证明6x^5+11x^4+5x^3+5x^2-x-6能被x^2+1整除?提示：判断6x^5+11x^4+5x^3+5x^2-x-6是否含因式（x+i)(x-i)
问题描述：
证明6x^5+11x^4+5x^3+5x^2-x-6能被x^2+1整除?
提示：判断6x^5+11x^4+5x^3+5x^2-x-6是否含因式（x+i)(x-i)
答
也就是是否存在x^2=-1
加入存在6x^5+11x^4+5x^3+5x^2-x-6=6x*(x^2)^2+11(x^2)^2+5x(x^2)+5x^2-x-6
=6x-11-5x-5-x-6=0
故在村x^2=-1使f(x)=6x^5+11x^4+5x^3+5x^2-x-6=0
故6x^5+11x^4+5x^3+5x^2-x-6能被x^2+1整除
答
6x^5+11x^4+5x^3+5x^2-x-6
=6x^5+6x^3+11x^4-x^3+5x^2-x-6
=6x^3(x^2+1)+11x^4+11x^2-x^3-6x^2-x-6
=6x^3(x^2+1)+11x^2(x^2+1)-(x^3+x)-(6x^2+6)
=6x^3(x^2+1)+11x^2(x^2+1)-x(x^2+1)-6(x^2+1)
所以,得证.