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. $\frac{d}{a}$
• C. $\frac{a}{d}$
• D. none of these

Answer 1

The correct option is A $-\frac{d}{a}$

As $x^2+x+1=0$ is factor of $a{x}^3+b{x}^2+cx+d=0$

Then

For $x=1;a+b+c+d=0$ ...(1)

For $x=\omega ;a{\omega }^3+b{\omega }^2+c\omega +d=0\Rightarrow a+b{\omega }^2+c\omega +d=0$ ...(2)

And for $x={\omega }^2;a{\omega }^6+b{\omega }^4+c{\omega }^2+d=0\Rightarrow a+b\omega +c{\omega }^2+d=0$ ...(3)

Solving (1), (2) and (3), we get

$a+d=0$ and $b+c=0$

Then equation can be factorize as

$(x^2+x+1)(ax+d)=0$

Now as roots of $(x^2+x+1)$ are $1,\omega ,{\omega }^2$

And $\omega ,{\omega }^2$ are imaginary and imaginary roots occur in conjugate pair

Hence, $x=-\frac{d}{a}$ is real root