Question

The sum of non-real roots of the polynomial equation $x^3+3x^2+3x+3=0$ :

• A. Equal $0$
• B. Lies between $0$ and $1$
• C. Lies between $-1$ and $0$
• D. Has absolute value bigger than $1$

Answer 1

The correct option is C Lies between $-1$ and $0$

For the cubic equation $f\left(x\right)=x^3+3x^2+3x+3=0$, sum of all the roots is equal to -3.

Also, $f^{\prime }\left(x\right)=3x^2+6x+3=3(x^2+2x+1)$

Equating $f^{\prime }\left(x\right)$ to zero gives only one root $x=-1$

So, the equation $f\left(x\right)=0$ has only one real root and two non-real roots.

$f(-2)=-8+12-6+3=1$ and $f(-3)=-6$

$\Rightarrow$ a root lies in the interval $(-3,-2)$

So, the sum of the non-real roots would be lying between $-1$ and $0$