SOLUTION: Let f be a cubic polynomial such that f(0) = 5, f(7) = -5, f(-3) = 8, and f(3)=13. What is the sum of the coefficients of f?

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Let f be a cubic polynomial such that f(0) = 5, f(7) = -5, f(-3) = 8, and f(3)=13. What is the sum of the coefficients of f?      Log On


   

Question 1209337: Let f be a cubic polynomial such that f(0) = 5, f(7) = -5, f(-3) = 8, and f(3)=13. What is the sum of the coefficients of f?
Answer by mananth(16946) About Me  (Show Source):
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The polynomial f(x) can be
f(x)=ax3+bx2+cx+d.
given : f(0)=5, f(7) = -5, f(-3) = 8, and f(3)=13.
1. f(0) = 5 Therefore d=5
2. f(7) = 343a+49b+7c+5= -5
343a+49b+7c =-10...............1

3. f(-3)= -27a+9b-3c+5= 8
-27a+9b-3c =3..................2

4. f(3)= 27a^3+9x+3x+5= 13
27a+9b+3c =8 ..................3
Add equation 2 &3 we get
18b=11
b=11/18
343a+49b+7c =-10...............1
-27a+9b-3c =3..................2
Multiply (1) by 3 and (2) by 7
1029a+147b+21c=-30
-189a+63b -21c= 21
Add the equations
we get
840a +210b=51
we have b= 11/18 plug b above
840a+210(11/18)= 51
a= -(103/630)
plug a & b find c
c=(242/105)
Knowing a,b,c add them up
a= -(103/630) , b=11/18 , c = (242/105)