Question 1209337
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)