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avatar+219 

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?

 Apr 12, 2019
 #1
avatar+111455 
+1

Since f(0)  = 47.....this must be the constant term

 

And since f(1) =  32

 

The sum of the coefficients  + 47  = 32        subtract 47 from both sides

 

Sum of the coefficients  =  32 - 47  =  -15

 

 

cool cool cool

 Apr 12, 2019
 #2
avatar+219 
0

-15 was the incorrect answer

Rudram592  Apr 12, 2019
 #3
avatar+12433 
+1

The sum of the coefficients is 32.

laugh

 Apr 12, 2019
edited by Omi67  Apr 12, 2019
edited by Omi67  Apr 12, 2019
 #4
avatar+111455 
+1

Mmmmm....

 

Let  f(x)  =  ax^m  + bx^n + cx^p  + .......+ constant term

 

f(0)  = a(0)^m + b(0)^n + c(0)^p  +  ......+ constant term  =  47

The terms with coefficients  all = 0....so the constant term must be 47

 

f(1)  = a(1)^m + b(1)^n + c(1)^p + .......+ 47  =  32

The coefficient terms will just sum to  the sum of their coefficients  ...i.e.,  a + b + c + ......

 

So....

 

sum of coefficients + constant term  =  32

 

sum of coefficients + 47 = 32

 

sum of coefficients  =  32 - 47  =  - 15

 

 

cool cool cool

 Apr 12, 2019
edited by CPhill  Apr 12, 2019
 #5
avatar+111455 
0

I agree with Omi IF the polynomial is a cubic....but.....we aren't given any info about its degree....

 

cool cool cool

 Apr 12, 2019
 #6
avatar+219 
+1

32 was the correct answer! Thanks CPhill and Omi

 Apr 12, 2019
 #7
avatar+111455 
0

OK...I'll relent.....!!!!!

 

 

cool coolcool

CPhill  Apr 12, 2019

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