Latest TCS Aptitude Question SOLUTION: If P(x) = ax^4 + bx^3 + cx^2 + dx + e has roots at x = 1; 2; 3 and 4, and P(0) = 48/ what is P(5)? a) 48 b) 24 c) 0 d) 50 e) none of these

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(#M40010783) TCS Logical Reasoning question Keep an EYE

If P(x) = ax^4 + bx^3 + cx^2 + dx + e has roots at x = 1; 2; 3 and 4,
and P(0) = 48/ what is P(5)?

a) 48
b) 24
c) 0
d) 50
e) none of these

Asked In TCS Srivastav Reddy (11 years ago)

Solved Aman Read Solution (4)

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Read Solution (4) :

Aman  (11 years ago)

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

Is this solution Helpfull? Yes (117) | No (9)  

shekhar pankaj  (11 years ago)

u rember Quad expresion?
sum of roots=-b/a
prod of rooot=c/a

similarly
x1+x2+x3+x4=-b/a(taken 1 at tme4c1=4 case)
x1*x2+x1*x3+.....4c2...=c/a(taken 2at a time 4c2 =6 case).....1)
x1*x2*x3+......4c3=-d/a........2)
x1*x2*x3*x4=e/a........3)
so
e/a=4.3.2.1..ie, e=24a
for p(0)=e as all term cancel aout ..so e=48...so a=2
from 1 2 and 3 eqn calculate b c d
b=-20
c=70
d=-100
so no wput in p(5)
all terms again cancel leaving e ..so 24 answer


i request u all if u all are here to crack TCS please solve urself because the solution u copying from other may be wrong as eg in lower u was just a step closer to loose ur marks.....

Is this solution Helpfull? Yes (5) | No (15)  

Deepak kumar singh  (8 years ago)

let A=1,B=2,C=3 nd D=4 are d roots of given eqn.
A+B+C+D=-b/a
AB+BC+CD+DA=c/a
ABC+BCD+CDA=-d/a
ABCD=e/a
since e=48
calculate d value of a,b,c,d
a=2, b=-20, c=48, d=-84
therefore
p(5)=-322
Answer=d) None of these

Is this solution Helpfull? Yes (1) | No (1)  

Prasun Bhattacharjee  (8 years ago)

As it has been given that 1,2,3,4 is the root of the solution.Therefore,we can conclude (x-1),(x-2),(x-3),(x-4) are the factors of the equation. so we could write (x-1)*(x-2)*(x-3)*(x-4)=ax^4+bx^3+cx^2+dx+e
=> x^4-10x^3+35x^2-50x+24=ax^4+bx^3+cx^2+dx+e
equating the coefficient we get a=1,b=-10,c=35,d=24.
therefore P(5)=(5^4)*1-10*(5^3)+35*(5^2)+24
after solving we get p(5)=24 so B should be the answer......
But here is a catch point P(0)=48. if you plug in the value of x=0 in the equation ax^4+bx^3+cx^2+dx+e
only e will be left ,other term will cancel out.e=constant. so,e=24 but in the Question e=48. if you solve ax^4+bx^3+cx^2+dx+e this putting x=5 we will get e as left. therefore e=48. The ans is option A

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