Consider the expansion of \({x^2}{\left( {3{x^2} + \frac{k}{x}} \right)^8}\). The constant term is \({\text{16 128}}\).

Find \(k\).

valid approach     (M1)

eg     \({\text{\(\left( \begin{array}{c}8\\r\end{array} \right)\)}}{\left( {3{x^2}} \right)^{8 - r}}{\left( {\frac{k}{x}} \right)^r}\),

\({\left( {3{x^2}} \right)^8} + {\text{\(\left( \begin{array}{c}8\\1\end{array} \right)\)}}{\left( {3{x^2}} \right)^7}\left( {\frac{k}{x}} \right) + {\text{\(\left( \begin{array}{c}8\\2\end{array} \right)\)}}{\left( {3{x^2}} \right)^6}{\left( {\frac{k}{x}} \right)^2} +  \ldots \), Pascal’s triangle to \({9^{{\text{th}}}}\) line

attempt to find value of r which gives term in \({x^0}\)     (M1)

eg     exponent in binomial must give \({x^{ - 2}},{\text{ }}{x^2}{\left( {{x^2}} \right)^{8 - r}}{\left( {\frac{k}{x}} \right)^r} = {x^0}\)

correct working     (A1)

eg     \(2(8 - r) - r =  - 2,{\text{ }}18 - 3r = 0,{\text{ }}2r + ( - 8 + r) =  - 2\)

evidence of correct term     (A1)

eg     \({\text{\(\left( \begin{array}{c}8\\2\end{array} \right)\), \(\left( \begin{array}{c}8\\6\end{array} \right)\)}}{\left( {3{x^2}} \right)^2}{\left( {\frac{k}{x}} \right)^2},{\text{ }}r = 6,{\text{ }}r = 2\)

equating their term and 16128 to solve for \(k\)     M1

eg     \({x^2}{\text{\(\left( \begin{array}{c}8\\6\end{array} \right)\)}}{\left( {3{x^2}} \right)^2}{\left( {\frac{k}{x}} \right)^6} = 16128,{\text{ }}{k^6} = \frac{{16128}}{{28(9)}}\)

\(k =  \pm 2\)     A1A1     N2

Note: If no working shown, award N0 for \(k = 2\).

Total [7 marks]