If `x^(2)+x+1` is a factor of `ax^(3)+bx^(2)+cx+d`, then the real root of `ax^(3)+bx^(2)+cx+d=0` is

Options

A. `-d//a`
B. `d//a`
C. `a//d`
D. none of these

Answer

To solve the problem, we need to find the real root of the cubic polynomial \( ax^3 + bx^2 + cx + d = 0 \) given that \( x^2 + x + 1 \) is a factor of it.

### Step-by-Step Solution:

1. **Identify the Roots of the Factor**:
The polynomial \( x^2 + x + 1 \) has roots that can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = 1 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
The roots are \( \omega = \frac{-1 + i\sqrt{3}}{2} \) and \( \omega^2 = \frac{-1 - i\sqrt{3}}{2} \).

2. **Understanding the Cubic Polynomial**:
Since \( x^2 + x + 1 \) is a factor of \( ax^3 + bx^2 + cx + d \), we can express the cubic polynomial as:
\[ ax^3 + bx^2 + cx + d = (x^2 + x + 1)(ax + k) \]
where \( k \) is some constant.

3. **Finding the Third Root**:
The roots of the cubic polynomial are \( \omega, \omega^2, \) and a real root \( \alpha \). By Vieta's formulas, the sum of the roots of the polynomial \( ax^3 + bx^2 + cx + d = 0 \) is given by:
\[ \omega + \omega^2 + \alpha = -\frac{b}{a} \]
Since \( \omega + \omega^2 = -1 \) (from the properties of the roots of unity), we have:
\[ -1 + \alpha = -\frac{b}{a} \implies \alpha = -\frac{b}{a} + 1 \]

4. **Product of the Roots**:
The product of the roots is given by:
\[ \omega \cdot \omega^2 \cdot \alpha = -\frac{d}{a} \]
Since \( \omega \cdot \omega^2 = 1 \), we have:
\[ 1 \cdot \alpha = -\frac{d}{a} \implies \alpha = -\frac{d}{a} \]

5. **Conclusion**:
The real root \( \alpha \) of the cubic polynomial \( ax^3 + bx^2 + cx + d = 0 \) is:
\[ \alpha = -\frac{d}{a} \]

### Final Answer:
The real root of \( ax^3 + bx^2 + cx + d = 0 \) is \( -\frac{d}{a} \).