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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

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    $\begingroup$ Have you solved this system? $\endgroup$ – Dr. Sonnhard Graubner Apr 13 '18 at 14:12
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    $\begingroup$ 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. $\endgroup$ – lulu Apr 13 '18 at 14:13
  • $\begingroup$ @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 $\endgroup$ – user444830 Apr 13 '18 at 14:16
  • $\begingroup$ I know this theorem $\endgroup$ – Dr. Sonnhard Graubner Apr 13 '18 at 14:19
  • $\begingroup$ 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? $\endgroup$ – user444830 Apr 13 '18 at 14:21
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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 $3$.

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  • $\begingroup$ $p(x)=x^3-x^2$ has exactly three roots! b) is wrong. $\endgroup$ – Dr. Sonnhard Graubner Apr 13 '18 at 14:14
  • $\begingroup$ @Dr.SonnhardGraubner If you count them with multiplicity. That's not mentioned here. $\endgroup$ – José Carlos Santos Apr 13 '18 at 14:15
  • $\begingroup$ And why not? I Can not your condition. Every polynomial of degree three has exactly three Zeros. $\endgroup$ – Dr. Sonnhard Graubner Apr 13 '18 at 14:17
  • $\begingroup$ Please correct your answer, since it is wrong! $\endgroup$ – Dr. Sonnhard Graubner Apr 13 '18 at 14:20
  • $\begingroup$ 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! $\endgroup$ – user444830 Apr 13 '18 at 14:20
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Your guesses are right for a,b,c and d.

For b, as many have noticed, showing an example is not enough. A suitable proof would use the fact that (as you stated) complex roots come in pairs.

For c, one possibility would be to use the Lagrange interpolation formula. The theorem actually assesses both existence and uniqueness and is just the right fit for this question.

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  • $\begingroup$ This one helps too! Thanks! $\endgroup$ – user444830 Apr 13 '18 at 14:50

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