If p(x)=a x^4+b x^3+c x^2+d x+e.....has roots at x=1,2,3,4......p(0)=48...then p(5)=?????

• P(x)=Q(x-1)(x-2)(x-3)(x-4)
  as x=1,2,3,4 are roots of equation...
  P(0)=48=Q(-1)(-2)(-3)(-4)
  by solving Q=2
  so, P(x)=2(x-1)(x-2)(x-3)(x-4)
  put x=5
  P(5)=2(5-1)(5-2)(5-3)(5-4)
  P(5)=48 is the ans.
• 12 years agoHelpfull: Yes(26) No(0)
• take p(x) = k(x-1)(x-2)(x-3)(x-4) , where k is a constant
  but given p(0) =48
  therefore 24 k =48
  k=2
  p(x) =2(x-1)(x-2)(x-3)(x-4)
  p(5)= 2*4*3*2*1 = 48
  ans:48
• 12 years agoHelpfull: Yes(5) No(0)
• yes its 48..
• 12 years agoHelpfull: Yes(2) No(0)
• how this formula is applied????
  p(x)=Q(x-1)(x-2)(x-3)(x-4)
• 12 years agoHelpfull: Yes(2) No(0)
• Please explain clearly not getting.....
• 12 years agoHelpfull: Yes(0) No(0)