试题

题目：
已知x2+x+3=0，求x5+3x4+2x3+2x2-10x的值．
答案
解：∵x2+x+3=0，
∴x2+x=-3，
x5+3x4+2x3+2x2-10x
=x3（x2+x）+2x4+2x3+2x2-10x
=-3x3+2x4+2x3+2x2-10x
=2x4-x3+2x2-10x
=2x2（x2+x）-3x3+2x2-10x
=-3x3-4x2-10x
=-3x（x2+x）-x2-10x
=9x-x2-10x
=-x2-x
=-（x2+x）
=3．
解：∵x2+x+3=0，
∴x2+x=-3，
x5+3x4+2x3+2x2-10x
=x3（x2+x）+2x4+2x3+2x2-10x
=-3x3+2x4+2x3+2x2-10x
=2x4-x3+2x2-10x
=2x2（x2+x）-3x3+2x2-10x
=-3x3-4x2-10x
=-3x（x2+x）-x2-10x
=9x-x2-10x
=-x2-x
=-（x2+x）
=3．
考点梳理
因式分解的应用．
要先提公因式，再根据x2+x+3=0得出x2+x=-3分别一步步代入，即可求出答案；
此题考查了因式分解的应用；解题的关键先提取公因式后进行分解，再把x2+x=-3代入即可，解题时要细心．
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