If x=i is one root of x^3 + x^2 + x + 1, find the other two roots.
Thank you!
Well the complex roots of a real polynomial occur in conjugate pairs, so if $\displaystyle x=i$ is a root so is $\displaystyle x=-i$, so $\displaystyle x^2+1$ is a factor of your cubic.
Now divide $\displaystyle x^3+x^2+x+1$ by $\displaystyle x^2+1$ to get the remaining linear factor and hence find the final root.
RonL
I don't get this. If $\displaystyle x=i$ and $\displaystyle x=-i$, that would mean $\displaystyle i=-i$, which I'm pretty sure is not true. It's been a while since pre-calc for me. What is this root lingo that allows both $\displaystyle x=i$ and $\displaystyle x=-i$ ?
no.
both i and -i being roots does not imply they are equal.
look at x^2 - 1 = 0
the roots are 1 and -1, does that mean 1 = -1? of course not
CaptainBlack called both x, but he was suggesting that these are the x-values that give roots for the equation, not that they are equal. an abuse of notation, i know, but it is understood in the context
complex roots come in conjugate pairs. we get complex roots when we take even roots of negative numbers, these roots therefore come in +/- form
Hello, antz215!
Another approach . . .
Multiply by $\displaystyle (x-1)\!:\;\;(x-1)(x^3+x^2+x+1) \:=\:0\quad\Rightarrow\quad x^4-1\:=\:0$If $\displaystyle x=i$ is one root of $\displaystyle x^3 + x^2 + x + 1 \:=\:0$, find the other two roots.
. . $\displaystyle x^4 \:=\:1\quad\Rightarrow\quad x^2 \:=\:\pm1 \quad\Rightarrow\quad x \:=\:\pm\sqrt{\pm1} \quad\Rightarrow\quad x \;=\;1,\,\text{-}1,\,i,\,\text{-}i$
We introduced $\displaystyle x = 1$; we were given $\displaystyle x = i$
Therefore, the other two roots are: .$\displaystyle x \;=\;\text{-}1,\,\text{-}i$