Solve the equation $2x^3 - 7x + 2 = 0$, leaving non-rational roots in the form $a \pm b\sqrt{2}$, where $a$ and $b$ are rational numbers.

1. Use the Factor Theorem to find an integer root of the polynomial.

Let $f(x) = 2x^3 - 7x + 2$.
Testing integer values:
$f(1) = 2(1)^3 - 7(1) + 2 = -3 \neq 0$
$f(-1) = 2(-1)^3 - 7(-1) + 2 = 7 \neq 0$
$f(2) = 2(2)^3 - 7(2) + 2 = 4 \neq 0$
$f(-2) = 2(-2)^3 - 7(-2) + 2 = -16 + 14 + 2 = 0$
Since $f(-2) = 0$, $x = -2$ is a root and $(x + 2)$ is a factor.

2. Factorise the cubic polynomial completely by finding the quadratic factor.

Let $2x^3 - 7x + 2 = (x + 2)(Ax^2 + Bx + C)$.
Comparing coefficients:
Coefficient of $x^3$: $A = 2$
Constant term: $2C = 2 \implies C = 1$
Coefficient of $x^2$: $B + 2A = 0 \implies B + 4 = 0 \implies B = -4$
So, $2x^3 - 7x + 2 = (x + 2)(2x^2 - 4x + 1) = 0$.

3. Solve the quadratic equation to find the remaining roots.

For $2x^2 - 4x + 1 = 0$, using the quadratic formula:
$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)}$
$x = \frac{4 \pm \sqrt{16 - 8}}{4}$
$x = \frac{4 \pm \sqrt{8}}{4}$

4. Simplify the surd to the required form $a \pm b\sqrt{2}$.

$x = \frac{4 \pm 2\sqrt{2}}{4}$
$x = 1 \pm \frac{1}{2}\sqrt{2}$
The roots are $x = -2$ and $x = 1 \pm \frac{1}{2}\sqrt{2}$.