If P(x)=Ax^4+Bx^3+Cx^2+Dx+E has roots at x=1,2,3,4and P(0)=48,what is P(5)?
a)48
b)24
c)0
d)50

• answer a) 48
  
  if 1 is a root then x-1 is a factor of p(x)
  similarly x-2 is a factor,x-3 ,x-4 are factors
  
  but p(x) is 4th degree polynomial therefore it can be in the form
  
  p(x) = k(x-1)(x-2)(x-3)(x-4) , where k is a constant
  
  but given p(0) =48
  
  therefore 24 k =48
  
  k=2
  
  p(x) =2(x-1)(x-2)(x-3)(x-4)
  
  p(5)= 2*4*3*2*1 = 48
• 11 years agoHelpfull: Yes(56) No(12)
• Both the solutions previously mentioned are correct and enough elegant.Now, consider another solution which tackles the given problem from a bit different point of view and also produces the result with a different approach compared to the previous methods..
  Given that,
  P(x)=Ax^4+Bx^3+Cx^2+Dx+E
  =A(x^4 -x)+B(x^3 -x)+C(x^2 -x)+Dx+Ax+Bx+Cx+E
  =x(x-1)[A(x^2 +x+1) +B(x+1)+C] + x(A+B+C+D) +E....[EQN(1)]
  Given P(0)=0; ==> E=48 and also since x=1 is a root of P(x) so,it implies that A+B+C+D+E=0 ,==> A+B+C+D=-E,==> A+B+C+D=-48;
  So, from [EQN(1)]...we get,
  P(x)=x(x-1)[A(x^2 +x+1) +B(x+1)+C] + x(-48) +48; [since A+B+C+D=-48 & E=48]
  =>P(x)=x(x-1)[A(x^2 +x+1) +B(x+1)+C] - 48(x-1)........[EQN(2)]
  Now, since x=2,3,4 are roots of [EQN(2)],
  SO, For x=2,3,4
  0=x(x-1)[A(x^2 +x+1) +B(x+1)+C] - 48(x-1)
  =>48 = x[A(x^2 +x+1) +B(x+1)+C] ;;{(x-1) cancels out since (x-1)is not=0}
  =>[A(x^2 +x+1) +B(x+1)+C] = (48/x);............[EQN(3)].
  Now,in [EQN(3)],Putting x=2,3,4 we get respectively
  7A + 3B + C =24....[EQN(4)]
  13A + 4B + C = 16....[EQN(5)]
  21A + 5B + C = 12....[EQN(6)]
  Now,from [EQN(2)] we have,
  P(x)=x(x-1)[A(x^2 +x+1) +B(x+1)+C] - 48(x-1)
  =>P(5)=20(31A + 6B + C) -192..........[EQN(7)]
  Now,
  at this point,observe a fact that {1*[EQN(4)] -3*[EQN(5)] +3*[EQN(6)]} gives,
  1*(7A + 3B + C =24)- 3*(13A + 4B + C) +3*(21A + 5B + C ) = 24-48+36
  ==>(31A + 6B + C)=12;
  So, from [EQN(7)], we get,
  P(5)=20(31A + 6B + C) -192
  =>P(5)= 20*12 -192
  =>P(5)=(240-192)=48
  =>P(5)=48..
  
  So, P(5)=48....OPTION (A)48 , is therefore correct..
  
  Kindly share ur feedback with us....
  Thanks to all..:-):-)
• 11 years agoHelpfull: Yes(23) No(7)
• sum of roots is a+b+c+d=-B/A
  ------ ----- ab+bc....(6 terms)=C/A
  ------------- abc+abd+acd+bcd=-D/A .........eq(1)
  product is abcd=E/A
  here a,b,c,d are roots
  abcd=24=E/A
  (E=48 by substuting 0 in eq)
  @@@@@@ A=2 @@@@
  a+b+c+d=-B/A==>10*2=-B
  @@@@@ B=-20 @@@@
  in eq(1)sub values 50=-D/A==>@@@@@ D= - 100 @@@@
  and A+B+C+D+E=0 (as 1 is a root)
  substuting we get C=70
  now 2x^4-20x^3+70x2+-100x+48(see if we substute 5 here all get cancelled excpt 48 )
  ans is 48 option is (a)
  
• 11 years agoHelpfull: Yes(4) No(11)