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).

• as x=1,2,3,4 are roots so (x-1)(x-2)(x-3)(x-4)will satisfy p(x) i.e
  K(x-1)(x-2)(x-3)(x-4)=ax4+bx3+cx2+dx+e=P(x)so put x=0 we get k*1*2*3*4=48 i.e k=2
  and now x=5 in same eqn p(5)=K(x-1)(x-2)(x-3)(x-4) i.e
  P(5)=2*4*3*2*1 => 48
• 11 years agoHelpfull: Yes(56) No(0)
• CEE 4!=24
  CEF 4!=24
  CEPE 3!=6
  CEPFERT(ANS)
• 11 years agoHelpfull: Yes(1) No(1)