多項式f(x),g(x)以x^+2x-1除之其餘式分別為x+1及x-1,則(1)xf(x)+(x^-1)g(x)除以x^+2x-1之餘式為?(2)f(x)*g(x)除以x^+2x-1之餘式為?

設多項式 f(x)、g(x) 以 (x² + 2x - 1) 除之,其商式分別為 P(x) 和 Q(x).
f(x) = (x² + 2x - 1)•P(x) + (x + 1)
g(x) = (x² + 2x - 1)•Q(x) + (x - 1)
(1) 因為 xf(x)+(x²-1)g(x) = x[(x² + 2x - 1)•P(x)]+x(x + 1) + (x²-1)[(x² + 2x - 1)•Q(x)]+(x²-1)(x - 1)
所以要求 xf(x)+(x²-1)g(x)除以x²+2x-1的餘式
即求[x(x+1) + (x²-1)(x-1)]除以 x²+2x-1 之後的餘式 = 5x-1
(2) 同理,因為 f(x)*g(x) = [(x² + 2x - 1)•P(x) + (x + 1)]*[(x² + 2x - 1)•Q(x) + (x - 1)]
所以,要求f(x)*g(x)除以x²+2x-1的餘式 , 就是求 (x + 1)*(x - 1) 除 x²+2x-1的餘式 = -2x

(1)
設多項式 f(x)、g(x) 以 (x² + 2x - 1) 除之，其商式分別為 P(x) 及 Q(x)。

f(x) = (x² + 2x - 1)•P(x) + (x + 1)
g(x) = (x² + 2x - 1)•Q(x) + (x - 1)

x•f(x) + (x² - 1)•g(x)
= x[(x² + 2x - 1)•P(x) + (x + 1)] + (x² - 1)[(x² + 2x - 1)•Q(x) + (x - 1)]
= (x² + 2x - 1)[x•P(x) + (x² - 1)•Q(x)] + x(x + 1) + (x² - 1)(x - 1)
= (x² + 2x - 1)[x•P(x) + (x² - 1)•Q(x)] + x² + x + x³ - x² - x + 1
= (x² + 2x - 1)[x•P(x) + (x² - 1)•Q(x)] + x³ + 1
= (x² + 2x - 1)[x•P(x) + (x² - 1)•Q(x)] + x(x² + 2x - 1) - x(2x - 1) + 1
= (x² + 2x - 1)[x•P(x) + (x² - 1)•Q(x) + x] - 2x² + x + 1
= (x² + 2x - 1)[x•P(x) + (x² - 1)•Q(x) + x] - 2(x² + 2x - 1) + 2(2x - 1) + x + 1
= (x² + 2x - 1)[x•P(x) + (x² - 1)•Q(x) + x - 2] + 4x - 2 + x + 1
= (x² + 2x - 1)[x•P(x) + (x² - 1)•Q(x) + x - 2] + 5x - 1

[x•f(x) + (x² - 1)•g(x)] 除以 (x² + 2x - 1) 的餘式 = 5x - 1


(2)
f(x)•g(x)
= [(x² + 2x - 1)•P(x) + (x + 1)] • [(x² + 2x - 1)•Q(x) + (x - 1)]
= (x² + 2x - 1)²•P(x)•Q(x) + (x² + 2x - 1)[P(x) + Q(x)] + (x + 1)(x - 1)
= (x² + 2x - 1)[(x² + 2x - 1)•P(x)•Q(x) + P(x) + Q(x)] + x² - 1
= (x² + 2x - 1)[(x² + 2x - 1)•P(x)•Q(x) + P(x) + Q(x)] + (x² + 2x - 1) - (2x - 1) - 1
= (x² + 2x - 1)[(x² + 2x - 1)•P(x)•Q(x) + P(x) + Q(x) + 1] - 2x + 1 - 1
= (x² + 2x - 1)[(x² + 2x - 1)•P(x)•Q(x) + P(x) + Q(x) + 1] - 2x

f(x)•g(x) 除以 (x² + 2x - 1) 的餘式 = -2x

Sol
(1)
f(x)=p(x)(x^2+2x－1)+x+1
g(x)=q(x)(x^2+2x－1)+x－1
h(x)=xf(x)+(x^2－1)g(x)
=xp(x)(x^2+2x－1)+x^2+x+(x^2－1)q(x)(x^2+2x-1)+(x^2－1)(x－1)
=x^2+x+(x^2－1)(x－1)
=[xp(x)+(x^2－1)q(x)](x^2+2x－1)+x^2+x+(x^3－x^2－x+1)
=[xp(x)(x^2－1)q(x)+x－2](x^2+2x－1)+5x－1
餘式=5x－1
(2)
k(x)=f*x)*g(x)
=[p(x)(x^2+2x－1)+x+1]*[q(x)(x^2+2x－1)+x－1]
=m(x)(x^2+2x－1)+x^2－1
=[m(x)+1]*(x^2+2x－1)+2x
餘式=2x