+753 Suppose f is a polynomial such that f(0) = 47, f(1) = 32, f(2) = -13, and f(3)=16. What is the sum of the coefficients of f?
+129852
There could be several possibilities....but let's assume that it's of the form Ax ^3 + Bx^2 + Cx + D
If f(0) = 47....then D = 47
And we have these 3 equations
A (1)^3 + B(1)^2 + C(1) + 47 = 32
A(2)^3 + B (2)^2 + C (2) + 47 = -13
A(3)^3 + B(3)^2 + C(3) + 47 = 16
Simplifying these, we have
A + B + C = -15
8A + 4B + 2C = - 60
27A + 9B + 3C = -31
Multiplying the first equation by -2 and adding it to the second we have that
6A + 2B = -30 (1)
And multiplying the first equation by -3 and adding it to the third gives us
24A + 6B = 14 (2)
Multiplying (1) by -3 and adding it to (2) produces
6A = 104
A = 104/6 = 52/3
And to find B ... 6(52/3) + 2B = -30 → 104 + 2B = -30 → 2B = -134 → B = -67
And to find C
52/3 + -67 + C = -15
-149/3 + C = -15
C = -15 + 149/3 = 104/3
So....the sum of A + B + C + D =
52/3 + -67 + 104/3 + 47 =
156/3 - 20 =
52 - 20 =
32
