How is it possible to construct two distinct third-degree polynomial equations with real coefficients and with solutions $2$ and $2+ i$?

Isn't the only possibility $p(x)=(x-2)(x-2-i)(x-2+i)=0$?

The equation $2p(x)=0$ also has the solutions $2$ and $2+i$, but I wouldn't call that a distinct equation from $p(x)=0$.

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    $\begingroup$ The equation $2p(x)=0$ is equivalent to $p(x)=0$ but not equal. $\endgroup$ – Américo Tavares Mar 6 '13 at 22:19
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    $\begingroup$ It appears to be a misunderstanding of the word "distinct" on my part. $\endgroup$ – Random Variable Mar 6 '13 at 22:42

If it has degree $3$ then it has exactly three roots (in $\Bbb C$) counting multiplicities.

If $2$ and $2+i$ are roots of the polynomial, then, since you require for the polynomial to have real coefficients, $2-i$ must be a root.

Combine both statements above to conclude that the polynomial must be $$\lambda(x-2)(x-(2+i))(x-(2-i))$$for some $\lambda\in \Bbb R\setminus \{0\}$.

Note that the use of the quantifier some is correct, even though all polynomials that look like that satisfy your conditions.

  • $\begingroup$ And this is valid even with coefficients in $\mathbb{C}$. $\endgroup$ – vonbrand Mar 6 '13 at 22:35
  • $\begingroup$ @vonbrand Only because the initial question was altered, and not by the OP, I must add. $\endgroup$ – Git Gud Mar 6 '13 at 22:37
  • $\begingroup$ @GitGud Sorry for editing that... but I thought imaginary roots only occurred in conjugate pairs, and thus (thought) the edit was trivial... $\endgroup$ – apnorton Mar 6 '13 at 22:44
  • $\begingroup$ @anorton It is, but it makes vonbrand's comment false. If we don't consider the conjugate root, then $(x-2)^2(x-(2+i))$ won't be valid, as vonbrand suggested. $\endgroup$ – Git Gud Mar 6 '13 at 22:47
  • $\begingroup$ @GitGud oh. I see now. thanks. $\endgroup$ – apnorton Mar 6 '13 at 22:52

If a polynomial $\displaystyle P=\sum_{k=0}^na_kx^k$ with real coefficients has pure complex root $z$ (not real), then its conjugate $\overline{z}$ is also a root of $P$. Indeed, with the fact $\overline{a_k}=a_k$, we have $$0=P(z)=\overline{P(z)}=P(\overline{z}).$$ Now, the result is clear.


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