Question

If $x^2+x+1$ is a factor of $a{x}^3+b{x}^2+cx+d$, then the real root of $a{x}^3+b{x}^2+cx+d=0$ is:

• A. $-\frac{d}{a}$
• B. $\frac{c}{a}$
• C. $\frac{b}{a}$
• D. None of these

Answer 1

The correct option is A

$-\frac{d}{a}$

Explanation for the correct option

Cube roots of unity:

Taking, $x^2+x+1=0$, then the roots will be

$$\begin{gathered} \frac{-1\pm \sqrt{1-4}}{2}\mathbf{}\left[\mathrm{by}\mathrm{roots}\mathrm{of}\mathrm{quadratic}\mathrm{equation}=\frac{-b\pm \sqrt{b^2-4\mathrm{ac}}}{2a}\right] \\ =\frac{-1\pm \sqrt{3}i}{2} \end{gathered}$$

So, the roots are $\omega ,\mathrm{and}{\omega }^2$. $As\omega =\frac{-1+\sqrt{3}i}{2},\mathrm{and}{\omega }^2=\frac{-1-\sqrt{3}i}{2}$

Let the roots of $a{x}^3+b{x}^2+cx+d$ be $\alpha ,\beta ,\mathrm{and}\gamma$.

So, $\alpha =\omega ,\beta ={\omega }^2$ and thus the real root is $\gamma$.

Product of roots $=-\frac{d}{a}$

$$\begin{gathered} \Rightarrow \alpha \beta \gamma =-\frac{d}{a} \\ \Rightarrow {\omega }^3\gamma =-\frac{d}{a} \\ \Rightarrow \gamma =-\frac{d}{a}\mathbf{[}\mathbf{a}\mathbf{s}\mathbf{}{\mathbf{\omega }}^{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{]} \end{gathered}$$

Hence, option A is correct.