Why is $-1$ not THE only real root of $f(x)=x^3+x+1$?

Question

Here is what I thought.

By the Descartes' Rule of Signs, $f(x)=x^3+x+1$ has no change in signs so it has no positive real root, and $f(-x)=-x^3-x+1$ has only one change in signs so it has exactly one negative real root.
Furthermore, because $f(x)$ is a rational polynomial, all its irrational roots appear in conjugate pairs and so its complex roots if there is any. Therefore the only negative real root can not be irrational or complex, which means it must be a negative rational root.
Then by the rational root test, all rational must be either $1$ or $-1$ if there is any. Hence $-1$ should be the only negative real root.

May anyone tell me where I was wrong because obvioulsy $f(-1)\ne 0$...?

“All it’s irrational roots come in conjugate pairs.” This is not true. It is true that all complex roots come in conjugate pairs, but real irrational roots do no necessarily come in pairs. — Thomas Andrews, Oct 21 '21 at 02:52
$-1$ is not the only real root, because it is not a root to the equation. The only real root would be approximately $-0.68233...$. — Graham Kemp, Oct 21 '21 at 03:07
@ThomasAndrews I remembered that the irrational roots of a rational polynomial come in pairs. It is not? — Andy Z, Oct 21 '21 at 04:32
@RobertIsrael haha so true but is it true that if all the complex roots of a rational polynomial are real, then the irrational roots must come in pairs? — Andy Z, Oct 21 '21 at 05:10
It is only the non-real complex roots that must come in pairs. — Robert Israel, Oct 21 '21 at 05:26
@AndyZ "irrational roots must come in pairs" $;-;$ That's true for quadratics with rational coefficients, and for square roots by The radical conjugate roots theorem, but neither applies here. — dxiv, Oct 21 '21 at 07:02
@dxiv Alright, it must be me that misremembered this. Thanks a lot. — Andy Z, Oct 22 '21 at 01:40

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