1.證明：若%7Cf(x)%2Cd(x)%7Cg(x))
,且)
為)
與)
的一個(gè)組合,則是與的一個(gè)最大公因式
證：
%7Cf(x)%2Cd(x)%7Cg(x))
)
是與的一個(gè)公因式
若)
是與的一個(gè)公因式
則可整除與的任一組合
%7Cd(x))
是與的一個(gè)最大公因式
2.證明：h(x)%2Cg(x)h(x))%3D(f(x)%2Cg(x))h(x))
(首項(xiàng)系數(shù)為1)
證：
%2Cg(x))%7Cf(x)%2Cg(x))
%2Cg(x))h(x)%7Cf(x)h(x)%2Cg(x)h(x))
%2Cg(x))h(x))
是h(x)%2Cg(x)h(x))
的一個(gè)公因式
設(shè)%2Cg(x))%3Du(x)f(x)%2Bv(x)g(x))
,則
%2Cg(x))h(x)%3Du(x)f(x)h(x)%2Bv(x)g(x)h(x))
是的一個(gè)最大公因式
)
首項(xiàng)系數(shù)為1?
首項(xiàng)系數(shù)為1?
h(x)%2Cg(x)h(x))%3D(f(x)%2Cg(x))h(x))
3.證明：若%2Cg(x))
不全為零,則%5Cover%20(f(x)%2Cg(x))%7D%2C%7Bg(x)%5Cover%20(f(x)%2Cg(x))%7D)%3D1)
證：
%2Cv(x))
使得?
f(x)%3Dv(x)g(x)%3D(f(x)%2Cg(x)))
%7Bf(x)%5Cover%20(f(x)%2Cg(x))%7D%2Bv(x)%7Bg(x)%5Cover%20(f(x)%2Cg(x))%7D%3D1)
%5Cover%20(f(x)%2Cg(x))%7D%2C%7Bg(x)%5Cover%20(f(x)%2Cg(x))%7D)%3D1)
4.證明：若不全為零,且,則%2Cv(x))%3D1)
證：
f(x)%3Dv(x)g(x)%3D(f(x)%2Cg(x)))
%5Cover%20(f(x)%2Cg(x))%7Du(x)%2B%7Bg(x)%5Cover%20(f(x)%2Cg(x))%7Dv(x)%3D1)
%2Cv(x))%3D1)
5.證明：若%2Cg(x))%3D1%2C(f(x)%2Ch(x))%3D1)
,則%2Cg(x)h(x))%3D1)
證：
%2Cg(x))%3D1%2C(f(x)%2Ch(x))%3D1)
%2Cv_1(x))
和%2Cv_2(x))
使得
f(x)%3Dv_1(x)g(x)%3D1)
f(x)%3Dv_2(x)h(x)%3D1)
兩式相乘可得?
u_2(x)f(x)f(x)%2Bu_1(x)v_2(x)f(x)h(x))
v_1(x)f(x)g(x)%2Bv_2(x)v_1(x)g(x)h(x))
u_2(x)f(x)%2Bu_1(x)v_2(x)h(x))
![+u_2(x)v_1(x)g(x)]f(x)+v_2(x)v_1(x)g(x)h(x)=1](https://math.jianshu.com/math?formula=%2Bu_2(x)v_1(x)g(x)%5Df(x)%2Bv_2(x)v_1(x)g(x)h(x)%3D1)
%2Cg(x)h(x))%3D1)
6.設(shè)![f_1(x),\cdots,f_m(x),g_1(x),\cdots,g_n(x)\in P[x]](https://math.jianshu.com/math?formula=f_1(x)%2C%5Ccdots%2Cf_m(x)%2Cg_1(x)%2C%5Ccdots%2Cg_n(x)%5Cin%20P%5Bx%5D)
,且%2Cg_j(x))%3D1(i%3D1%2C2%2C%5Ccdots%2Cm%3Bj%3D1%2C2%2C%5Ccdots%2Cn))
證明：f_2(x)%5Ccdots%20f_m(x)%2Cg_1(x)g_2(x)%5Ccdots%20g_n(x))%3D1)
證：
若不然
即%3D(f_1(x)f_2(x)%5Ccdots%20f_m(x)%2Cg_1(x)g_2(x)%5Ccdots%20g_n(x))%5Cneq%201)
則有一個(gè)不可約因式,設(shè)為)
%7Cf_1(x)f_2(x)%5Ccdots%20f_m(x))
%7Cf_s(x)(1%5Cle%20s%5Cle%20m))
同理可得
%7Cg_t(x)(1%5Cle%20t%5Cle%20n))
%7C(f_s(x)%2Cg_t(x)))
,矛盾
f_2(x)%5Ccdots%20f_m(x)%2Cg_1(x)g_2(x)%5Ccdots%20g_n(x))%3D1)
7.證明：若%2Cg(x))%3D1)
,則g(x)%2Cf(x)%2Bg(x))%3D1)
證：
%2Cg(x))%3D1)
%2Cf(x)%2Bg(x))%3D1%2C(g(x)%2Cf(x)%2Bg(x))%3D1)
g(x)%2Cf(x)%2Bg(x))%3D1)
8.求下列多項(xiàng)式的公共根：
%3Dx%5E3%2B2x%5E2%2B2x%2B1%2Cg(x)%3Dx%5E4%2Bx%5E3%2B2x%5E2%2Bx%2B1)
解：
%5Cqquad%5Cqquad%5Cqquad%5Cqquad%20g(x))
%3D%7B1%5Cover%202%7Dx%2B%7B1%5Cover%202%7D%26x%5E3%2B2x%5E2%2B2x%2B1%26x%5E4%2Bx%5E3%2B2x%5E2%2Bx%2B1%26q_1(x)%3Dx-1%5C%5C%20%26x%5E3%2Bx%5E2%2Bx%26x%5E4%2Bx%5E3%2B2x%5E2%2Bx%2B1%26%20%5C%5C%20%5Chline%20%26x%5E2%2Bx%2B1%26-x%5E3%2B1%5C%5C%20%26x%5E2%2Bx%2B1%26-x%5E3-2x%5E2-2x-1%26%20%5C%5C%20%5Chline%20%26r_2(x)%3D0%26r_1(x)%3D2x%5E2%2B2x%2B2%5C%5C%20%5Cend%7Barray%7D)
%2Cg(x))%3Dx%5E2%2Bx%2B1)
)
與的公共根為
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