Let f(x) be a quadratic polynomial such that f(-4) = -22, f(-1)=2, 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. (For example, the sum of the coefficients of the terms in -7x^3 + 4x^2 + 10x - 5 that have even degree is (4) + (-5) = -1.)
+128408 I worked through this
f(x) = (-3/2)x^2 + (1/2)x + 4
g (x) = (f(x))^16
Here is the expansion of this
(43046721 x^32)/65536 - (14348907 x^31)/4096 - (157837977 x^30)/8192 + (518154975 x^29)/4096 + (4068180855 x^28)/16384 - (8827766451 x^27)/4096 - (14523632991 x^26)/8192 + (94257891351 x^25)/4096 + (220712269635 x^24)/32768 - (706381131105 x^23)/4096 - (23495103567 x^22)/8192 + (3940745861637 x^21)/4096 - (1959573354921 x^20)/16384 - (16925275164105 x^19)/4096 + (6150417128055 x^18)/8192 + (57033142395909 x^17)/4096 - (166621058538623 x^16)/65536 - (19011047465303 x^15)/512 + (683379680895 x^14)/128 + (626862043115 x^13)/8 - (24192263641 x^12)/4 - 129736489272 x^11 - 1031335136 x^10 + 165371348480 x^9 + 17223697920 x^8 - 156919300096 x^7 - 32238338048 x^6 + 104507375616 x^5 + 32107397120 x^4 - 43620761600 x^3 - 17716740096 x^2 + 8589934592 x + 4294967296
Adding the coefficients on the even powered terms would be very tedious....as you can imagine !!!
+128408 I WILL be glad to show you how I got f(x)....but....adding those co-efficients.....forget it !!!!
+128408 We have this system of equations
a(-4)^2 + b(-4) + c = -22
a(-1)^2 + b(-1) + c = 2
a(2)^2 + b(2) + c = -1 simplify these
16a -4b + c = -22
a - b + c = 2
4a + 2b + c = -1
Subtract the second and third equations from the first and we get that
15a - 3b = -24 ⇒ 5a -b = -8 ⇒ -10a + 2b = 16
12a -6b = -21 ⇒ 4a - 2b = -7 ⇒ 4a - 2b = -7
Add the last two equations and we have that
-6a = 9 divide both sides by -6
a =9/-6 = -3/2
And
5(-3/2) - b = -8
-15/2 - b = -16/2
-b = -16/2 + 15/2
-b = -1/2
b = 1/2
And
a - b + c = 2
(-3/2) - (1/2) + c = 2
-2 + c = 2
c = 4
So
f (x) = ax^2 + bx + c = (-3/2)x^2 + (1/2)x + 4
