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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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1997 Paper 2 Q4
D: 1600.0 B: 1484.0
polynomials remainder theorem factor theorem polynomial division Lagrange interpolation system of equations

Show that, when the polynomial \({\rm p} (x)\) is divided by \((x-a)\), where \(a\) is a real number, the remainder is \({\rm p}(a)\).

  1. When the polynomial \({\rm p}(x)\) is divided by \(x-1,\,x-2,\,x-3\) the remainders are 3,1,5 respectively. Given that $${\rm p}(x)=(x-1)(x-2)(x-3){\rm q}(x)+{\rm r} (x),$$ where \({\rm q}(x)\) and \({\rm r}(x)\) are polynomials with \({\rm r}(x)\) having degree less than three, find \({\rm r}(x)\).
  2. Find a polynomial \({\rm P}(x)\) of degree \(n+1\), where \(n\) is a given positive integer, such that for each integer \(a\) satisfying \(0\le a\le n\), the remainder when \({\rm P}_n(x)\) is divided by \(x-a\) is \(a\).

Solution: Notice by polynomial division, we can write \(p(x) = (x-a)q(x) + r(x)\) where degree \(r(x) < 1\), ie \(r(x)\) is a constant. Evaluating at \(x = a\), we have \(p(a) = (a-a)q(a) + r(a) = r(a)\). Therefore \(r(a) = p(a)\) and since \(r(x)\) is a constant, it is always \(p(a)\).

  1. \(\,\) \begin{align*} && p(x) &= (x-1)(x-2)(x-3)q(x) + r(x) \\ && p(1) &= r(1) = 3 \\ && p(2) &= r(2) = 1 \\ && p(3) &= r(3) = 5 \end{align*} Therefore \(r(x)\) is a polynomial of degree \(2\) or less through \((1,3),(2,1), (3, 5)\) we can write this as \begin{align*} && r(x) &= 5\frac{(x-1)(x-2)}{(3-1)(3-2)} + 1\frac{(x-1)(x-3)}{(2-1)(2-3)} + 3\frac{(x-2)(x-3)}{(1-2)(1-3)} \\ &&&= \frac52(x^2-3x+2)-(x^2-4x+3) + \frac32(x^2-5x+6) \\ &&&= 3x^2-11x+11 \end{align*}
  2. Let \(P_n(x) = x(x-1)\cdots(x-n) + x\), then \(P_n(a) = a\)

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