If p(x)=ax^4+bx^3+cx^2+dx+e has roots at x=1,2,3,4 and p(0)=48, what is p(5)
A.48
B.24
C.0
D.50

• p(x)=ax^4+bx^3+cx^2+dx+e
  
  it can be written as
  
  p(x)=a*(x-1)*(x-2)*(x-3)*(x-4) ----(1)
  
  put x=0 in(1), we get
  p(0)=a*(-1)(-2)(-3)(-4)
  => 48=a*24
  => a=2
  
  so, p(x)=2*(x-1)*(x-2)*(x-3)*(x-4)
  => p(5)=2*(5-1)*(5-2)*(5-3)*(5-4)=2*4*3*2*1=48
  
  p(5) = 48
• 10 years agoHelpfull: Yes(40) No(6)
• p(x)=(x-1)*(x-2)*(x-3)*(x-4)
  p(5)=(5-1)*(5-2)*(5-3)*(5-4)
  =24
• 10 years agoHelpfull: Yes(6) No(9)
• p(x)=(x-1)*(x-2)*(x-3)*(x-4)
  p(5)=(5-1)*(5-2)*(5-3)*(5-4)
  =24
• 10 years agoHelpfull: Yes(3) No(2)
• ans=48 as coefficient 'a' value is needed and is 2
• 10 years agoHelpfull: Yes(1) No(0)
• if the root is 1,2,3,4 and p(0)=48
  then equation becomes p(x)=(x-1)(x-2)(x-3)(x-4)
  when we put p(0)=24 it means some term will be multiplied in equation ,
  so, the final equation is= 2*(x-1)(x-2)(x-3)(x-4)
  therefore,p(5)=48 Ans
• 10 years agoHelpfull: Yes(1) No(0)
• how we write p(x)=a*(x-1)*(x-2)*(x-3)*(x-4) ----(1)
  plz explain i do'n understand plz
• 10 years agoHelpfull: Yes(0) No(0)
• how we write p(x)=a*(x-1)*(x-2)*(x-3)*(x-4) ----(1)
  plz explain...!!!!!!!!
• 10 years agoHelpfull: Yes(0) No(0)
• it is 48 as we find the equation by
  (x-1)(x-2)(x-3)(x-4)=0
  we get eqation as x^4-10x^3+35x^2-50x+24
  and we need to multiply throughout by 2 as p(0)= 48
• 10 years agoHelpfull: Yes(0) No(0)