题目内容
若(x2+mx+n)(x2-2x-3)的乘积中不含x3、x2项,则m=
2
2
,n=7
7
.分析:把两个多项式相乘,合并同类项后使结果的x3与x2项的系数为0,求解即可.
解答:解:∵(x2+mx+n)(x2-2x-3)
=x4-2x3-3x2+mx3-2mx2-3mx+nx2-2nx-3n,
=x4+(-2+m)x3+(-3-2m+n)x2+(-3m-2n)x-3n,
∴要使(x2+mx+n)(x2-2x-3)的乘积中不含x3与x2项,
则有
,
解得
.
故答案为:2,7.
=x4-2x3-3x2+mx3-2mx2-3mx+nx2-2nx-3n,
=x4+(-2+m)x3+(-3-2m+n)x2+(-3m-2n)x-3n,
∴要使(x2+mx+n)(x2-2x-3)的乘积中不含x3与x2项,
则有
|
解得
|
故答案为:2,7.
点评:本题主要考查了多项式乘多项式的运算,由不含x3与x2项,让这两项的系数等于0,列方程组是解题的关键.
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