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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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