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2023 Paper 2 Q4
D: 1500.0 B: 1500.0
polynomials algebraic manipulation minimal polynomial surds integer coefficients substitution expansion

  1. Show that, if \((x-\sqrt{2})^2 = 3\), then \(x^4 - 10x^2 + 1 = 0\). Deduce that, if \(\mathrm{f}(x) = x^4 - 10x^2 + 1\), then \(\mathrm{f}(\sqrt{2}+\sqrt{3}) = 0\).
  2. Find a polynomial \(\mathrm{g}\) of degree 8 with integer coefficients such that \(\mathrm{g}(\sqrt{2}+\sqrt{3}+\sqrt{5}) = 0\). Write your answer in a form without brackets.
  3. Let \(a\), \(b\) and \(c\) be the three roots of \(t^3 - 3t + 1 = 0\). Find a polynomial \(\mathrm{h}\) of degree 6 with integer coefficients such that \(\mathrm{h}(a+\sqrt{2}) = 0\), \(\mathrm{h}(b+\sqrt{2}) = 0\) and \(\mathrm{h}(c+\sqrt{2}) = 0\). Write your answer in a form without brackets.
  4. Find a polynomial \(\mathrm{k}\) with integer coefficients such that \(\mathrm{k}(\sqrt[3]{2}+\sqrt[3]{3}) = 0\). Write your answer in a form without brackets.

Solution:

  1. \(\,\) \begin{align*} && 3 &= (x-\sqrt2)^2 \\ &&&= x^2 - 2\sqrt2 x + 2 \\ \Rightarrow && 2\sqrt2 x &= x^2-1 \\ \Rightarrow && 8x^2 &= x^4 - 2x^2 + 1 \\ \Rightarrow && 0 &= x^4 - 10x^2 + 1 \end{align*} Noticing that \((\sqrt2+\sqrt3-\sqrt2)^2 = 3\) we note that \(\sqrt2 + \sqrt3\) is a root of our quartic.
  2. Suppose \(x = \sqrt2 + \sqrt3 + \sqrt5\) then \begin{align*} && 0 &= (x - \sqrt5)^4 - 10(x-\sqrt5)^2 + 1 \\ &&&= x^4 - 4\sqrt5x^3 + 30x^2-20\sqrt5 x +25 - 10x^2+20\sqrt5x -50 + 1\\ &&&= (x^4+20x^2- 24) - 4\sqrt5 x^3 \\ \Rightarrow && 80x^6 &= (x^4+20x^2-24)^2 \\ &&&= x^8 + 40x^6 + 352x^4 - 960x^2+576 \\ \Rightarrow && 0 &= x^8-40x^6 + 352x^4-960x^2+576 \end{align*} So take \(g(x) = x^8-40x^6 + 352x^4-960x^2+576\).
  3. Notice that if \(p(t) = t^3-3t+1\) then \(p(t -\sqrt2) = 0\) for \(t = a,b,c\) so \begin{align*} && 0 &= (t - \sqrt2)^3 -3(t - \sqrt2) + 1 \\ &&&= t^3-3\sqrt2 t^2 + 6t - 2\sqrt2 - 3t + 3\sqrt 2 + 1 \\ &&&= (t^3+3t+1) - \sqrt2 (3t^2+1) \\ \Rightarrow && 2(3t^2+1)^2 &= (t^3+3t+1)^2 \\ \Rightarrow && 2(9t^4+6t^2+1) &= t^6 + 6t^4+2t^3+9t^2+6t+1 \\ \Rightarrow && 0 &= t^6-12t^4+2t^3-3t^2+6t-1 \end{align*}
  4. \(\,\) \begin{align*} && t &= \sqrt[3]{2} + \sqrt[3]{3} \\ \Rightarrow && t^3 &= 2 + 3\sqrt[3]{12} + 3\sqrt[3]{18} + 3 \\ &&&= 5 + 3 \sqrt[3]{6}(\sqrt[3]{2} + \sqrt[3]{3}) \\ &&&= 5 + 3\sqrt[3]{6}t \\ \Rightarrow && 162t^3 &= (t^3-5)^3 \\ &&&= t^9-15t^6+75t^3 -125 \\ \Rightarrow && 0 &= t^9-15t^6-87t^3-125 \end{align*} so \(k(x) = x^9 - 15x^6-87x^3-125\)

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2018 Paper 1 Q5
D: 1484.0 B: 1516.0
polynomials remainder theorem factor theorem integer coefficients divisibility degree 4 polynomial combinatorics proof by contradiction

  1. Write down the most general polynomial of degree 4 that leaves a remainder of 1 when divided by any of \(x-1\,\), \(x-2\,\), \(x-3\,\) or \(x-4\,\).
  2. The polynomial \(\P(x)\) has degree \(N\), where \(N\ge1\,\), and satisfies \[ \P(1) = \P(2) = \cdots = \P(N) =1\,. \] Show that \(\P(N+1) \ne 1\,\). Given that \(\P(N+1)= 2\,\), find \(\P(N+r)\) where \(r\) is a positive integer. Find a positive integer \(r\), independent of \(N,\) such that \(\P(N+r) = N+r\,\).
  3. The polynomial \({\rm S}(x)\) has degree 4. It has integer coefficients and the coefficient of \(x^4\) is 1. It satisfies \[ {\rm S}(a) = {\rm S}(b) = {\rm S}(c) = {\rm S}(d) = 2001\,, \] where \(a\), \(b\), \(c\) and \(d\) are distinct (not necessarily positive) integers.
    • Show that there is no integer \(e\) such that \({\rm S}(e) = 2018\,\).
    • Find the number of ways the (distinct) integers \(a\), \(b\), \(c\) and \(d\) can be chosen such that \({\rm S}(0) = 2017\) and \(a < b< c< d\,.\)

Solution:

  1. \(p(x) = C(x-1)(x-2)(x-3)(x-4)+1\)
  2. Suppose \(P(N+1) = 1\) them we could consider \(f(x) = P(x) - 1\) to be a polynomial of degree \(N\) with at least \(N+1\) roots, which would be a contradiction. Therefore \(P(N+1) \neq 1\). Since \(P(x) = C(x-1)(x-2)\cdots(x-N) + 1\) and \(P(N+1) = 2\) we must have \(C \cdot N! + 1 = 2 \Rightarrow C = \frac{1}{N!}\), hence \(P(x) = \binom{x-1}{N} + 1\) ie \(P(N+r) = \binom{N+r-1}{N}+1\) so \(P(N+2) = \binom{N+1}{N} +1= N+2\), so we can take \(r=2\).
    1. Suppose consider \(p(x) = S(x) - 2001\), then \(p(x)\) has roots \(a,b,c,d\) and suppose we can find \(e\) such that \(p(e) = 17\) then we must have \((e-a)(e-b)(e-c)(e-d) = 17\) but the only possible factors of \(17\) are \(-17,-1,1,17\) and we cannot have all \(4\) of them. Hence this is not possible.
    2. Now we have \(abcd = 16\), so we can have factors \(-16,-8,-4, -2, -1, 1, 2, 4,8,16\) (and we need to have \(4\) of them). If we have \(0\) negatives, the smallest product is \(1 \cdot 2 \cdot 4 \cdot 8 > 16\) If we have \(2\) negatives we must have \(1\) and \(-1\) (otherwise we have the same problem of being too large. So \(\{-1,1,-2,8\},\{-1,1,2,-8\},\{-1,1,-4,4\},\) If we have \(4\) negatives that's the same issue as with \(0\) negatives.

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