Let's represent in the Cartesian coordinate system those pairs of real numbers $(a ; b)$ for which the two-variable polynomial

$$
x(x+4)+a\left(y^{2}-1\right)+2 b y
$$

can be factored into the product of two first-degree polynomials.
I. solution. Suppose the polynomial can be factored as

$$
x(x+4)+a\left(y^{2}-1\right)+2 b y=(A x+B y+C)(D x+E y+F)
$$

where $A, B, C, D, E$ and $F$ are real numbers. Since two polynomials are identical if and only if their corresponding coefficients are equal, and the coefficient of $x^{2}$ in the original polynomial is 1, we have $A D=1$. Dividing the first linear factor by $A$ and the second by $D$, we still get the factorization of the original polynomial into two first-degree polynomials, so it is sufficient to give the condition for the factorization into the form

$$
x(x+4)+a\left(y^{2}-1\right)+2 b y=(x+u y+v)(x+p y+q)
$$

By comparing the corresponding coefficients, we get the following equations:

$$
\begin{aligned}
& y^{2}: a=u p \\
& x y: 0=u+p \\
& x: 4=v+q \\
& y: 2 b=u q+v p
\end{aligned}
$$

$$
\text { constant: }-a=v q
$$

If $a=0$, then from (1) and (2) we have $u=p=0$, and from (4) it follows that the factorization is only possible if $b=0$ (in this case, however, the original polynomial is not bivariate). If $a \neq 0$, then from (1) we have $u \neq 0$. From (2) we get $p=-u$, substituting this into (1) and (4) we get $a=-u^{2}$ and $2 b=u(q-v)$. From the latter, $q-v=2 b / u$ follows, and from (3) we have $q=2+b / u$ and $v=2-b / u$. Finally, substituting these into (5) we get

$$
-a=\left(2-\frac{b}{u}\right)\left(2+\frac{b}{u}\right)=4-\frac{b^{2}}{u^{2}}=4-\frac{b^{2}}{-a}
$$

that is, $a^{2}=-4 a-b^{2}$, or $(a+2)^{2}+b^{2}=4$. This is the equation of a circle $k$ with center $(-2,0)$ and radius 2. Therefore, if the polynomial can be factored, then the point $(a ; b)$ lies on $k$.

Conversely, if we take any point $(c ; d)$ on $k$, then $c \leq 0$ holds. If $c=0$, then $d=0$ also holds. The pair $(0 ; 0)$ corresponds to the obviously factorable polynomial $x(x+4)$. If $c \neq 0$, then let

$$
u=\sqrt{-c}, \quad p=-\sqrt{-c}, \quad v=2-\frac{d}{\sqrt{-c}} \quad \text { and } \quad q=2+\frac{d}{\sqrt{-c}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_72b532f69540f24eb475g-1.jpg?height=393&width=425&top_left_y=1592&top_left_x=839)

A simple calculation shows that

$$
(x+u y+v)(x+p y+q)=x^{2}+4 x+c y^{2}+2 d y-c
$$

that is, the original polynomial factors into the product of two first-degree polynomials.

Therefore, the desired geometric locus is the circle with center $(-2 ; 0)$ and radius 2.

II. solution. Using coordinate geometric methods, the necessary condition given in the first solution can also be shown with slightly less calculation. However, with the following reasoning, we cannot prove the sufficiency of the condition.

Apply the linear substitution $x=w-2, y=z+1$. Then

$$
P(x, y)=x(x+4)+a\left(y^{2}-1\right)+2 b y=w^{2}+a z^{2}+2(a+b) z+2 b-4=Q(w, z)
$$

Since a first-degree polynomial is obtained from a first-degree polynomial by a linear substitution, $P(x, y)$ can be factored into the product of two first-degree polynomials if and only if $Q(w, z)$ can be factored.

Consider the points in the plane that satisfy the equation $Q(w, z)=0$. If

$$
Q(w, z)=(A w+B z+C)(D w+E z+F)
$$

then these points form the lines

$$
A w+B z+C=0 \quad \text { and } \quad D w+E z+F=0
$$

Since the coefficient of $w^{2}$ in $Q(w, z)$ is 1, we have $A D=1$ for the coefficients in the factorization, so neither of the lines is parallel to the $w$ coordinate axis.

If $a=0$ and $b \neq 0$, then

$$
z=-\frac{1}{2 b} w^{2}-1+\frac{2}{b}
$$

which is the equation of a parabola, so in this case the polynomial cannot be factored into linear factors. If, however, $a=b=0$, then $Q(w, z)=w^{2}$, which can clearly be factored, for example, as $w \cdot w$. In this case, the two lines coincide.

If $a \neq 0$, then $Q(w, z)$ contains a $z^{2}$ term, so both linear factors must contain a term of the form $c z$ where $c \neq 0$ is a real number. This means that neither of the lines corresponding to the factors is parallel to the $z$ coordinate axis, so for every real number $d$, the line $z=d$ intersects the set of points defined by $Q(w, z)=0$. Rearranging the equation $Q(w, z)=0$, we get

$$
w^{2}=-a z^{2}-2(a+b) z-2 b+4
$$

Since $w^{2} \geq 0$, the quadratic polynomial on the right-hand side (since $a \neq 0$) cannot take a negative value for any $z=d$, because that would mean that the line $z=d$ does not intersect the set of points defined by $Q(w, z)=0$. On the other hand, the line $w=0$ is not parallel to the lines forming the set of points, so it intersects the set of points, i.e., the equation

$$
-a z^{2}-2(a+b) z-2 b+4=0
$$

has a root. These two conditions are satisfied simultaneously if and only if the discriminant of equation (6) is 0. Therefore, if $Q(w, z)$ can be factored into linear factors, then

$$
(-2(a+b))^{2}-4(-a)(-2 b+4)=0, \quad \text { i.e., } \quad(a+2)^{2}+b^{2}=4
$$

Remark. The sufficiency of the condition can also be obtained from the second solution, but for this, a deeper knowledge of conic sections given by quadratic equations is necessary. Write the polynomial $P(x, y)$ in matrix form as follows:

$$
(x+4)+a\left(y^{2}-1\right)+2 b y=(x, y, 1) M(x, y, 1)^{T}
$$

where

$$
M=\left(\begin{array}{ccc}
1 & 0 & 2 \\
0 & a & b \\
2 & b & -a
\end{array}\right)
$$

It is a known theorem that the polynomial can be factored over the complex numbers if and only if $\det M=0$. The coefficients in the factors are real if and only if the determinant of the upper-left $2 \times 2$ submatrix of $M$ is non-positive.