+218 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?
+128406 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
+128406 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
