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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?

 

(pls show work)

Guest Feb 18, 2018
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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 $f$?

 

Guess that we have a third degree polynomial of the form  ax^3 + bx^2 + cx + d

If  f(0)  =  47, then d =  47

 

And we have this system

 

a + b + c + 47  = 32  ⇒   a +  b + c   =  -15      (1)

8a + 4b + 2c + 47  = -13  ⇒  8a + 4b + 2c  = - 60   (2)

27a + 9b + 3c + 47  = 16  ⇒  27a + 9b + 3c  =  -31    (3)

 

Multiply (1)  by -2  add to (2)  we get

 

6a + 2b =  -30    ⇒  3a + b  = -15 ⇒  -9a - 3b = 45    (4)

 

Multiply (1)  by -3   add to (3)  we get

 

24a + 6b  = 14  ⇒  12a + 3b = 7   (5)

 

Add  (4)  and (5)  we get

 

3a =  52    ⇒  a    =  52/3

 

3(52/3) + b  = - 15

52 + b  =  -15

b =  -67

 

52/3   -  67 + c  = -15

52/3 + c  =  52

52 + 3c  = 156

3c  =  104

c  =  104/3

 

So

 

a + b + c + d  =

 

52/3  - 67  + 104/3  + 47

 

156/3  - 20  =

 

52  -  20   =  32

 

 

 

cool cool cool

CPhill  Feb 18, 2018
edited by CPhill  Feb 18, 2018

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