# Third degree Polynomial has the properties

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?

• Have you solved this system? Apr 13 '18 at 14:12
• For $b$, the question asks what is possible, not inevitable. Therefore showing a single example for which the property does not hold does not show that $b$ is false.
– lulu
Apr 13 '18 at 14:13
• @Dr. Sonnhard Graubner: $A^{-1}b$ is a solution where $A$ is the coefficient matrix and $b$ is a column vector on the right hand side
– user444830
Apr 13 '18 at 14:16
• I know this theorem Apr 13 '18 at 14:19
• Dr. Sonnhard Graubner: I know how to solve this system sir. My question is whether this argument is valid or not? If not, what I'm doing wrong?
– user444830
Apr 13 '18 at 14:21

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 $3$.

• $p(x)=x^3-x^2$ has exactly three roots! b) is wrong. Apr 13 '18 at 14:14
• @Dr.SonnhardGraubner If you count them with multiplicity. That's not mentioned here. Apr 13 '18 at 14:15
• And why not? I Can not your condition. Every polynomial of degree three has exactly three Zeros. Apr 13 '18 at 14:17
• My reason for d) means if $i-1$ is a root then $-i-1$ is also a root and similarly for $i+1$ , so it has four roots .That's what you are saying!