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# Help with Polynomials

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357
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+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
+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

Apr 12, 2019
#2
+219
0

Rudram592  Apr 12, 2019
#3
+12433
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The sum of the coefficients is 32.

Apr 12, 2019
edited by Omi67  Apr 12, 2019
edited by Omi67  Apr 12, 2019
#4
+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

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

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

Apr 12, 2019
#6
+219
+1

32 was the correct answer! Thanks CPhill and Omi

Apr 12, 2019
#7
+111455
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OK...I'll relent.....!!!!!

CPhill  Apr 12, 2019