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 p(5)?

• p(x)=k(x-1)(x-2)(x-3)(x-4)
  put x=0
  p(0)=k(-1)(-2)(-3)(-4)
  48=k(24)
  k=2
  now p(5)=2(4)(3)(2)(1)
  p(5)=48
• 11 years agoHelpfull: Yes(26) No(0)
• when roots r given then the expression can b written as:
  k(x-1)(x-2)(x-3)(x-4). now p(0)=48
  so, k(0-1)(0-2)(0-3)(0-4)=48 so, 24k =48 so, k=2
  so p(x)=2(x-1)(x-2)(x-3)(x-4). now p(5)= 2*4*3*2*1=48.
  ans=48
• 11 years agoHelpfull: Yes(3) No(0)
• 48
  .......
  
• 11 years agoHelpfull: Yes(0) No(0)