Transformation of a fourth degree polynomial 
Given $$ P (x) = ax ^ 4 + bx ^ 3 + cx ^ 2 + dx + e = a (x ^ 2 + p_1x + q_1) (x ^ 2 + p_2x + q_2) $$ for some $ a, b, c, d, e, p_1, q_1, p_2, q_2  \in \mathbb R$, prove that $ P (x) $ can be reduced to the form $$ \frac {(M_1 + N_1t ^ 2) (M_2 + N_2t ^ 2)} {(\gamma t + 1) ^ 4} $$ by replacing $$ x = x (t) = \frac {\alpha t + \beta} {\gamma t + 1} $$

I could prove it only for the case when $ P (x) $ does not have all real roots. If there is at least one complex root $ z $, that is, its pair is the conjugate root $ \bar {z} $ and then if $$ x ^ 2 + px + q = (x-z) (x-z_1) $$ where $ p $ and $ q $ are real numbers, then $ z_1 = \bar {z} $. This implies that $ p_1, q_1, p_2, q_2 $ are numbers uniquely determined by $ a, ..., e $. This greatly simplifies the situation, so by extracting the inequality $ p_1 ^ 2-4q_1 <0 $, one can prove the required. If we have 4 real roots $ x_1, x_2, x_3, x_4 $, then in the equality $ (x-x_i) (x-x_j) = M_1 + N_1t ^ 2 $ the numbers $ i $ and $ j $  can be any of the set $ \left \lbrace1,2,3,4 \right \rbrace $. How to be in this case, I do not know.
 A: We homogenize the polynomial in question, and obtain quadratic forms $x^2 + p_1 xy + p_2 y^2, x^2 + q_1 xy + q_2 y^2$. We are done if we can find an element in $\text{GL}_2(\mathbb{R})$ which diagonalizes both of them. 
We can certainly transform the first quadratic form to $x^2 \pm y^2$ via an element of $\text{GL}_2(\mathbb{R})$. The first case occurs precisely when the quartic has a non-real root, and the question is reduced to finding an element in the standard orthogonal group
$$\displaystyle O_2(\mathbb{R}) = \left\{\begin{pmatrix} \cos(t) & \sin(t) \\ \pm \sin(t) & \mp \cos(t) \end{pmatrix} : t \in [0, 2\pi) \right\},$$
say $T(t)$, which diagonalizes $x^2 + q_1 xy + q_2 y^2$. If $T(t) = \left(\begin{smallmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{smallmatrix}\right)$ say, then we have
$$\displaystyle (\cos(t) x + \sin(t) y)^2 + q_1 (\cos(t) x + \sin(t) y)(-\sin(t)x + \cos(t)y) + q_2 (-\sin(t)x + \cos(t)y)^2$$
whose $xy$-coefficient is given by
$$\displaystyle \sin(2t) + q_1 \cos(2t) - q_2 \sin(2t).$$
We can always find a zero for this function, hence it is always possible to diagonalize $x^2 + q_1 xy + q_2 y^2$. 
Next we assume that $F(x,y) = y^4f(x/y)$ is totally real. It is standard that there exists an $A \in \mathbb{R}$ such that $F$ is $\text{GL}_2(\mathbb{R})$-equivalent to $F_A(x,y) = x^4 + Ax^2 y^2 + y^4$. We can factor $F_A(x,y) = (x^2 - uy^2)(x^2 - vy^2$ for $u,v > 0$ over $\mathbb{R}$, as desired. 
