Question

若(x²+px+3)(x²-2x+q)的乘积中不含x²,x³项,求p,q的值

要过程x²就是x的平方x³就是x的立方

Answer 1

(x²+px+3)(x²-2x+q)的乘积中含x²项为:(q-2p+3)x²
(x²+px+3)(x²-2x+q)的乘积中含x³项为:(p-2)x³
(x²+px+3)(x²-2x+q)的乘积中不含x²,x³项
即q-2p+3=0
p-2=0
p=2
q-2*2+3=0
q=1
所以p=2,q=1

Answer 2

(x²+px+3)(x²-2x+q)
=x^4-2x³+qx²+px³-2px²+pqx+3x²-6x+3q
=x^4+(-2+p)x³+(q-2p+3)x²+(pq-6)x+3q
乘积中不含x²,x³项，则
-2+p=0
q-2p+3=0
解得p=2，q=1

Answer 3

解：(x2+px+3)(x2-2x+q)
=x^4-2x^3+qx^2+px^3-2px^2+pqx+3x^2-6x+3q
=x^4+(p-2)x^3+(q-2p+3)x^2+(pq-6)x+3q
由于x是任意数，且乘积中不含x2,x3项，所以x2,x3项的系数为零。
所以p=2 q=1