IF p(x)=a*(x^4)+b*(x^3)+c*(x^2+d*x+e,
x has roots at x=1,2,3,4 and p(0)=48, what is the value of p(5)?

• since the equation has 4 roots we can write it has
  p(x)=q(x-1)(x-2)(x-3)(x-4) where q is constant.
  for x=0; p(0)=q(0-1)(0-2)(0-3)(0-4)=48 as given in the question after solving u will get q=2,so p(5)=2*(5-1)(5-2)(5-3)(5-4)=2*24=48 should be the answer....
• 11 years agoHelpfull: Yes(38) No(0)
• ANS:48
  the equation has roots of 1,2,3,4
  so form the equation with roots
  (x-1)(x-2)(x-3)(x-4)+k=P(x)
  now P(0)=48.put in above eqn.
  -1*-2*-3*-4+k=48.
  k=24. P(5)=5-1*5-2*5-3*5-4+24
  P(5)=4*3*2*1+24
  P(5)=48.
  
  
• 11 years agoHelpfull: Yes(9) No(1)
• (x-1)(x-2)(x-3)(x-4)+36=p(x)
  so,answer is 60
• 11 years agoHelpfull: Yes(0) No(10)