Question

A monic polynomial f(x) of degree four satisfies f(1)=10, f(2)=20, and f(3)=30. Determine
f(12)+f(−8)−19000.

Answer 1

In a monic polynomial, the leading coefficient is 1
So let
f(x) = x^4 + ax^3 + bx^2 + cx + d
f(1) = 1 +a+b+c+d = 10
a+b+c+d = 9 , #1

f(2) = 16 + 8a + 4b + 2c + d = 20
8a + 4b + 2c + d = 4 , #2

f(3) = 81 + 27a + 9b + 3c + d = 30
27a + 9b + 3c + d = -51 , #3

#2-#1:
7a + 3b + c = -5, #4
#3-#1:
26a + 8b + 2c = -60
13a + 4b + c = -30 , #5

#5-#4:
6a + b = -25

we cannot find explicit values of a, b, c, and d, since we have only 3 equations with 4 unknowns.

f(12) + f(-8) - 19000
= 12^4 + 12^3 a + 12^2 b + 12c + d + (-8)^4 + (-8)^3 a + (-8)^2 b - 8c + d - 19000
= 20736 + 1728a + 144b + 12c + d + 4096 - 512a + 64b - 8c + d - 19000
= 5832 + 1216a + 208b + 4c + 2d
= 5232 + 2(608a + 104b + 2c + d) ----- check my arithmetic.

I was hoping for one of the above relations to show up , but no such luck.
we could try to back-substitute,
e.g.
b = -25-6a
but that back into #4 and get c in terms of a
etc.
Give it a try.

Answer 2

Yes, it works!

If we solve a,b,c in terms of d, we get,
a=-(d+36)/6,
b=d+11,
c=-(11d-24)/6
from which we can define
f(x)=x^4 -(d+36)x^3/6 +(d+11)x^2 -(11d-24)x/6 +d
and based on this,
f(1)=10
f(2)=20
f(3)=30
f(12)+f(-8)-19000
=-608(d+36)/3 +208(d+11)+2d+2(24-11*d)/3 +5832
=840

Answer 3

840 is the correct answer.