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?
(pls show work)
+130071 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 $f$?
Guess that we have a third degree polynomial of the form ax^3 + bx^2 + cx + d
If f(0) = 47, then d = 47
And we have this system
a + b + c + 47 = 32 ⇒ a + b + c = -15 (1)
8a + 4b + 2c + 47 = -13 ⇒ 8a + 4b + 2c = - 60 (2)
27a + 9b + 3c + 47 = 16 ⇒ 27a + 9b + 3c = -31 (3)
Multiply (1) by -2 add to (2) we get
6a + 2b = -30 ⇒ 3a + b = -15 ⇒ -9a - 3b = 45 (4)
Multiply (1) by -3 add to (3) we get
24a + 6b = 14 ⇒ 12a + 3b = 7 (5)
Add (4) and (5) we get
3a = 52 ⇒ a = 52/3
3(52/3) + b = - 15
52 + b = -15
b = -67
52/3 - 67 + c = -15
52/3 + c = 52
52 + 3c = 156
3c = 104
c = 104/3
So
a + b + c + d =
52/3 - 67 + 104/3 + 47
156/3 - 20 =
52 - 20 = 32
