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Suppose f is a polynomial such that f(0) = 47f(1) = 32f(2) = -13, and f(3)=16. What is the sum of the coefficients of f?

Guest Dec 4, 2014

Best Answer 

 #1
avatar+81004 
+10

Suppose f is a polynomial such that f(0) = 47f(1) = 32f(2) = -13, and f(3)=16. What is the sum of the coefficients of f?

This polynomial has -  at least - two real zeroes.....(there may be more).....let's suppose that it's a cubic

So we have

f(x) =  ax^3 + bx^2 + cx + d    ....Note, "d" has to be equal to 47

So we have this system

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(2) + 47 = 16

We can simplify these to:

a + b + c  = -15

8a + 4b + 2c = -60

27a + 9b + 3c = -31

I used WolframAlpha solve this one ....  (I'm lazy.....)

a = 52/3   b = -67 c = 104/3  d = 47.....    I'll let you sum these.......!!!!

 

Here's the graph.........https://www.desmos.com/calculator/gnbrhqhxdr

(Yep...that works !!!!)

Note that - if this is a cubic - there had to be another real zero...also, this solution may not be unique....other functions might be possible......

 

CPhill  Dec 4, 2014
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2+0 Answers

 #1
avatar+81004 
+10
Best Answer

Suppose f is a polynomial such that f(0) = 47f(1) = 32f(2) = -13, and f(3)=16. What is the sum of the coefficients of f?

This polynomial has -  at least - two real zeroes.....(there may be more).....let's suppose that it's a cubic

So we have

f(x) =  ax^3 + bx^2 + cx + d    ....Note, "d" has to be equal to 47

So we have this system

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(2) + 47 = 16

We can simplify these to:

a + b + c  = -15

8a + 4b + 2c = -60

27a + 9b + 3c = -31

I used WolframAlpha solve this one ....  (I'm lazy.....)

a = 52/3   b = -67 c = 104/3  d = 47.....    I'll let you sum these.......!!!!

 

Here's the graph.........https://www.desmos.com/calculator/gnbrhqhxdr

(Yep...that works !!!!)

Note that - if this is a cubic - there had to be another real zero...also, this solution may not be unique....other functions might be possible......

 

CPhill  Dec 4, 2014
 #2
avatar+91451 
0

That looks good - I wish I had more time.  I'd love to play with more of these polynomial problems.

There is never enough time when you are having fun        

Melody  Dec 4, 2014

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