Show that $x+2$ is a solution of the equation $2x^3 + 5x^2 + x - 2 = 0$ and hence solve the equation completely.

1. Show that $x = -2$ is a root by substituting it into the polynomial

Let $P(x) = 2x^3 + 5x^2 + x - 2$.
Substitute $x = -2$:
$P(-2) = 2(-2)^3 + 5(-2)^2 + (-2) - 2$
$P(-2) = 2(-8) + 5(4) - 2 - 2$
$P(-2) = -16 + 20 - 4 = 0$
Since $P(-2) = 0$, $x = -2$ is a solution, which means $(x + 2)$ is a factor.

2. Find the quadratic factor by dividing the polynomial by $(x + 2)$

Using inspection or long division:
$2x^3 + 5x^2 + x - 2 = (x + 2)(Ax^2 + Bx + C)$
Comparing coefficients:
$x^3$ term: $A = 2$
Constant term: $2C = -2 \implies C = -1$
$x^2$ term: $2A + B = 5 \implies 2(2) + B = 5 \implies B = 1$
So, the quadratic factor is $2x^2 + x - 1$.

3. Factorise the quadratic expression and solve the equation completely

$2x^2 + x - 1 = (2x - 1)(x + 1)$
Therefore, the equation becomes:
$(x + 2)(2x - 1)(x + 1) = 0$
$x = -2$, $x = \frac{1}{2}$, or $x = -1$