Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Isn't the only possibility $p(x)=(x-2)(x-2-i)(x-2+i)=0$? Am I losing my mind?

$2p(x)=0$ has the same roots, but that's not a distinct equation.

share|improve this question
3  
The equation $2p(x)=0$ is equivalent to $p(x)=0$ but not equal. –  Américo Tavares Mar 6 '13 at 22:19
1  
It appears to be a misunderstanding of the word "distinct" on my part. –  Random Variable Mar 6 '13 at 22:42

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
And this is valid even with coefficients in $\mathbb{C}$. –  vonbrand Mar 6 '13 at 22:35
    
@vonbrand Only because the initial question was altered, and not by the OP, I must add. –  Git Gud Mar 6 '13 at 22:37
    
@GitGud Sorry for editing that... but I thought imaginary roots only occurred in conjugate pairs, and thus (thought) the edit was trivial... –  anorton Mar 6 '13 at 22:44
    
@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. –  Git Gud Mar 6 '13 at 22:47
    
@GitGud oh. I see now. thanks. –  anorton 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.