F4 Math一問~~~~~

將 x^2+4x+5 除 x - 1
得商 x + 5, 餘數 10

故 x^2+4x+5 = (x-1)(x+5) + 10
= (x-1)((x-1) + 6) + 10
g (x-1) = (x-1)((x-1) + 6) + 10

設 y = x-1
g (y) = (y)(y + 6) + 10

故 g(a) = (a)(a + 6) + 10
= a^2 + 6a + 10

g(x - 1) = x2 + 4x + 5

g(x) = (x + 1)2 + 4(x + 1) + 5

g(x) = x2 + 6x + 10

g(a) = a2 + 6a + 10