If $ 3x^2 + 2\alpha xy + 2y^2 + 2ax - 4y + 1 $ can be resolved into two linear factors, then prove the following. Prove that : $ \alpha $ is a root of the equation $ x^2 + 4ax + 2a^2 + 6 = 0 $. 
What does it mean by "can be resolved into two linear factors"?  If it means $( ax + b ) ( cx + d )$ , is it necessary that $ a,b,c,d \in Q $ . 
The solving says that: For the condition to be true , the roots of the equation must be rational. 
$ \implies { \displaystyle\frac{ -2\alpha y + 2a \pm \sqrt {D} }{2} } $ must give rational roots. 
For the above expression to give rational roots, $D$ must be a perfect square of a rational number. 
$\implies 4 { ( \alpha^2 - 2 ) y^2 + ( 2a\alpha + 4 ) y + a^2 - 1 } $ 
should be a perfect square. 
$\implies $ The discriminant of above expression $= 0$. By doing that, we get our required result. 
**My Doubt ** : 
Cant it be resolved into linear factors if $D$ is not a perfect square, i.e., $D > 0$, giving irrational roots? 
Is it necessary that for $ax + b$ to be called linear, $a,b \in Q$?
 A: Let $\alpha \in \mathbb{R}$ (or $\alpha \in \mathbb{C}$?). The set of points $(x,y) \in \mathbb{R}^2$ satisfying
$$ C:  3x^2 + 2\alpha xy + 2y^2 +2ax - 4y + 1 = 0 $$
is a conic. When $C$ factors into a product of two linear factors, we say that $C$ is degenerate. Degeneracy can be tested by computing the determinant of its matrix of coefficients of its homogeneous form. So let's homogenize $C$ first:
$$ \widetilde{C} : 3X^2 + 2\alpha XY + 2Y^2 + 2a XZ - 4YZ + Z^2 = 0. $$
Just "stick a $Z$ onto the factors until all the degrees become the same". The original conic $C$ may be recovered from $\widetilde{C}$ by setting $Z=1$. The matrix of coefficients of $\widetilde{C}$ is given by
$$ M := \begin{bmatrix}
3 & \alpha & a\\
\alpha & 2 & -2 \\
a& -2 & 1
 \end{bmatrix}. $$
The determinant of this matrix is
$$ \det M = \det \begin{bmatrix}
3 & \alpha & a\\
\alpha & 2 & -2 \\
a& -2 & 1
 \end{bmatrix} = 6 - 2a\alpha - 2a \alpha - 2a^2 - 12 - \alpha^2 = -(\alpha^2  +4a\alpha + 2a^2+6). $$
Hence, the conic $C$ is degenerate if and only if $\det M = 0$, if and only if $\alpha^2 + 4a\alpha + 2a^2 + 6 = 0$, if and only if $\alpha$ is a root of the equation $x^2 + 4ax + 2a^2 + 6 = 0$.
A: The polynomial
$$P(x,y)=ax^2+by^2+cxy+dx+ey+f=0$$
geometrically represents conic sections. When it is said that it can be resolved into linear factors, then it is meant that there is a factorization such that
$$P(x,y)=(Ax+By+C)(Ex+Fy+D)=0$$
Which geometrically represents two crossing or coinciding lines! For such a factorization to exist some relations must hold between the constants $a,b,c,d,e,f$. You can see this link for more information.
A: 
What does it mean by "can be resolved into two linear factors"? If it means (ax+b)(cx+d) , is it necessary that  a,b,c,d∈Q

Geometrically, it means that the second degree equation must represent a pair of straight line.
The general equation of a second degree is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$
Given equation is: $ 3x^2 + 2\alpha xy + 2y^2 + 2ax - 4y + 1 $
On comparing you get $a=3, b=2, c=1, f=-2, g=a, h= \alpha$
For any second-degree equation to represent a straight line, determinant of:
$$\begin{pmatrix}
a&h&g\\h&b&f\\g&f&c    
\end{pmatrix}.$$
should be zero
Thus, determinant of:
$$\begin{pmatrix}
3& \alpha &a\\\alpha&2&-2\\a&-2&1    
\end{pmatrix}.$$
should be zero
Upon solving you get, $\alpha^2 + 4a\alpha + 2a^2 + 6 = 0$
Replace $\alpha$ with x and you get your solution.
