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Question: \({{x}^{2}}+x+1\) is a factor of \(a{{x}^{3}}+b{{x}^{2}}+cx+d=0,\) then the real root of above equat...

x2+x+1{{x}^{2}}+x+1 is a factor of ax3+bx2+cx+d=0,a{{x}^{3}}+b{{x}^{2}}+cx+d=0, then the real root of above equation is (a,b,c,d∈R)\left( a,b,c,d\in R \right)
(a) −da\dfrac{-d}{a}
(b) da\dfrac{d}{a}
(c) (b−a)a\dfrac{\left( b-a \right)}{a}
(d) (a−b)a\dfrac{\left( a-b \right)}{a}

Explanation

Solution

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 −da\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=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}. The cube roots of unity are 1,ω,ω21,\omega ,{{\omega }^{2}} and the values of ω,ω2\omega ,{{\omega }^{2}} are given by −1+3i2,−1−3i2\dfrac{-1+\sqrt{3}i}{2},\dfrac{-1-\sqrt{3}i}{2}. Also, we know that −1=i\sqrt{-1}=i.
Also, we know the algebraic identity (a3+b3)=(a−b)(a2+ab+b2)\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 x2+x+1{{x}^{2}}+x+1 by using x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
The roots of the equation will be,
x=−1±(−1)2−4(1)(1)2(1) ⇒x=−1±1−42 ∴x=−1±−32 \begin{aligned} & 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{aligned}
And we know that −1=i\sqrt{-1}=i
So, x=−1±3i2x=\dfrac{-1\pm \sqrt{3}i}{2}
We also know that the cube roots of unity are 1,ω,ω21,\omega ,{{\omega }^{2}}
And the values of ω,ω2\omega ,{{\omega }^{2}}are −1+3i2,−1−3i2\dfrac{-1+\sqrt{3}i}{2},\dfrac{-1-\sqrt{3}i}{2}
As x2+x+1{{x}^{2}}+x+1 is the factor of equation ax3+bx2+cx+d=0,a{{x}^{3}}+b{{x}^{2}}+cx+d=0,

So, we can say that the roots of the equation ax3+bx2+cx+d=0,a{{x}^{3}}+b{{x}^{2}}+cx+d=0,are α,β,γ\alpha ,\beta ,\gamma and their values are −1+3i2,−1−3i2\dfrac{-1+\sqrt{3}i}{2},\dfrac{-1-\sqrt{3}i}{2} and γ\gamma
We know that the product of the roots of a cubic equation is −da\dfrac{-d}{a}
So, we will have, α×β×γ=−da\alpha \times \beta \times \gamma =\dfrac{-d}{a}
Now, substitute the values of roots in the equation.
α×β×γ=−da\alpha \times \beta \times \gamma =\dfrac{-d}{a}
On substituting values, we gte
−1+3i2×−1−3i2×γ=−da\dfrac{-1+\sqrt{3}i}{2}\times \dfrac{-1-\sqrt{3}i}{2}\times \gamma =\dfrac{-d}{a}
On simplification, we get
−1+32×γ=−da\dfrac{-1+3}{2}\times \gamma =\dfrac{-d}{a}
On solving, we get
22×γ=−da\dfrac{2}{2}\times \gamma =\dfrac{-d}{a}
γ=−da\gamma =\dfrac{-d}{a}
As our other roots are imaginary, therefore our third root i.e. γ\gamma is the real root of our equation.
So, the real root of our equation is −da\dfrac{-d}{a}

So, the correct answer is “Option A”.

Note:
Remember that the value of cube roots of unity is 1,ω,ω21,\omega ,{{\omega }^{2}}and their values are 1, −1+3i2,−1−3i2\dfrac{-1+\sqrt{3}i}{2},\dfrac{-1-\sqrt{3}i}{2}. Also, remember that the product of roots of a cubic equation is −da\dfrac{-d}{a}and general formula to solve the quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 is x=−b±b2−4ac2ax=\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.