Question

若(x-3)(3x+2)=3x^2+mx+n,则m=__ n=___ 若(x-8)(x^2-x+m)中不含一次项，则m的值为

Answer 1

(x-3)(3x+2)
=3x²+2x-9x-6
=3x²-7x-6
=3x^2+mx+n
∴盯塌液m=-7, n=-6
(x-8)(x^2-x+m)
=x³-x²凯物+mx-8x²+8x-8m
=x³-9x²+(m+8)x-8m
∵(x-8)(x^2-x+m)中不含衫余一次项
∴m+8=0
m=-8

Answer 2

(x-3)(3x+2)=3x^2+mx+n

3x^2+2x-9x-6=3x^2+mx+n
3x^2-7x-6=3x^2+mx+n

m=-7
n=-6

(x-8)(x^2-x+m)

=x^3-x^2+mx-8x^2+8x-8m
=x^3-9x^2+(m+8)x-8m

式中含一次项祥橘氏伍含为：m+8
不含一次谨散项即：m+8=0

m=-8

Answer 3

若(x-3)(3x+2)=3x^2+mx+n,则饥樱带m=_-7_ n=-6___
3x²-7x-6=3x²+mx+n
若(x-8)(x^2-x+m)中不烂芦含一次项，则m的颂兆值为-8
x³-x²+mx-8x²+8x-8m;m+8=0;m=-8;