求:(x²+px+q)(x²-3x+2)的乘积中不含x²和x项,则p,q的值是多少?
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![求:(x²+px+q)(x²-3x+2)的乘积中不含x²和x项,则p,q的值是多少?](/uploads/image/z/10755055-55-5.jpg?t=%E6%B1%82%EF%BC%9A%EF%BC%88x%26%23178%3B%2Bpx%2Bq%EF%BC%89%EF%BC%88x%26%23178%3B-3x%2B2%EF%BC%89%E7%9A%84%E4%B9%98%E7%A7%AF%E4%B8%AD%E4%B8%8D%E5%90%ABx%26%23178%3B%E5%92%8Cx%E9%A1%B9%2C%E5%88%99p%2Cq%E7%9A%84%E5%80%BC%E6%98%AF%E5%A4%9A%E5%B0%91%3F)
求:(x²+px+q)(x²-3x+2)的乘积中不含x²和x项,则p,q的值是多少?
求:(x²+px+q)(x²-3x+2)的乘积中不含x²和x项,则p,q的值是多少?
求:(x²+px+q)(x²-3x+2)的乘积中不含x²和x项,则p,q的值是多少?
(x²+px+q)(x²-3x+2)
=x^4+(p-3)x³+(2+q-3p)x²+(2p-3q)x+2q
∵ 乘积中不含x²和x项
∴2+q-3p=0
2p-3q=0
∴p=6/7
q=4/7
2-3p+q=0
2p-3q=0
p=6/7
q=4/7
解:(x²+px+q)(x²-3x+2)=x^4-3x^3+2x²+px^3-3px²+2px+qx²-3qx+2q
合并同类项
x^4-(3-p)x^3+(2-3p+q)x²+(2p-3q)x+2q
所以2-3p+q=0
2p-3q=0
所以p=4/3
q=2