The same remainder is found when \(2{x^3} + k{x^2} + 6x + 32\) and \({x^4} - 6{x^2} - {k^2}x + 9\) are divided by \(x + 1\) . Find the possible values of k .

let \(f(x) = 2{x^3} + k{x^2} + 6x + 32\)

let \(g(x) = {x^4} - 6{x^2} - {k^2}x + 9\)

\(f( - 1) =  - 2 + k - 6 + 32( = 24 + k)\)     A1

\(g( - 1) = 1 - 6 + {k^2} + 9( = 4 + {k^2})\)     A1

\( \Rightarrow 24 + k = 4 + {k^2}\)     M1

\( \Rightarrow {k^2} - k - 20 = 0\)

\( \Rightarrow (k - 5)(k + 4) = 0\)     (M1)

\( \Rightarrow k = 5,\, - 4\)     A1A1

[6 marks]

Candidates who used the remainder theorem usually went on to find the two possible values of k. Some candidates, however, attempted to find the remainders using long division. While this is a valid method, the algebra involved proved to be too difficult for most of these candidates.