Question

wully

想請問這一題

Y=1
已知多項式 f(x)除以x-1的商式為g(x),餘式為3; g(x)除以x-2的商式為h(x),餘式為2;h(x)除以x-3的餘式為1,
則 xf(x)除以(x-1)(x-2)(x-3)的餘式為何?
f(x) = (x-1) [ (x-2)(x) +²] +3
f(x) = (x-1) g(x) + 3
g(x) =/(x-2) h(x) +2
= (x-1) (X-2)h(X)+2x+1)
h(x) = (x-3) Q(x) +1
= (x-1)(x-2)[X-30(X) + 1] +28+}
= (x-1)(x-²)(x-³) Q(x) + x²-x +3
X(X+)(x-2)(X-~3) Q(x)
(x-1)(x-2)(x-3)

Answer 1

✨ ベストアンサー ✨

靜

如圖