If P x ax 4 bx 3 cx 2 dx e has roots at x 1 2 3 4 and

The problem is not wrong!!
However it is lengthy and in parts tricky!!
There are two ways to arrive at the answer:

1st Method:
The solution can be derived as:
P(0)=48, so putting x=0 in the above equation we get,p(0)=e=48.
The P(x)=ax^4+bx^3+cx^2+dx+48.
Now put, x=1,2,3,4.
Since,these are roots so, P(1)=P(2)=P(3)=P(4)=0
We get 4 equations...
a+b+c+d+48=0 ..................1
16*a+8*b+4*c+2*d+48=0..........2
81*a+27*b+9*c+3*d+48=0.........3
256*a+64*b+16*c+4*d+48=0.......4

Solving we get,a=2,b=-20,c=70,d=-100.
Substitute them,
P(x)=2*x^4-20*x^3+70*x^2-100*x+48

Put x=5, we get P(5)=2*5^4-20*5^3+70*5^2-100*5+48 = 48.

So, Answer is 48.

Method 2: (Use Chinese Remainder Theorem!!)
(Read that in the Wikipedia article and try to solve it!! ) As, a hint I will just say, it has something to do with the remainder.

If you know the theorem, this question can be solved in 5 seconds...
11 years agoHelpfull: Yes(9) No(2)
f(x) = k(x-p)(x-q)(x-r)(x-s) where p q r s are roots of equation
Now calculate k;;
it is coming 2
f(5)= 2(5-1)(5-2)(5-3)(5-4)
f(5)=48
11 years agoHelpfull: Yes(8) No(0)
Given problem is wrong.P(0) should be 24.
then P(5)=24.
11 years agoHelpfull: Yes(1) No(3)
(x-1)(x-2)(x-3)(x-4)=p(x)

p(5)=(4*3*2*1)=24
11 years agoHelpfull: Yes(0) No(4)
48
11 years agoHelpfull: Yes(0) No(0)

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).