Show that a polynomial p can't have both roots 1 and i

Question

I have to show that a polynomial $p\in P_3(\mathbb{R})$ where $p\neq0$, can't have both roots, $1$ and $i$, where $i^2=-1$. The polynomial has degree 2 or less.

$p(\alpha)=c_0+c_1\alpha+c_2\alpha^2$

I know that, when we want to find the roots of a function, we have to solve the function for $x$ so that $f(x)=0$

I don't know how to solve this kind of question, when $p$ cant be equal to $0$ and I have to show that the polynomial $p\in P_3(\mathbb{R})$ can't have both roots 1 and $i$.

Thanks in advance

Answer 1

I don't understand the fuss... Just write $p(x)=(x-1)(x-i)$ and expand it to get $p(x)=x^2-(1+i)x+i$, which clearly does not belong to $P_{3}(\mathbb{R})$. Or is it?

Answer 2

If we have a polynomial with real coefficients, with $z$ as a root, one can check that $\overline{z}$ must also be a root of the polynomial. Thus, if $i$ is a root of a polynomial $p$, then so is $-i$. If we also require that $1$ be a root of $p$, then $p$ has at least $3$ roots. But how many roots can $p$ have, given that it's of degree at most $2$?

Answer 3

There is a very good theorem I suggest you to know:

Let $P(x)$ a polynomial of degree $n$ (with real coefficents) of the form $P(x)=a_nx^n+\cdots a_1x+x_0$ has:

A)An even number of real solutions (counted with their multiplicity) if $n$ is even;

B)An odd number of real solutions (counted with their multiplicity) if $n$ is odd.

This is a powerful theorem and sowe understood that $P(\alpha)$ can't have zeros $i$ and $1$ because the degree of $P$ is $2$, we have $0$ or $2$ real solutions (counted with their multiplicity).

Answer 4

If $p(x)=c_0+c_1x+c_2x^2 $ then $p(1) = c_0+c_1+c_2 $ and $p(i) = c_0+ic_1-c_2 $.

If $p(1) = p(i) = 0$ then, adding these, $0 =2c_0+(1+i)c_1 =2c_0+c_1+ic_1 $. Since all the $c_j$ are real, this implies that $c_1 = 0$ and $2c_0+c_1 = 0$ so $c_0 = 0$. Since $c_0+c_1+c_2 0$, $c_2 = 0$ so the polynomial is the zero polynomial.