If P (x) = ax4 + bx3 + cx2 + dx + e has roots at x= 1,2,3 and 4, and P(0) = 48, what is P(5) ?

• ans-48
  (x-1) * (x-2) *(x-3) *(x-4) the four root
  so p(0)= -1*-2*-3*-4 = 24 so we need 24
  that p(x) =(x-1) * (x-2) *(x-3) *(x-4) +24
  p(5)= (5-1) * (5-2) *(5-3) *(5-4)+24 = 48
  
• 8 years agoHelpfull: Yes(9) No(4)
• -b/a=1+2+3+4=10
  c/a=(1*2)+(1*3)(1*4)+(2*3)+(2*4)+(3*4)=35
  -d/a=(1*2*3)+(1*3*4)+(2*3*4)+(3*4*1)=54
  e/a=1*2*3*4=24
  p(0)=e=48
  a=e/24=2
  b=-20
  c=70
  d=-108
  p(5)=2(5^4)-20(5^3)+70(5^2)-108(5)+48=8
  
• 8 years agoHelpfull: Yes(3) No(4)
• p(5)=48 by binomial
• 8 years agoHelpfull: Yes(0) No(1)
• ans is 48...
• 8 years agoHelpfull: Yes(0) No(2)