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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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2016 Paper 2 Q2
D: 1600.0 B: 1516.0
polynomials factor theorem factorisation symmetric functions substitution cubic equation algebraic manipulation

Use the factor theorem to show that \(a+b-c\) is a factor of \[ (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2) +8(a^3+b^3+c^3) \,. \tag{\(*\)} \] Hence factorise (\(*\)) completely.

  1. Use the result above to solve the equation \[ (x+1)^3 -3 (x+1)(2x^2 +5) +2(4x^3+13)=0\,. \]
  2. By setting \(d+e=c\), or otherwise, show that \((a+b-d-e)\) is a factor of \[ (a+b+d+e)^3 -6(a+b+d+e)(a^2+b^2+d^2+e^2) +8(a^3+b^3+d^3+e^3) \, \] and factorise this expression completely. Hence solve the equation \[ (x+6)^3 - 6(x+6)(x^2+14) +8(x^3+36)=0\,. \]

Solution: Suppose \(c = a+b\) then \begin{align*} (a+b+c)^3 &-6(a+b+c)(a^2+b^2+c^2) +8(a^3+b^3+c^3) \\ &= (2(a+b))^3-6(2(a+b))(a^2+b^2+(a+b)^2) + 8(a^3+b^3+(a+b)^3) \\ &=16(a+b)^3 - 24(a+b)(a^2+b^2+ab)+8(a^3+b^3) \\ &= 8(a+b)(2(a+b)^2-3(a^2+b^2+ab)+(a^2-ab+b^2)) \\ &= 0 \end{align*} Therefore \(a+b-c\) is a factor. By symmetry \(a-b+c\) and \(-a+b+c\) are also factors. Since our polynomial is degree \(3\) it must be \(K(a+b-c)(b+c-a)(c+a-b)\) for some \(K\). Since the coefficient of \(a^3\) is \(3\), \(K = 3\). so we have: \(3(a+b-c)(b+c-a)(c+a-b)\)

  1. We want \(x + a + b = x+1\), \(x^2 + a^2 + b^2 = x^3+\frac52, x^3 + a^3 + b^3 = x^3+ \frac{13}{4}\). \(a+b = 1, a^2 + b^2 = 5/2\) so \(a = \frac32, b = -\frac12\) \begin{align*} 0 &= (x+1)^3 - 3(x+1)(2x^2+5)+2(4x^3+13) \\ &= 3(x +\frac{3}{2}+\frac{1}{2})(x - \frac{3}{2} - \frac{1}{2})(-x + \frac{3}{2} - \frac{1}{2}) \\ &= 3(x+2)(x-2)(1-x) \end{align*} and so the roots are \(x = 1, 2, -2\)
  2. Letting \(c = d+e\) we have \begin{align*} (a+b+d+e)^3 &-6(a+b+d+e)(a^2+b^2+d^2+e^2) +8(a^3+b^3+d^3+e^3) \\ &= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2-2de) +8(a^3+b^3+c^3 - 3cde) \\ &= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)+12(a+b+c)de - 24cde \\ &= \underbrace{(a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)}_{\text{has a factor of }a+b-c} + 12(a+b-c)de \end{align*} Therefore there is a factor of \(a+b-c\) or \(a+b-d-e\). By symmetry we must have the factors: \((a+b-d-e)(a-b-d+e)(a-b+d-e)\) and so the final expression must be: \(K(a+b-d-e)(a-b-d+e)(a-b+d-e)\) The coefficient of \(a^3\) is \(3\), therefore \(K = 3\) We want \(x+a+b+c = x + 6\), \(x^2+a^2+b^2+c^2 = 14\) and \(x^3 + a^3+b^3+c^3 = 36\), ie \(a = 1,b=2,c=3\) would work, so \begin{align*} 0 &= (x+6)^3 - 6(x+6)(x^2+14) +8(x^3+36) \\ &= 3(x+1-2-3)(x-1+2-3)(x-1-2+3) \\ &= 3x(x-4)(x-2) \end{align*} ie the roots are \(x = 0, 2, 4\)

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2009 Paper 2 Q4
D: 1600.0 B: 1500.0
polynomials divisibility repeated roots derivative degree 9 symmetric functions factor theorem

The polynomial \(\p(x)\) is of degree 9 and \(\p(x)-1\) is exactly divisible by \((x-1)^5\).

  1. Find the value of \(\p(1)\).
  2. Show that \(\p'(x)\) is exactly divisible by \((x-1)^4\).
  3. Given also that \(\p(x)+1\) is exactly divisible by \((x+1)^5\), find \(\p(x)\).

Solution: \(p(x) = q(x)(x-1)^5 + 1\) where \(q(x)\) has degree \(4\).

  1. \(p(1) = q(1)(1-1)^5 + 1 = 1\).
  2. \(p'(x) = q'(x)(x-1)^5 + 5(x-1)^4q(x) + 0 = (x-1)^4((x-1)q'(x) + 5q(x))\) so \(p'(x)\) is divisible by \((x-1)^4\)
  3. \(p(x)+1\) divisible by \((x+1)^5\) means that \(p(-1) = -1\) and \(p'(x)\) is divisible by \((x+1)^4\). Since \(p'(x)\) is degree \(8\) it must be \(c(x+1)^4(x-1)^4 = c(x^2 - 1)^4\). Expanding and integrating, we get \(p(x) = c(\frac{1}{9}x^9 -\frac{4}{7}x^7 + \frac{6}{5}x^5 - \frac{4}{3}x^3 + x) + d\). When \(x = 1\) we get \(c \frac{128}{315} + d = 1\) and when \(x = -1\) we get \(-c \frac{128}{315} + d = -1\) so \(2d = 0 \Rightarrow d = 0, c = \frac{315}{128}\) and \[ p(x) =\frac{315}{128} \l \frac{1}{9}x^9 -\frac{4}{7}x^7 + \frac{6}{5}x^5 - \frac{4}{3}x^3 + x\r \]

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2004 Paper 1 Q3
D: 1500.0 B: 1500.0
polynomials factorisation factor theorem multivariable polynomial algebraic manipulation

  1. Show that \(x-3\) is a factor of \begin{equation} x^3-5x^2+2x^2y+xy^2-8xy-3y^2+6x+6y \;. \tag{\(*\)} \end{equation} Express (\( * \)) in the form \((x-3)(x+ay+b)(x+cy+d)\) where \(a\), \(b\), \(c\) and \(d\) are integers to be determined.
  2. Factorise \(6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10\) into three linear factors.

Solution:

  1. Let \(f(x,y) = x^3-5x^2+2x^2y+xy^2-8xy-3y^2+6x+6y\), then \begin{align*} f(3,y) &= 27 - 5 \cdot 9 +18y + 3y^2-24y-3y^2+18 + 6y \\ &= 0 \end{align*}, therefore \(x-3\) is a factor of \(f(x,y)\). \begin{align*} f(x,y) &= x^3-5x^2+6x+y(2x^2-8x+6) + y^2(x-3) \\ &= (x-3)(x^2-2x)+y(x-3)(2x-2)+y^2(x-3) \\ &= (x-3)(x^2-2x+2y(x-1)+y^2) \\ &= (x-3)(x+y)(x+y-2) \end{align*}
  2. Let \(g(x,y) = 6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10\), notice that \(g(x,-2) = 0\), so \(y+2\) is a factor, \begin{align*} g(x,y) &= 6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10 \\ &= x^2(2+y) + x(12-4y-5y^2) + 6y^3-y^2-21y+10 \\ &= x^2(y+2) + x(y+2)(6-5y) + (y+2)(6y^2-13y+5) \\ &= (y+2)(x^2+(6-5y)x+(6y^2-13y+5)) \\ &= (y+2)(x-2y +1)(x-3y+5) \end{align*}

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2000 Paper 2 Q2
D: 1600.0 B: 1503.8
polynomials repeated roots differentiation factor theorem proof quartic factor

Prove that if \({(x-a)^{2}}\) is a factor of the polynomial \(\p(x)\), then \(\p'(a)=0\). Prove a corresponding result if \((x-a)^4\) is a factor of \(\p(x).\) Given that the polynomial $$ x^6+4x^5-5x^4-40x^3-40x^2+32x+k $$ has a factor of the form \({(x-a)}^4\), find \(k\).

Solution: First notice that \(p(x) = (x-a)^2q(x)\) so \(p'(x) = 2(x-a)q(x) + (x-a)^2q'(x) = (x-a)(2q(x)+(x-a)q'(x))\), in particular \(p'(a) = 0\) so \(x-a\) is a root of \(p'(x)\). If \((x-a)^4\) is a root of \(p(x)\) then \(p^{(3)}(a)= 0\). The proof is similar. Differentiating \(3\) times we obtain: \(6 \cdot 5 \cdot 4 x^3 + 4 \cdot 5 \cdot 4 \cdot 3 x^2 - 5\cdot4 \cdot 3 \cdot 2 x-40 \cdot 3 \cdot 2 \cdot 1 = 5!(x^3+2x^2-x-2) = 5!(x+2)(x^2-1)\). So our possible (repeated) roots are \(x=-2,-1,1\). We can check \(p'(x) = 6x^5+20x^4-20x^3-120x^2-80x+32\), and see \(p'(1) = 36 - 200 \neq 0\), \(p'(-1) = -6+20+20-120+80+32 \neq 0\), therefore \(a = -2\)

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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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