The number of triangle formed by the curves `x^(3)-x^(2)-x-2=0` and `4x^(2)+x^(2)y^(2)+2y^(2)-2xy^(2)-12x+8=0` is equal to

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem of finding the number of triangles formed by the curves given by the equations \(x^3 - x^2 - x - 2 = 0\) and \(4x^2 + x^2y^2 + 2y^2 - 2xy^2 - 12x + 8 = 0\), we will follow these steps:

### Step 1: Analyze the first equation
The first equation is:
\[ x^3 - x^2 - x - 2 = 0 \]
We will factor this polynomial.

### Step 2: Factor the first equation
To factor \(x^3 - x^2 - x - 2\), we can use the Rational Root Theorem or trial and error to find a root. Testing \(x = 2\):
\[ 2^3 - 2^2 - 2 - 2 = 8 - 4 - 2 - 2 = 0 \]
Thus, \(x = 2\) is a root. We can factor the polynomial as:
\[ x^3 - x^2 - x - 2 = (x - 2)(x^2 + x + 1) \]
The quadratic \(x^2 + x + 1\) has no real roots (discriminant \(= 1^2 - 4 \cdot 1 \cdot 1 = -3 < 0\)). Therefore, the only real solution is \(x = 2\).

### Step 3: Analyze the second equation
The second equation is:
\[ 4x^2 + x^2y^2 + 2y^2 - 2xy^2 - 12x + 8 = 0 \]
We will substitute \(x = 2\) into this equation to check for \(y\).

### Step 4: Substitute \(x = 2\) into the second equation
Substituting \(x = 2\):
\[ 4(2^2) + (2^2)y^2 + 2y^2 - 2(2)y^2 - 12(2) + 8 = 0 \]
This simplifies to:
\[ 16 + 4y^2 + 2y^2 - 4y^2 - 24 + 8 = 0 \]
Combining like terms:
\[ 0 = 0 \]
This indicates that the equation holds for all \(y\), meaning that when \(x = 2\), we have a line in the \(y\)-direction.

### Step 5: Conclusion about triangles
Since the first equation gives us a single point (the line \(x = 2\) does not intersect the second curve at distinct points), there are no triangles formed by the intersection of these two curves.

### Final Answer
The number of triangles formed by the curves is:
\[ \text{0 triangles} \]