急 ! F4數about Polynomica #111

111(a). f(x + 1/x) = (2x)^3 + 3
Put x = 1, we get,
f(1 + 1/1) = (2*1)^3 + 3 = 11
therefore, f(2) = 11

111(b). By using the remainder theorem, the remainder is :
f(2)
= 11

112(a). f(1) = 2, g(2) = 1, let h(x) = f(x) - 2g(x+1) + 1, therefore,
the remainder when h(x) is divided by (x - 1) is :
h(1)
= f(1) - 2g(1+1) + 1
= f(1) - 2g(2) + 1
= 2 - 2x1 + 1
= 1

112(b). [f(g(2))]^2
= [f(1)]^2
= 2^2
= 4

2014-03-05 17:41:06 補充：
Put x = 1，可以做到 f(2)。
如果 put x = 2，x + 1/x = 2 + 1/2 = 2.5，只做到 f(2.5)，做不到 f(2)。

2014-03-05 17:42:53 補充：
若 x + 1/x = 2，可得
x^2 + 1 = 2x
==> x^2 - 2x + 1 = 0
==> (x - 1)^2 = 0
==> x = 1