Tìm đa thức H ( x ) sao cho H ( x ) + 2 K ( x ) = P ( x ) .

c) Ta có: \(H\left( x \right) + 2K\left( x \right) = P\left( x \right)\) hay \(H\left( x \right) = P\left( x \right) - 2K\left( x \right)\).

Do đó, \(H\left( x \right) = 3{x^5} + 5{x^3} - x - 9 - 2\left( {3{x^5} - 3{x^3} + x + 5} \right)\)

           \(H\left( x \right) = 3{x^5} + 5{x^3} - x - 9 - 6{x^5} + 6{x^3} - 2x - 10\)

          \(H\left( x \right) = \left( {3{x^5} - 6{x^5}} \right) + \left( {5{x^3} + 6{x^3}} \right) + \left( { - x - 2x} \right) + \left( { - 9 - 10} \right)\)

         \(H\left( x \right) = - 3{x^5} + 11{x^3} - 3x - 19\).

Vậy \(H\left( x \right) = - 3{x^5} + 11{x^3} - 3x - 19\).