Let f(x) be a quadratic polynomial such that \(f(-4) = -22,f(-1)=2 \text { and } f(2)=-1. \) Let \(g(x) = f(x)^{16}.\) Find the sum of the coefficients of the terms in g(x) that have even degree.
P.S. I'd like to know how to get the equation for f(x), I'm still stuck here... (it would be really helpful to know the steps).
+33614 You don't need to find the explicit form of f(x)16 here. See the following:
Thanks so much! do you mind telling me how you got the values for a, b and c? I was really stumped there.
+33614 Write the three equations as:
-22 = 16a - 4b + c. (1)
2 = a - b + c. (2)
-1 = 4a + 2b + c. (3)
(3) - (2): 3a + 3b = -3. (4)
(1) - (2): 15a - 3b = -24 (5)
(4) + (5): 18a = -27. Divide by 9: 2a = -3. a = -3/2
Substitute for a in (4) or (5) to find b, then substitute both into any of (1), (2) or (3) to find c.
+33614 Let g(x) = f(x)16. Write g(x) as g(x) = g(x)eventerms + g(x)oddterms.
g(1) = g(1)eventerms + g(1)oddterms ...(1)
g(-1) = g(-1)eventerms + g(-1)oddterms
g(-1) = g(1)eventerms - g(1)oddterms ...(2)
(1) + (2): g(1) + g(-1) = 2*g(1)eventerms
so: g(1)eventerms = (g(1) + g(-1))/2
or g(1)eventerms = (f(1)16 + f(-1)16)/2
