show that there exist 2 polynomials $F(x,y,z)$ ang $G(x,y,z)$ such that $F(x,y,z)^2+G(x,y,z)^2=A(x,y)z^2+B(x,y)z+C(x,y)$ 
Let $A(x,y), B(x,y)$, and $ C(x,y) $ are three homogeneous real-coefficient polynomials of x and y with degree 2, 3, and 4 respectively. we know that there is a real-coefficient polinimial R(x,y) such that $$B(x,y)^2-4A(x,y)C(x,y)=-R(x,y)^2$$show that there exist 2 polynomials $F(x,y,z)$ ang $G(x,y,z)$ such that $$F(x,y,z)^2+G(x,y,z)^2=A(x,y)z^2+B(x,y)z+C(x,y)$$ if for any x, y, z real numbers $A(x,y)z^2+B(x,y)z+C(x,y)\ge 0$

Does this help? $$A(x,y)z^2+B(x,y)z+C(x,y) = \frac{1}{4A(x,y)}[(2A(x,y)z + B(x,y))^2 + R(x,y)^2]$$
 A: The key fact to realize is that $4AC=B^2+R^2$, so that $A$ divides $B^2+R^2$
in ${\mathbb R}[x,y]$, so that the $\mathbb C[x,y]$-irreducible factors of $A$
will divide either $B+iR$ or $B-iR$ in $\mathbb C[x,y]$.
Let $P(x,y,z)=A(x,y)z^2+B(x,y)z+C(x,y)$. The nonegativity hypothesis on $P$
forces $A$ to be nonnegative on ${\mathbb R}^2$. By the classical theory of quadratic
forms, $A$ is the sum of zero, one or two (linearly independent) squares. Since $A$ is homogeneous of degree $2$,
$A$ cannot be zero.
If $A$ consists of just one square, using a change of variables we may assume without loss that $A=x^2$. Then $x^2$ divides $B^2+R^2=(B-Ri)(B+Ri)$, so $x$ divides one of $B-Ri$ or $B+Ri$ in ${\mathbb C}[x,y]$. This forces $x$ to divide both $B$ and $R$ in ${\mathbb R}[x,y]$ and then we have $P=F^2+G^2$ with $F=xz+\frac{B}{2x}, G=\frac{R}{2x}$.
If $A$ is a sum of two (linearly independent) squares,  using a change of variables we may assume without loss that $A=x^2+y^2$. Then $x+yi$ divides one of $B-Ri$ or $B+Ri$ in ${\mathbb C}[x,y]$, suppose for example that it divides $x+yi$. We have real homogeneous polynomials $U,V$ such that $B+iR=(x+iy)(U+iV)$, whence $B=xU-yV, R=xV+yU$, and then we have $P=F^2+G^2$ with $F=xz+\frac{U}{2}, G=yz-\frac{V}{2}$, which finishes the proof.
