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?

MIRB16
Aug 5, 2017

#1**+3 **

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

CPhill
Aug 5, 2017