(3X+2)(2X-5)=AX^2+ KX+N;
THEN WHAT IS THE VALUE OF A-N+K?

• (3x+2)(2x-5)=6x^2-11x-10
  So,A=6, K=-11, N=-10
  Finally,
  A-N+K=6-(-10)+(-11)
  =6+10-11
  =5
  
• 13 years agoHelpfull: Yes(7) No(2)
• (3x-2)(2x-5)= 6x2-11x-10
  A=6, K=-11, n= -10
  A-N+K= 6+11-10= 5
• 13 years agoHelpfull: Yes(3) No(1)
• 6x^2-11x-10
  
  A=6 K=-11 N=-10
  
  A-N+K = 5
• 13 years agoHelpfull: Yes(3) No(2)
• Simplify LHS..
  (3x+2)(2x−5)=6x2−11x−10=ax2−n+k
  (
  3
  �
  +
  2
  )
  (
  2
  �
  −
  5
  )
  =
  6
  �
  2
  −
  11
  �
  −
  10
  =
  �
  �
  2
  −
  �
  +
  �
  ...
  equate the coefficients...
  a=6 ; k = -11 and n = -10..
  so a−n+k=6−(−10)+(−11)=6+10−11=5
• 1 year agoHelpfull: Yes(0) No(0)