<s>[INST] If $x^2+x+1$ is a factor of $f (x)= ax^3+bx^2+cx+d$ , then the real root of $f (x)=0$ is If $x^2+x+1$  is a factor of  $f (x)= ax^3+bx^2+cx+d$  , then the real root of $f (x)$ is  
A) $-d/a$
B) $d/a$
C) $a/d $
D) none of these 
My try 
I take $f (x) = ( x^2+x+1)(x+1)  = x^3+2x^2+2x+1$
Real root $x=  -d/a =-1$
And also  
$f (x) = ( x^2+x+1)(x-2)  = x^3-x^2-x-2$
Real root $x =  -d/a =2$
But how can I prove it in general? 
 [/INST] The only way $x^2+x+1$ can be a factor of $ax^3+\cdots+d$ is if the
complementary factor is $ax+d$. As $x^2+x+1>0$ for all real $x$, the real solutions of $ax^3+\cdots+d=0$ must be the real solutions of $ax+d=0$.
</s>