Given A = x2 & 3 1 & 3x and B = 3 & 6 2 & x .

Given $A = \begin{pmatrix} x^2 & 3 \\ 1 & 3x \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 6 \\ 2 & x \end{pmatrix}$.

$$AB = \begin{pmatrix} x^2 & 3 \\ 1 & 3x \end{pmatrix} \begin{pmatrix} 3 & 6 \\ 2 & x \end{pmatrix}$$ $$AB = \begin{pmatrix} (x^2)(3) + (3)(2) & (x^2)(6) + (3)(x) \\ (1)(3) + (3x)(2) & (1)(6) + (3x)(x) \end{pmatrix}$$ $$AB = \begin{pmatrix} 3x^2 + 6 & 6x^2 + 3x \\ 3 + 6x & 6 + 3x^2 \end{pmatrix}$$

$$BA = \begin{pmatrix} 3 & 6 \\ 2 & x \end{pmatrix} \begin{pmatrix} x^2 & 3 \\ 1 & 3x \end{pmatrix}$$ $$BA = \begin{pmatrix} (3)(x^2) + (6)(1) & (3)(3) + (6)(3x) \\ (2)(x^2) + (x)(1) & (2)(3) + (x)(3x) \end{pmatrix}$$ $$BA = \begin{pmatrix} 3x^2 + 6 & 9 + 18x \\ 2x^2 + x & 6 + 3x^2 \end{pmatrix}$$

$$\begin{pmatrix} 3x^2 + 6 & 6x^2 + 3x \\ 3 + 6x & 6 + 3x^2 \end{pmatrix} = \begin{pmatrix} 3x^2 + 6 & 9 + 18x \\ 2x^2 + x & 6 + 3x^2 \end{pmatrix}$$

Equating the top-left and bottom-right elements gives identities ($3x^2+6 = 3x^2+6$ and $6+3x^2 = 6+3x^2$), which do not help find $x$. We need to equate the other two elements:

From the top-right elements:

$$6x^2 + 3x = 9 + 18x$$

From the bottom-left elements:

$$3 + 6x = 2x^2 + x$$

Let's solve the equation from the top-right elements:

$$6x^2 + 3x - 18x - 9 = 0$$ $$6x^2 - 15x - 9 = 0$$

Divide the entire equation by 3:

$$2x^2 - 5x - 3 = 0$$

We can solve this quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}$$ $$x = \frac{5 \pm \sqrt{25 + 24}}{4}$$ $$x = \frac{5 \pm \sqrt{49}}{4}$$ $$x = \frac{5 \pm 7}{4}$$

This gives two possible values for $x$:

$$x_1 = \frac{5 + 7}{4} = \frac{12}{4} = 3$$ $$x_2 = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2}$$

Now, let's check the equation from the bottom-left elements:

$$3 + 6x = 2x^2 + x$$ $$2x^2 + x - 6x - 3 = 0$$ $$2x^2 - 5x - 3 = 0$$

This is the same quadratic equation, so it yields the same solutions.

The possible values of $x$ are $\boxed{3, -\frac{1}{2}}$.