Let $p(x)$ be a polynomial of degree $3$ with real coefficients. Which of the following is possible ?
a) $p(x)$ has no real roots
b) $p(x) $ has exactly two real roots
c) $p(1)=-1, \; p(2)=1,\; p(3)=11,\; p(4)=35$
d) $i-1$ and $i+1$ are roots of $p(x)$, where $i=\sqrt{-1}$
Every odd degree real polynomial has at least one real root, so a) is false
b) is false as $x^3-1$ serve an example
d) is also false as complex roots are occur in pairs.
so c) is true.
My try for c): take $p(x)=a_0+a_1x+a_2x^2+a_3x^3$ where $a_3 \neq 0$
$p(1)=-1 \Rightarrow a_0+a_1+a_2+a_3=-1$
$p(2)=1 \Rightarrow a_0+2a_1+2^2a_2+2^3a_3=1$
$p(3)=11 \Rightarrow a_0+3a_1+3^2a_2+3^3a_3=11$
$p(4)=35 \Rightarrow a_0+4a_1+4^2a_2+4^3a_3=35$
So this can be written as $$\begin{align*}\left[\begin{matrix}1&1 &1 &1\\1&2 & 2^2 & 2^3\\ 1 & 3 & 3^2 &3^3\\ 1 &4 &4^2 &4^3\end{matrix}\right]\left[\begin{matrix}a_0\\a_1\\a_2\\a_3\end{matrix}\right] &= \left[\begin{matrix}-1\\1\\11\\35\end{matrix}\right] \end{align*}$$
Since the coefficient matrix is Vandermonde, so the determinant is not zero and hence this system has a unique solution. So such a unique polynomial exist.
Is my argument correct? or any other method to show this?
Thanks in advance
d) is no impssible indeed, but the reason is the if $z\in\mathbb C$ is a root of a polynomial with real coefficients, then $\overline z$ is also a root. Therefore, $p(x)$ would have four roots: $\pm i-1$ and $\pm i+1$. That's impossible, since the polynomial has degree $3$.