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

• Roots are 1,2,3,4
  So we can write this
  equation As p(x)=
  a(x-1)(x-2)(x-3)(x-4).....eq1
  Put x=0here we p(0)=48=24a
  TheN a=2
  Now put a in eq1
  And get the answer...
  P(5)=a.4.3.2.1=48
• 9 years agoHelpfull: Yes(8) No(0)
• 48 IS THE ANSWER
  
• 9 years agoHelpfull: Yes(2) No(1)
• can u explain how the ans is 48?
• 9 years agoHelpfull: Yes(2) No(0)
• 48 is the answer
• 9 years agoHelpfull: Yes(0) No(0)
• A. 48,
  since 1,2,3,4 are the roots of p(x) so it can be equation p(x) i.e for p(x)=0 we get the value of x.so p(5)= 48 because only constant value "e" is evaluated and all other are zero.
• 9 years agoHelpfull: Yes(0) No(0)