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2011 Paper 3 Q2
D: 1700.0 B: 1516.0
polynomials rational root theorem proof by contradiction irrationality integer roots divisibility monic polynomial

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.

  1. Show that the \(n\)th root of \(2\) is irrational for \(n\ge2\).
  2. Show that the cubic equation \[ x^3- x +1 =0 \] has no rational roots.
  3. Show that the polynomial equation \[ x^n- 5x +7 =0 \] has no rational roots for \(n\ge2\).

Solution: 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.

  1. 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 \(n \geq 2\)
  2. 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.
  3. 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.

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2002 Paper 1 Q5
D: 1484.0 B: 1516.0
roots of polynomials symmetric functions factorization polynomial evaluation integer roots Vieta's formulas

Let \[ \f(x) = x^n + a_1 x^{n-1} + \cdots + a_n\;, \] where \(a_1\), \(a_2\), \(\ldots\), \(a_n\) are given numbers. It is given that \(\f(x)\) can be written in the form \[ \f(x) = (x+k_1)(x+k_2)\cdots(x+k_n)\;. \] By considering \(\f(0)\), or otherwise, show that \(k_1k_2 \ldots k_n =a_n\). Show also that $$(k_1+1)(k_2+1)\cdots(k_n+1)= 1+a_1+a_2+\cdots+a_n$$ and give a corresponding result for \((k_1-1)(k_2-1)\cdots(k_n-1)\). Find the roots of the equation \[ x^4 +22x^3 +172x^2 +552x+576=0\;, \] given that they are all integers.

Solution: \begin{align*} && f(0) &= 0^n + a_1\cdot 0^{n-1} + \cdots + a_n \\ &&&= a_n \\ && f(0) &= (0+k_1)(0+k_2) \cdots (0+k_n) \\ &&&= k_1 k_2 \cdots k_n \\ \Rightarrow && a_n &= k_1 k_2 \cdots k_n \\ \\ &&f(1) &= 1^n + a_1 \cdot 1^{n-1} + \cdots + a_n \\ &&&= 1 + a_1 + a_2 + \cdots + a_n \\ && f(1) &= (1 + k_1) (1 + k_2) \cdots (1+k_n) \\ \Rightarrow && (k_1+1)\cdots (k_n+1) &= 1 + a_1 + \cdots + a_n \\ \\ && f(-1) &= (-1)^{n} + a_1 \cdot (-1)^{n-1} + \cdots + a_n \\ &&&= a_n - a_{n-1} + \cdots + (-1)^{n-1} a_1 + (-1)^{n} \\ && f(-1) &= (-1+k_1)(-1+k_2) \cdots (-1+k_n) \\ &&&= (k_1-1)(k_2-1)\cdots(k_n-1) \\ \Rightarrow && (k_1-1)\cdots(k_n-1) &= a_n - a_{n-1} + \cdots + (-1)^{n-1} a_1 + (-1)^{n} \end{align*} \(576 = 2^6 \cdot 3^2\). Notice that \(1 - 22 + 172 -552 + 576 = 175 = 5^2 \cdot 7\) and \(1+22 + 172+552+576 = 1323 = 3^3 \cdot 7^2\). \(k_i = 2, 6, 6, 8\) therefore the roots are \(-2, -6, -6, -8\)

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