The sum of the coefficients of f is 1 .
We know that f is a cubic polynomial of the form f(x)=ax^3+bx^2 +cx+d, so we can write down the following system of equations from the given information:
a(0^3 )+b(0^2 )+c(0)+d=5
a(2^3 )+b(2^2 )+c(2)+d=8
a(3^3 )+b(3^2 )+c(3)+d=13
a(5^3 )+b(5^2 )+c(5)+d=−5
Simplifying the equations, we get:
d=5
8a+4b+2c+d=8
27a+9b+3c+d=13
125a+25b+5c+d=−5
We can now solve for the coefficients of f. From Equation 1, we know that d=5. Substituting this into Equation 2, we get:
8a + 4b + 2c + 5 = 8
8a + 4b + 2c = 3
Similarly, substituting d=5 into Equation 3 and Equation 4, we get:
27a + 9b + 3c + 5 = 13
27a + 9b + 3c = 8
125a + 25b + 5c + 5 = -5
125a + 25b + 5c = -10
We now have three equations with three unknowns: 8a+4b+2c=3, 27a+9b+3c=8, and 125a+25b+5c=−10. Solving this system of equations using elimination, we find:
a=− 10/3
b= 23/3
c=− 11/3
Therefore, the polynomial f(x)=− 10/3 x^3+ 23/3 x^2 − 11/3 x+5 satisfies the given conditions.
The sum of the coefficients of f is − 10/3 + 23/3 − 11/3+5= 1 .