Question

If $(x^2-x+1)$ is a factor of $f\left(x\right)=a{x}^3+b{x}^2+cx+d$, then the real root of $f\left(x\right)=0$, is

• A. $\frac{a}{d}$
• B. $\frac{d}{a}$
• C. $-\frac{a}{d}$
• D. $-\frac{d}{a}$

Answer 1

Answer: (D)

$-\frac{d}{a}$

Roots of $x^2-x+1=0$ are $-\omega ,-{\omega }^2$ and it is a factor of $f\left(x\right)=a{x}^3+b{x}^2+cx+d$.

So let the third root of $f\left(x\right)$ be $\alpha$

So $\alpha \times (-\omega )\times (-{\omega }^2)=-\frac{d}{a}$

$\alpha =-\frac{d}{a}$