Year: 2011
Paper: 3
Question Number: 2
Course: LFM Stats And Pure
Section: Polynomials
The percentages attempting larger numbers of questions were higher this year than formerly. More than 90% attempted at least five questions and there were 30% that didn't attempt at least six questions. About 25% made substantive attempts at more than six questions, of which a very small number indeed were high scoring candidates that had perhaps done extra questions (well) for fun, but mostly these were cases of candidates not being able to complete six good solutions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
The polynomial $\f(x)$ is defined by
\[
\f(x) = x^n + a_{{n-1}}x^{n-1}
+ \cdots + a_{2} x^2+ a_{1} x + a_{0}\,,
\]
where $n\ge2$ and the coefficients $a_{0}$, $\ldots,$ $a_{{n-1}}$ are integers, with $a_0\ne0$. Suppose that the equation $\f(x)=0$ has a rational root $p/q$, where $p$ and $q$ are integers with no common factor greater than $1$, and $q>0$. By considering $q^{n-1}\f(p/q)$, find the value of $q$ and deduce that any rational root of the equation $\f(x)=0$ must be an integer.
\begin{questionparts}
\item Show that the $n$th root of $2$ is irrational for $n\ge2$.
\item Show that the cubic equation
\[
x^3- x +1 =0
\]
has no rational roots.
\item
Show that the polynomial equation
\[
x^n- 5x +7 =0
\]
has no rational roots for $n\ge2$.
\end{questionparts}Let $\f(x) = x^n + a_{{n-1}}x^{n-1}+ \cdots + a_{2} x^2+ a_{1} x + a_{0}$, and suppose $f(p/q) = 0$ with $(p,q) = 1$, the consider
\begin{align*}
&& 0 &= q^{n-1}f(p/q) \\
&&&= \frac{p^n}{q} + \underbrace{a_{n-1}p^{n-1} + a_{n-2}p^{n-2}q + \cdots + a_0q^{n-1}}_{\in \mathbb{Z}} \\
\end{align*}
But $p^n/q \not \in \mathbb{Z}$ unless $q = 1$ therefore $p/q$ must be an integer, ie all rational roots are integers.
\begin{questionparts}
\item Note that $\sqrt[n]2$ is a root of $x^n - 2 =0$, but this has no integer solutions. (We can try all factors of $2$). Therefore all its roots must be irrational, ie $\sqrt[n]2$ is irrational for $n \geq 2$
\item If $n$ is a root of $x^3 - x+1$ then it must be $1$ or $-1$ by the rational root theorem, ie $1-1+1 \neq 0$ and $-1 + 1 +1 \neq 0$, therefore no integer roots, therefore no rational roots.
\item Suppose $m$ is an integer root of $x^n - 5x + 7 = 0$ then by considering parity we must have $m^n - 5m + 7 \equiv 1 \pmod{2}$ therefore we cannot have any rational roots.
\end{questionparts}
This was quite a popular question, being attempted by 70% of candidates. Scores were polarized, though overall the mean score was below half marks, much the same as half of the questions on the paper. Most candidates successfully dealt with the stem. Attempts at part (i) were in equal proportions, applying the stem or a variant of the standard proof of the irrationality of the square root of 2, though some of the latter overlooked the fact that it was the nth root being discussed. Parts (ii) and (iii) saw three methods employed. One method was to consider the location of the real roots then apply the stem, the second being to re-arrange the expression to equal the integer and consider factors (again applying the stem). In both these cases, failure to consider all cases lost marks, and there were frequent lacks of rigour. However, considering x being odd or even, when used, was particularly slick and successful.