Latest TCS Aptitude Question SOLUTION: 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

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..

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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)