Year: 2019
Paper: 3
Question Number: 4
Course: UFM Pure
Section: Roots of polynomials
There was a significant rise in the total entry this year with an increase of nearly 8.5% on 2018. One question was attempted by over 90%, two others were very popular, and three further questions were attempted by 60% or more. No question was generally avoided and even the least popular attracted more than 10% of the candidates. 88% restricted themselves to attempting no more than 7 questions, and only a handful, but not the very best, scored strongly attempting more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The $n$th degree polynomial P$(x)$ is said to be \textit{reflexive} if:
\begin{enumerate}
\item[(a)] P$(x)$ is of the form $x^n - a_1x^{n-1} + a_2x^{n-2} - \cdots + (-1)^na_n$ where $n \geq 1$;
\item[(b)] $a_1, a_2, \ldots, a_n$ are real;
\item[(c)] the $n$ (not necessarily distinct) roots of the equation P$(x) = 0$ are $a_1, a_2, \ldots, a_n$.
\end{enumerate}
\begin{questionparts}
\item Find all reflexive polynomials of degree less than or equal to 3.
\item For a reflexive polynomial with $n > 3$, show that
$$2a_2 = -a_2^2 - a_3^2 - \cdots - a_n^2.$$
Deduce that, if all the coefficients of a reflexive polynomial of degree $n$ are integers and $a_n \neq 0$, then $n \leq 3$.
\item Determine all reflexive polynomials with integer coefficients.
\end{questionparts}\begin{questionparts}
\item Suppose $n = 1$, then all polynomials are reflexive (since $x - a_1$ has the root $a_1$.
Suppose $n = 2$, then we want
\begin{align*}
&& x^2-a_1x+a_2 &= (x-a_1)(x-a_2) \\
&&&= x^2-(a_1+a_2)x+a_1a_2 \\
\Rightarrow && a_2 &= 0 \\
\end{align*}
So all polynomials of the form $x^2-a_1x$ work and no others.
Suppose $n = 3$ then we want
\begin{align*}
&& x^3-a_1x^2+a_2x-a_3 &= (x-a_1)(x-a_2)(x-a_3) \\
&&&= x^3-(a_1+a_2+a_3)x+(a_1a_2+a_1a_3+a_2a_3)x-a_1a_2a_3 \\
\Rightarrow && a_2+a_3 &= 0 \\
&& a_2a_3 &= a_2 \\
\Rightarrow && -a_2^2 &= a_2 \\
\Rightarrow && a_2 &= 0, -1 \\
&& -a_1a_2^2 &= -a_2 \\
\Rightarrow && a_2 &= 0, a_2 = 1/a_1
\end{align*}
So we need either $x^3-a_1x$ or $(x+1)^2(x-1) = x^3+x^2-x-1$
\item Suppose $n > 3$ then
\begin{align*}
&& \sum a_i^2 &= \left (\sum a_i \right)^2 - 2 \sum_{i < j} a_i a_j \\
&& &= a_1^2 - 2a_2 \\
\Rightarrow && 2a_2 &= a_1^2 - \sum a_i^2 \\
&&&= -a_2^2 - a_3^2 - \cdots - a_n^2
\end{align*}
So $(a_2+1)^2 = 1-a_3^2 -\cdots -a_n^2$ so if $a_n > 0$ (or any other $a_i, i > 2$ for that matter) then we must have $a_n = \pm 1, a_{3}, \ldots a_{n-1} = 0$, but if $a_n = \pm 1$ $x = 0$ is not a root. Therefore we must have $a_0$ and $a_i = 0$ for all $i > 3$
\item The only reflexive polynomials therefore must be $x^n - kx^{n-1}$ and $x^{n+3}+x^{n+2}-x^{n+1}-x^n$
\end{questionparts}
Two thirds of candidates attempted this scoring, on average about half marks. The first part was often well answered by those that used the Vieta equations, though a common error was to divide by a potential zero and therefore omit one of the solutions. Those that substituted the three roots into the polynomial equation encountered equations that were more difficult to solve, and the method yielded additional solutions which were often not rejected, so generally those taking this approach did much less well. Part (ii) had many good answers from squaring the sum of roots equation, though a common error was made with inaccurate summation notation. Many argued the deduction correctly but likewise many others assumed the a's were non‐zero without a reason. Part (iii) was found more difficult. Those candidates that did not realise the significance of the previous deduction were rarely successful and others gave the correct answer, but with no explanation. Those making the inductive argument to factorise powers of x rarely justified that the remaining polynomial was likewise reflexive.