+0

+7
114
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+359

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
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+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
edited by CPhill  Aug 5, 2017

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