1

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?

user373141
  • 2,503
  • 3
    You must have $f(x)=(x^2+x+1)(ux+v)$. But then you get $u=a,v=d$, and the root is $-v/u=-d/a$. – Thomas Andrews Aug 03 '17 at 12:05
  • Terminology nitpicking: there is no root of $f(x) = 0$. That expression is equation, some $x$'s might solve it, some might not. If $\alpha$ is a solution of the equation, then we say that $\alpha$ is a root of $f$. – Ennar Aug 03 '17 at 12:12
  • @Ennar fyi: this is a very common abuse of language in English. – Bill Dubuque Aug 03 '17 at 14:29
  • @Bill Dubuque, the only question is whether OP is aware of it or not. Since this is an elementary question, isn't it better to point it out? I'm not trying to be obnoxious. – Ennar Aug 03 '17 at 14:45

2 Answers2

6

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 $ax+d=0$.

Angina Seng
  • 158,341
2

If you divide $f (x)= ax^3+bx^2+cx+d$ to $x^2+x+1$ : you will have $$f (x)= ax^3+bx^2+cx+d=(x^2+x+1)(ax+(b-a))+x(a+c-b+a)+(a-b+d)$$ when remainder is $0 $ so $$ax+(b-a)=0 \to x=\frac{a-b}{a} \tag{1}\\ \forall x:\begin{cases}a+c-b+a=0 \\a-b+d=0\end{cases}$$ now from 3rd equation we have $d=-(a-b)$ put into (1) $$x=\frac{a-b}{a}=x=\frac{-d}{a}$$

Khosrotash
  • 24,922