Question

If $x^3+1$ is a factor of $a{x}^2+b{x}^3+c{x}^2+dx+e,$ a, b, c, d, e $ϵRanda\ne 0,$ then the real root of $a{x}^4+b{x}^3+c{x}^2+dx+e=0$ other than -1 is

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

Answer 1

Answer: (A), (C)

The correct options are
A $-\frac{e}{a}$
C $-\frac{e}{d}$

Given $a{x}^4+b{x}^3+c{x}^2+dx+e=0$

$\Rightarrow a{x}^4+b{x}^3+c{x}^2+dx+e=(x^3+1)\times a\times (x-\alpha )$

$\Rightarrow x=-1,-\omega ,-{\omega }^2$

Sum of roots $=\frac{-b}{a}=-1-\omega -{\omega }^2+\alpha$

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

Products of roots $\frac{e}{a}=(-1)(-\omega )(-{\omega }^2)\alpha$

$=-\alpha$

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

Sum of product of roots taken $3$ at a time

$\frac{-d}{a}=(-1)(-\omega )(-{\omega }^2)+(-\omega )(-{\omega }^2)\left(\alpha \right)+(-1)(-\omega )\left(\alpha \right)+(-{\omega }^2)\left(\alpha \right)(-1)$

$=(-1)+\alpha +\alpha \omega +\alpha {\omega }^2$

$-1+\alpha (1+\omega +{\omega }^2)=-1$

$\frac{d}{a}=1$

$d=a$

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

Sum of product of roots taken $2$ at a time

$\frac{c}{a}=(-1)(-\omega )+(-1)(-{\omega }^2)+(-1)\left(\alpha \right)+(-\omega )(-{\omega }^2)+(-\omega )\left(\alpha \right)+(-{\omega }^2)\left(\alpha \right)$

$=\omega +{\omega }^2-\alpha +1-\omega \alpha -{\omega }^2\alpha$

$(-1)-\alpha (1+\omega +{\omega }^2)+1$

$=0$

$c=0$