YLL討論網

1.設f(x)與g(x)都是多項式，若x²+2x+3為3f(x)+4g(x)與f(x)+2g(x)的最高公因式，求f(x)與g(x)的最高公因式.
2.已知a是正整數,b.c是整數，若x-a是f(x)=x³+6x²+bx+6與g(x)=x³+cx²-x-2的最高公因式.
3.設a.b是常數，且兩多項式f(x)=2x³-x²+ax+1與g(x)=x³-4x²+bx-3的最高公因式為二次式，求a.b的值.

1.
令h(x) = x²+2x+3
利用Euclidean algorithm

3f(x)+4g(x) = 3(f(x)+2g(x))+(-2g(x))
因3f(x)+4g(x)和f(x)+2g(x)的最高公因式為h(x)
則f(x)+2g(x)和(-2g(x))且最高公因式為h(x)

f(x)+2g(x) = -1(-2g(x))+f(x)
因f(x)+2g(x)和(-2g(x))的最高公因式為h(x)
則(-2g(x))和f(x)的最高公因式為h(x)

則f(x)和g(x)的最高公因式為h(x) = x²+2x+3


2.
設f(x) = (x-a)(x²+ix+j)
0次項係數: -aj = 6
                => j = -6/a [因a不等於0] //若j是整數, a的值就是-6的正整因數 ----(1)
1次項係數: -ai + j = b //因b是整數, 若得i是整數就能確定j是整數 -----(2)
2次項係數: -a + i = 6
                => i是整數
                => j是整數 [因(2)]
                => a = 1, 2, 3, 6 [因(1)] -----(3)

同樣道理求g(x)可能的a值範圍
設g(x) = (x-a)(x²+i'x+j')
0次項係數: -aj' = 2
                => j' = -2/a
1次項係數: -ai' + j' = -1
2次項係數: -a + i' = c
                => i'是整數
                => j'是整數
                => a = 1, 2 -----(4)
因(3), (4)
=> a = 1, 2

假設 a = 1
f(1) = 1 + 6 + b + 6 = 0 => b = -13
g(1) = 1 + c - 1 - 2 = 0 => c = 2
f(x) = (x-1)(x²+7x-6)
g(x) = (x-1)(x²+3x+2) = (x-1)(x+2)(x+1)
最高公因式為(x-1), 合理.

假設 a = 2
f(2) = 8 + 24 + 2b + 6 = 0 => b = -19
g(2) = 8 + 4c - 2 - 2 = 0 => c = -1
f(x) = (x-2)(x²+8x-3)
g(x) = (x-2)(x²+x+1)
最高公因式為(x-2), 合理.

兩組解, a=1, b=-13, c=2; a=2, b=-19, c=-1

3.
令h(x)為f(x)和g(x)的最高公因式且deg(h(x)) = 2

去首項, f(x)-2g(x) = 7x²+(a-2b)x+7
因h(x)整除f(x)-2g(x)且deg(h(x)) = deg(f(x)-2g(x)) = 2
則c*h(x) = 7x²+(a-2b)x+7 for some scalar c ≠ 0 -----(1)

去末項, 3f(x)+g(x) = 7x³-7x²+(3a+b)x = x(7x²-7x+3a+b)
因h(x)整除3f(x)+g(x)且h(x)不能整除x
則h(x)整除(7x²-7x+3a+b)
又因deg(h(x)) = deg(7x²-7x+3a+b) = 2
則c'*h(x) = 7x²-7x+3a+b for some scalar c' ≠ 0 -----(2)

(1)/(2), 可得c/c'= 7/7 = (a-2b)/(-7) = 7/(3a+b)
=> a-2b = -7, 3a+b = 7
=> a=1, b=4



有誤請糾正, 謝謝!