If `x^(2)+x+1` is a factor of `ax^(3)+bx^(2)+cx+d` the real root of `ax^(3)+bx^(2)+cx+d=0` is

Options

A. `(d)/(a)`
B. `-(d)/(a)`
C. `-(b)/(a)`
D. `-(c)/(a)`

Answer

B

Explanation

It is given that `x^(2) + x + 1` is a factor of `ax^(3) + bx^(2) + cx + d`. Therefore, roots of `x^(2) + x + 1 = 0` are also the roots of the equation `ax^(3) + bx^(2) + cx + d = 0`. But, `x^(2) + x + 1 = 0` has its roots as `omega and omega^(2)`. So, two imaginary roots of `ax^(3) + bx^(2) + cx + d = 0` are `omega and omega^(2)`. Let the third real root be `alpha`. Then, `omega omega^(2) alpha = (-1)^(3) (d)/(a) rArr alpha = - (d)/(a)`
Hence, the real root of `ax^(3) + bx^(2) + cx + d = 0 "is" -(d)/(a)`.