For what value of 'k', the G.C.D. of x^2+x-(2k+2) and 2x^2+kx-12 is x+4?

• i think k=5
• 10 years agoHelpfull: Yes(0) No(0)
• k = 5
  
  Since (x+4) is the G.C.D of x^2+x-(2k+2) and 2x^2+kx-12
  Remainder of x^2+x-(2k+2)/ (x+4) and 2x^2+kx-12/(x+4) = 0
  
  ie, Remainder of x^2+x-(2k+2)/ (x+4) = 12 - (2k+2)
  ====> 12 - (2k+2) = 0
  2k + 2 =12
  2k = 10
  k = 5
  
• 10 years agoHelpfull: Yes(0) No(0)