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 : (a) `-d/a ` (B) `d/a ` (C) `a/b` (D)none of these

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Identify the Roots of the Factor**:
The polynomial \( x^2 + x + 1 = 0 \) has roots which can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = 1 \). Thus, the roots are:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
These roots are complex: \( \omega = \frac{-1 + i\sqrt{3}}{2} \) and \( \omega^2 = \frac{-1 - i\sqrt{3}}{2} \).

2. **Assume the Third Root**:
Let \( \alpha \) be the third root of the cubic polynomial \( ax^3 + bx^2 + cx + d = 0 \). The roots of the cubic polynomial are \( \omega, \omega^2, \) and \( \alpha \).

3. **Use the Product of Roots**:
According to Vieta's formulas, the product of the roots of the polynomial \( ax^3 + bx^2 + cx + d = 0 \) is given by:
\[ \text{Product of roots} = -\frac{d}{a} \]
Therefore, we have:
\[ \alpha \cdot \omega \cdot \omega^2 = -\frac{d}{a} \]

4. **Calculate \( \omega \cdot \omega^2 \)**:
We know that \( \omega \cdot \omega^2 = \omega^3 \). Since \( \omega^3 = 1 \) (as \( \omega \) is a cube root of unity), we have:
\[ \alpha \cdot 1 = -\frac{d}{a} \]
Thus, we can conclude that:
\[ \alpha = -\frac{d}{a} \]

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

### Final Answer:
The real root of the equation is \( -\frac{d}{a} \), which corresponds to option (A).

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