Question

${{x}^{2}}+x+1$ is a factor of $a{{x}^{3}}+b{{x}^{2}}+cx+d=0,$ then the real root of above equation is $\left( a,b,c,d\in R \right)$
(a) $\dfrac{-d}{a}$
(b) $\dfrac{d}{a}$
(c) $\dfrac{\left( b-a \right)}{a}$
(d) $\dfrac{\left( a-b \right)}{a}$

Answer 1

Hint: First, we will find roots of the quadratic equation as the quadratic equation is the factor of the cubic equation and we know that a cubic equation has three roots. If a quadratic equation is a factor of a cubic equation then that means that the roots of the quadratic will be the 2 roots among the total of three roots. So, to find the third root we will use the concept, which is the product of the roots of a cubic equation is $\dfrac{-d}{a}$. Using this we will find the third root of the cubic equation.

Complete step by step answer:
The roots of any general quadratic equation are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. The cube roots of unity are $1,\omega ,{{\omega }^{2}}$ and the values of $\omega ,{{\omega }^{2}}$ are given by $\dfrac{-1+\sqrt{3}i}{2},\dfrac{-1-\sqrt{3}i}{2}$. Also, we know that $\sqrt{-1}=i$.
Also, we know the algebraic identity $\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$.

So, first of all we will find roots of this equation ${{x}^{2}}+x+1$ by using $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
The roots of the equation will be,
$\begin{align}
  & x=\dfrac{-1\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( 1 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{-1\pm \sqrt{1-4}}{2} \\
 & \therefore x=\dfrac{-1\pm \sqrt{-3}}{2} \\
\end{align}$
And we know that $\sqrt{-1}=i$
So, $x=\dfrac{-1\pm \sqrt{3}i}{2}$
We also know that the cube roots of unity are $1,\omega ,{{\omega }^{2}}$
And the values of $\omega ,{{\omega }^{2}}$are $\dfrac{-1+\sqrt{3}i}{2},\dfrac{-1-\sqrt{3}i}{2}$
As ${{x}^{2}}+x+1$ is the factor of equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0,$

So, we can say that the roots of the equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0,$are $\alpha ,\beta ,\gamma $ and their values are $\dfrac{-1+\sqrt{3}i}{2},\dfrac{-1-\sqrt{3}i}{2}$ and $\gamma $
We know that the product of the roots of a cubic equation is $\dfrac{-d}{a}$
So, we will have, $\alpha \times \beta \times \gamma =\dfrac{-d}{a}$
Now, substitute the values of roots in the equation.
$\alpha \times \beta \times \gamma =\dfrac{-d}{a}$
On substituting values, we gte
$\dfrac{-1+\sqrt{3}i}{2}\times \dfrac{-1-\sqrt{3}i}{2}\times \gamma =\dfrac{-d}{a}$
On simplification, we get
$\dfrac{-1+3}{2}\times \gamma =\dfrac{-d}{a}$
On solving, we get
$\dfrac{2}{2}\times \gamma =\dfrac{-d}{a}$
$\gamma =\dfrac{-d}{a}$
As our other roots are imaginary, therefore our third root i.e. $\gamma $is the real root of our equation.
So, the real root of our equation is $\dfrac{-d}{a}$

So, the correct answer is “Option A”.

Note:
Remember that the value of cube roots of unity is $1,\omega ,{{\omega }^{2}}$and their values are 1, $\dfrac{-1+\sqrt{3}i}{2},\dfrac{-1-\sqrt{3}i}{2}$. Also, remember that the product of roots of a cubic equation is $\dfrac{-d}{a}$and general formula to solve the quadratic equation $a{{x}^{2}}+bx+c=0$ is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Also, the product of cube roots of unity is 1. Try not to do any calculation mistakes while solving the question.