急 ! F4數about Polynomica #113

113a) f(2x + 3) = x^2 + 2x + 3, so f(2y + 3) = y^2 + 2y + 3Let y = (x - 3)/2, therefore,f(x) = [(x - 3)/2]^2 + 2(x - 3)/2 + 3= (x^2 - 6x + 9)/4 + x= (x^2 - 2x + 9)/4
113b) f(x^2) = (x^4 - 2x^2 + 9)/4, let g(x) = f(x^2), therefore,when f(x^2) is divided by (2x - 3), the remainder is :g(3/2)= [(3/2)^4 - 2*(3/2)^2 + 9]/4= (81/16 - 9/2 + 9)/4= 153/64
114ai) As f(1) = 1, therefore,(1 + 1)(1 - m)(1 - n) - 5 = 1==> 2(m - 1)(n - 1) = 6==> (m - 1)(n - 1) = 3
114aii) As m > n > 0 and 3*1 = 3, thereforem - 1 = 3 and n - 1 = 1==> m = 4 and n = 2
114b) f(x) = x^2 - x - 7==> (x + 1)(x - 4)(x - 2) - 5 = x^2 - x - 7==> (x + 1)(x - 4)(x - 2) - (x^2 - x - 2) = 0==> (x + 1)(x - 4)(x - 2) - (x + 1)(x - 2) = 0==> (x + 1)(x - 2)(x - 4 - 1) = 0==> x = -1, 2, 5

2014-03-05 17:38:25 補充：
不是計，是 let y = (x - 3)/2。
(即假設 y = (x - 3)/2 )