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.
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.
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
840 is the correct answer.
To determine the value of f(12) + f(-8) - 19000, we first need to find the monic polynomial f(x) of degree 4 that satisfies the given conditions. Since f(x) is monic, it means that the coefficient of the highest power of x is 1.
Let's express the polynomial in the form f(x) = ax^4 + bx^3 + cx^2 + dx + e, where a, b, c, d, and e are constants that we need to determine.
Using the given information, we can set up a system of equations based on the polynomial evaluation:
f(1) = a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e = 10 -- (Equation 1)
f(2) = a(2)^4 + b(2)^3 + c(2)^2 + d(2) + e = 20 -- (Equation 2)
f(3) = a(3)^4 + b(3)^3 + c(3)^2 + d(3) + e = 30 -- (Equation 3)
Now, let's solve this system of equations to find the values of a, b, c, d, and e.
Plugging in the values from Equation 1, we get:
a + b + c + d + e = 10 -- (Equation 4)
Plugging in the values from Equation 2, we get:
16a + 8b + 4c + 2d + e = 20 -- (Equation 5)
Plugging in the values from Equation 3, we get:
81a + 27b + 9c + 3d + e = 30 -- (Equation 6)
Now, let's solve this system of equations to obtain the values of a, b, c, d, and e.
Subtracting Equation 4 from Equation 5, we get:
15a + 7b + 3c + d = 10 -- (Equation 7)
Subtracting Equation 4 from Equation 6, we get:
80a + 26b + 8c + 2d = 20 -- (Equation 8)
Next, subtracting Equation 7 from Equation 8, we get:
65a + 19b + 5c = 10 -- (Equation 9)
From here, we have three equations (Equations 7, 9, and 10) with three variables (a, b, and c). We can solve this system of equations to determine the values of a, b, and c.
After obtaining the values of a, b, and c, we can substitute them back into Equation 4 to find the value of d:
a + b + c + d + e = 10
Finally, once we have determined the values of a, b, c, d, and e, we can evaluate f(12) + f(-8) - 19000 by substituting x = 12 and x = -8 into the polynomial f(x), then performing the respective calculations.
In conclusion, to determine f(12) + f(-8) - 19000, we need to solve a system of equations to find the monic polynomial f(x) of degree four that satisfies the given conditions. Then, substitute the respective values of x into the polynomial and perform the calculations.