若(x2+mx+n)(x2-2x-3)的乘积中不含x3、x2项,则m=______,n=______.
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/08 22:03:19
若(x2+mx+n)(x2-2x-3)的乘积中不含x3、x2项,则m=______,n=______.
![若(x2+mx+n)(x2-2x-3)的乘积中不含x3、x2项,则m=______,n=______.](/uploads/image/z/16522290-18-0.jpg?t=%E8%8B%A5%EF%BC%88x2%2Bmx%2Bn%EF%BC%89%EF%BC%88x2-2x-3%EF%BC%89%E7%9A%84%E4%B9%98%E7%A7%AF%E4%B8%AD%E4%B8%8D%E5%90%ABx3%E3%80%81x2%E9%A1%B9%EF%BC%8C%E5%88%99m%3D______%EF%BC%8Cn%3D______%EF%BC%8E)
∵(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
−2+m=0
−3−2m+n=0,
解得
m=2
n=7.
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
−2+m=0
−3−2m+n=0,
解得
m=2
n=7.
故答案为:2,7.
已知(x2+mx+n)(x2-3x+2)的展开式中不含x2项和x项,则m=______,n=______.
已知(x3+mx+n)(x2-5x+3)的乘积中不含x3和x2项,求m、n的值.
若x2+mx-15=(x+3)(x+n),则m=______,n=______.
如果关于x的多项式(x2+mx+8)(x2-3x+n)展开后不含x2和x3项,则(-m)3n=--------
若(x2+mx-8)(x2-3x+n)的展开式中不含x2和x3项,求m和n的值.
若(x+5)(2x-n)=2x2+mx-15,则m的值为______.
若(x+3)(x+n)=x2+mx-15,则m的值为______.
若n(n≠0)是关于x的方程x2+mx+2n=0的根,则2m+n的值为______.
多项式x2+mx+5因式分解得(x+5)(x+n),则m=______,n=______.
要使(x2+mx+8)(x2-3x+n)的展开式中不含x3项和x2项,求m,n的值.
要使多项式mx3-2x2+3x-4x3+5x2-nx不含三次项及一次项,则m=______,n=______.
若n(n≠0)是关于x的方程x2-mx-4n=0的根,则m-n的值为______.