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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 \(f \)?

 Feb 19, 2019
 #1
avatar+99301 
+2

 

 

\(f(x)=ax^4+bx^3+cx^2+dx+e\\ e=47\\ f(x)=ax^4+bx^3+cx^2+dx+47\\ f(1)=a+b+c+d+47=32\\ a+b+c+d=32-47\\ a+b+c+d=-15\\\)

 

Ths sum of the coefficients is -15

 

This would be true even if it the leading power was not 4.

 Feb 19, 2019
 #3
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0

I'm sorry but that is incorrect. Thanks for trying anyone else please help???

Guest Feb 19, 2019
 #5
avatar+99301 
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It is fine to say it is incorrect but please give the reason that you think so. 

Melody  Feb 20, 2019
 #2
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+1

Why just  a + b + c + d ? , isn't e a coefficient as well ?

 Feb 19, 2019
 #4
avatar+99301 
0

No e is not a coefficient. e is the constant.

 

Coefficients are the numbers in front of the pronumerals

 

(if the pronumeral is 1 then it is invisable but it is still there - that is not relevant here though.)

Melody  Feb 19, 2019

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