已知(x²+px+q)(x²-3x+q)的乘积中不含x²和x³项,求p,q的值
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![已知(x²+px+q)(x²-3x+q)的乘积中不含x²和x³项,求p,q的值](/uploads/image/z/7295283-27-3.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%88x%26%23178%3B%2Bpx%2Bq%29%28x%26%23178%3B-3x%2Bq%29%E7%9A%84%E4%B9%98%E7%A7%AF%E4%B8%AD%E4%B8%8D%E5%90%ABx%26%23178%3B%E5%92%8Cx%26%23179%3B%E9%A1%B9%2C%E6%B1%82p%2Cq%E7%9A%84%E5%80%BC)
已知(x²+px+q)(x²-3x+q)的乘积中不含x²和x³项,求p,q的值
已知(x²+px+q)(x²-3x+q)的乘积中不含x²和x³项,求p,q的值
已知(x²+px+q)(x²-3x+q)的乘积中不含x²和x³项,求p,q的值
(x²+px+q)(x²-3x+q)
=x的4次方+(p-3)x³+(2q-3p)x²+(pq-3q)x+pq
∵乘积中不含x²和x³项
∴p-3=0
2q-3p=0
∴p=3
q=9/2
解
(x²+px+q)(x²-3x+q)
=x²(x²-3x+q)+px(x²-3x+q)+q(x²-3x+q)
不含x³即x³前面的系数为0
-3x³+px³=(p-3)x³
则p-3=0
不含x²即x²前面的系数为0
qx²-3px²+qx²=(2q-3p)x²
则2q-3p=0
∴p=3,q=9/4