利用整式和因式分解的知识说明（x²-4）（x²-10x+21)+100的值一定是非负数
问题描述：
利用整式和因式分解的知识说明（x²-4）（x²-10x+21)+100的值一定是非负数
答
(x^2-4)(x^2-10x+21)+100
=(x^2-4)*[(x-5)^2-4]+100
=x^2*(x-5)^2-4*(x-5)^2-4*(x^2-4)+16+100 化简得
=x^2*(x-5)^2-8*x*(x-5)+16 这步是关键
=[x*(x-5)-4]^2 >=0
（x²-4）（x²-10x+21)+100的值一定是非负数=x^2*(x-5)^2-8*x*(x-5)+16这步怎么算的(x^2-4)(x^2-10x+21)+100 =(x^2-4)*[(x-5)^2-4]+100 =x^2*(x-5)^2-4*(x-5)^2-4*(x^2-4)+100 =x^2*(x-5)^2-4x^2+40x-100-4x^2+16+100=x^2*(x-5)^2-8x^2+40x+16=x^2*(x-5)^2-8*x*(x-5)+16=[x*(x-5)-4]^2 >=0