Consider the polynomial \(p(x) = {x^4} + a{x^3} + b{x^2} + cx + d\), where a, b, c, d \( \in \mathbb{R}\).

Given that 1 + i and 1 − 2i are zeros of \(p(x)\), find the values of a, b, c and d.

METHOD 1

1 + i is a zero \( \Rightarrow \) 1 – i is a zero     (A1)

1 – 2i is a zero \( \Rightarrow \) 1 + 2i is a zero     (A1)

\(\left( {x - (1 - {\text{i}})} \right)\left( {x - (1 + {\text{i}})} \right) = ({x^2} - 2x + 2)\)     (M1)A1

\(\left( {x - (1 - 2{\text{i}})} \right)\left( {x - (1 + 2{\text{i}})} \right) = ({x^2} - 2x + 5)\)     A1

\(p(x) = ({x^2} - 2x + 2)({x^2} - 2x + 5)\)     M1

\( = {x^4} - 4{x^3} + 11{x^2} - 14x + 10\)     A1

\(a = - 4,{\text{ }}b = 11,{\text{ }}c = - 14,{\text{ }}d = 10\)

[7 marks]

METHOD 2

\(p(1 + {\text{i}}) = - 4 + ( - 2 + 2{\text{i}})a + (2{\text{i}})b + (1 + {\text{i}})c + d\)     M1

\(p(1 + {\text{i}}) = 0 \Rightarrow \left\{ {\begin{array}{*{20}{c}}
  { - 4 - 2a + c + d = 0} \\
  {2a + 2b + c = 0}
\end{array}} \right.\)     M1A1A1

\(p(1 - 2{\text{i}}) = - 7 + 24{\text{i}} + ( - 11 + 2{\text{i}})a + ( - 3 - 4{\text{i}})b + (1 - 2{\text{i}})c + d\)

\(p(1 - 2{\text{i}}) = 0 \Rightarrow \left\{ {\begin{array}{*{20}{c}}
  { - 7 - 11a - 3b + c + d = 0} \\
  {24 + 2a - 4b - 2c = 0}
\end{array}} \right.\)     A1

\(\left( {\begin{array}{*{20}{c}}
  a \\
  b \\
  c \\
  d
\end{array}} \right) = {\left( {\begin{array}{*{20}{c}}
  { - 2}&0&1&1 \\
  2&2&1&0 \\
  { - 11}&{ - 3}&1&1 \\
  2&{ - 4}&{ - 2}&0
\end{array}} \right)^{ - 1}}\left( {\begin{array}{*{20}{c}}
  4 \\
  0 \\
  7 \\
  { - 24}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 4} \\
  {11} \\
  { - 14} \\
  {10}
\end{array}} \right)\)     M1A1

\(a = - 4,{\text{ }}b = 11,{\text{ }}c = - 14,{\text{ }}d = 10\)

[7 marks]

Most candidates attempted this question, using different approaches. The most successful approach was the method of complex conjugates and the product of linear factors. Candidates who used this method were in general successful whereas candidates who attempted direct substitution and separation of real and imaginary parts to obtain four equations in four unknowns were less successful because either they left the work incomplete or made algebraic errors that led to incorrect answers.