:

$$f(-2) = (-2)^3 + p(-2)^2 + 2(-2) + q = 0$$ $$-8 + 4p - 4 + q = 0$$ $$4p + q - 12 = 0$$ $$4p + q = 12 \quad \text{(Equation 2)}$$

We have the system:

1. $p + q = -3$
2. $4p + q = 12$

Subtract Equation 1 from Equation 2:

$$(4p + q) - (p + q) = 12 - (-3)$$ $$4p - p + q - q = 12 + 3$$ $$3p = 15$$ $$p = \frac{15}{3}$$ $$p = 5$$

$$p + q = -3$$ $$5 + q = -3$$ $$q = -3 - 5$$ $$q = -8$$

The values are $\mathbf{p=5, q=-8}$.