Number of transformed equations of `x^3 +2x^2 +x+1=0` by eliminating third term is

Options

A. 4
B. 3
C. 2
D. 1

Answer

To solve the problem of finding the number of transformed equations of the polynomial \( x^3 + 2x^2 + x + 1 = 0 \) by eliminating the third term, we will follow these steps:

### Step 1: Write the given equation
The given equation is:
\[ x^3 + 2x^2 + x + 1 = 0 \]

### Step 2: Substitute \( x \) with \( x + h \)
To eliminate the third term (the linear term \( x \)), we substitute \( x \) with \( x + h \):
\[ (x + h)^3 + 2(x + h)^2 + (x + h) + 1 = 0 \]

### Step 3: Expand the equation
Now we expand the equation:
\[ (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \]
\[ 2(x + h)^2 = 2(x^2 + 2xh + h^2) = 2x^2 + 4xh + 2h^2 \]
\[ (x + h) = x + h \]
Combining these, we have:
\[ x^3 + 3x^2h + 3xh^2 + h^3 + 2x^2 + 4xh + 2h^2 + x + h + 1 = 0 \]

### Step 4: Combine like terms
Now we combine all the like terms:
\[ x^3 + (3h + 2)x^2 + (3h^2 + 4h + 1)x + (h^3 + 2h^2 + h + 1) = 0 \]

### Step 5: Set the coefficient of \( x \) to zero
To eliminate the third term (the coefficient of \( x \)), we set the coefficient of \( x \) to zero:
\[ 3h^2 + 4h + 1 = 0 \]

### Step 6: Solve the quadratic equation
Now we solve the quadratic equation \( 3h^2 + 4h + 1 = 0 \) using the quadratic formula:
\[ h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 3, b = 4, c = 1 \):
\[ h = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \]
\[ h = \frac{-4 \pm \sqrt{16 - 12}}{6} \]
\[ h = \frac{-4 \pm \sqrt{4}}{6} \]
\[ h = \frac{-4 \pm 2}{6} \]
This gives us two solutions:
1. \( h = \frac{-2}{6} = -\frac{1}{3} \)
2. \( h = \frac{-6}{6} = -1 \)

### Step 7: Conclusion
Since we have two distinct values for \( h \), we can conclude that there are **two transformed equations** obtained by eliminating the third term from the original equation.

Thus, the answer is:
\[ \text{Number of transformed equations} = 2 \]