Why?? As I see it $\displaystyle (x_1+2x_2)^2$ fulfill the positivity condition because $\displaystyle (x_1+2x_2)^2 \geq 0 \forall x_1, x_2$ and $\displaystyle (x_1+2x_2)^2 = 0$ if the vectors are zero so $\displaystyle x_1, x_2$ are zero.

Not really: try the vector $\displaystyle (-2,1)$ ... Tonio
Now, can you check my work for the homogeneity property:

Let $\displaystyle k$ be any scalar.

$\displaystyle v=\binom {x_1}{x_2}\,,\,\,u=(y_1\,\,y_2)$

$\displaystyle \left\langle ku,v \right\rangle =ku^TAv=(ky_1 \,\, ky_2) \binom {x_1+2x_2}{2x_1+4x_2} \binom {x_1}{x_2}$

$\displaystyle = (ky_1\,\,ky_2)\binom {x_1+2x_2}{2x_1+4x_2}$

$\displaystyle =ky_1 (x_1+2x_2) + ky_2 (2x_1+4x_2)$

$\displaystyle =ky_1x_1+ky_12x_2+ky_22x_1+ky_24x_2$

$\displaystyle = k(y_1x_1+y_12x_2+y_22x_1+y_24x_2) = k \left\langle u,v \right\rangle$

So it satisfies the homogeneity condition