Show how to do FFT by hand
Suppose we use fourth roots of unity, which corresponds to substituting $1,i,-1,-i$ for $x$. We also use decimation-in-time rather than decimation-in-frequency in the FFT algorithm. (We also apply a bit-reversal operation seamlessly.)

In order to compute the transform of the first polynomial, we start by writing the coefficients:
$$ 3,1,0,0. $$
The Fourier transform of the even coefficients $3,0$ is $3,3$, and of the odd coefficients $1,0$ is $1,1$. (This transform is just $a,b \mapsto a+b,a-b$.) Therefore the transform of the first polynomial is 
$$ 4,3+i,2,3-i. $$
This is obtained using $X_{0,2} = E_0 \pm O_0$, $X_{1,3} = E_1 \mp i O_1$. ( From twiddle factor calculation ).

Let's do the same for the second polynomial. The coefficients are
$$2,0,2,0.$$
The even coefficients $2,2$ transform to $4,0$, and the odd coefficients $0,0$ transform to $0,0$. Therefore the transform of the second polynomial is
$$ 4,0,4,0. $$

We obtain the Fourier transform of the product polynomial by multiplying the two Fourier transforms pointwise:
$$ 16, 0, 8, 0. $$
It remains to compute the inverse Fourier transform. The even coefficients $16,8$ inverse-transform to $12,4$, and the odd coefficients $0,0$ inverse-transform to $0,0$. (The inverse transform is $x,y \mapsto (x+y)/2,(x-y)/2$.) Therefore the transform of the product polynomial is
$$6,2,6,2.$$
This is obtained using $X_{0,2} = (E_0 \pm O_0)/2$, $X_{1,3} = (E_1 \mp i O_1)/2$.
We have obtained the desired answer
$$ (3 + x)(2 + 2x^2) = 6+2x+6x^2+2x^3. $$