p(X)=ax^4+bx^3+cx^2+dx+e has a root at x=1,2,3and4 and p(0)=48 what is p(5)?(a)48(b)24(c)0(d)50

• Ans: 48
  if roots are 1,2,3,4 than
  p(x)=(x-1)(x-2)(x-3)(x-4)+C
  x=0
  p(0)=>48=24+c
  c=24
  p(5)=4*3*2*1+24=48
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• ans:48
  p(x)=k(x+1)(x+2)(x+3)(x+4)[as 1,2,3,4 are factors)
  now, p(0)=48
  so k=2
  so, p(5)=2(-4)(-3)(-2)(-1)
  =48
  
• 12 years agoHelpfull: Yes(1) No(2)
• 1)sum of roots= -(b/a)= -48,
  2)product of each two roots= (c/a)=35,
  3)product of each three roots = -(d/a) = -50,
  4)product of roots = (e/a) = 48,
  5)on solving a=2, b= -20, c=70, d=-100, e= 48,
  6)on substituting we get ans p(5)= 48.
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