Question

If polynomial P(x) with leading coefficient 1, of degree 4, is such that P(1)=1, P(2)=2, P(3)=3, P(4)=4. Then P(5)=? Note P(5) is not equal to 5

Answer 1

P(x) = a x^4 + b x^3 + c  x^2 + d x + e

P(1) = a + b + c + d + e = 1
P(2) = 16 a + 8 b + 4 c + 2 d + e = 2
P(3) = 81 a + 27 b + 9 c + 3 d + e = 3
P(4) = 256 a + 64 b + 16 c + 4 d + e = 4

P(5) = 625 a + 12 5 b + 25 c + 5 d + e = ??

Working on the four given equations, we get the following

15 a + 7 b + 3 c + d = 1
175 a + 37 b + 7 c+ d = 1

85 a + 21 b + 5 c + d = 1
40 a + 13 b + 4 c + d = 1

from the above four equations we get:
80 a + 15 b + 2 c = 0
45 a + 8 b + c = 0

Solving them we get    45 a - 80 a + c = 0

     c = 35 a,      b = -10 a,        d = - 50 a,         e = 1 + 24 a

P(5) = 625 a - 1250 a + 875 a - 250 a + 1 + 24 a
         = 1 + 24 a  = e = P(0)

P(5) = P(0)

Answer 2

Answer:

29

Step-by-step explanation:

you can write the polynomial as

p(x) = x + (x-1)(x-2)(x-3)(x-4)

Notice the term (x-1)(x-2)(x-3)(x-4) becomes 0 when x=1,2,3,4, so you get answer as 1,2,3,4 respectively.

p(5) = 5 + 4*3*2*1

      = 5+24

      =29