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P(1) = a + b + c + d + 1 = p(2) = 16 + 8a + 4b + 2 c + d

7 a + 3 b + c + 15 = 0 -- eq 1

c = - 7 a - 3b - 15 -- eq 4

p(3) = 81 + 27 a + 9 b + 3c +d = 0

27 a + 9 b + 3 c + d = - 81 -- eq 2

4 * eq 1 - equ 2 gives a + 3b + c - 21 = d -- eq 3 substitute value of c.

d = a + 3b -7a - 3b - 15 - 21 = - 6 a - 36 -- eq 5

p(4)+p(0) = (256 + 64 a + 16 b + 4 c + d) + ( d)

= 64 a + 16 b + 4 c + 2 d + 256

substituting values of c and d,

= 24 a + 4b + 124

7 a + 3 b + c + 15 = 0 -- eq 1

c = - 7 a - 3b - 15 -- eq 4

p(3) = 81 + 27 a + 9 b + 3c +d = 0

27 a + 9 b + 3 c + d = - 81 -- eq 2

4 * eq 1 - equ 2 gives a + 3b + c - 21 = d -- eq 3 substitute value of c.

d = a + 3b -7a - 3b - 15 - 21 = - 6 a - 36 -- eq 5

p(4)+p(0) = (256 + 64 a + 16 b + 4 c + d) + ( d)

= 64 a + 16 b + 4 c + 2 d + 256

substituting values of c and d,

= 24 a + 4b + 124

substitute the values of x=2 and 3

equate them. you will get P(4 )+ p(0) = 64a+16b+4c+2d+256

since, you have got the values of c and d, substitute them. you will get the final equation as 24a+4b+124