如果a,b,m都是整数,且(x+a)(x+b)=x*2+mx+12,则m的值可能是哪些?
1个回答

(x + a)(x + b) = x^2 + ( a + b)x + ab

因为:(x + a)( x + b) = x^2 + mx + 12

所以,x^2 + (a + b)x + ab = x^2 + mx + 12

因此 ab = 12,m = a + b

因为a,b,m都是整数

所以,由ab = 12得,a,b的可能取值有:

a =1,b=12; a=-1,b= -12; a=3,b=4; a = -3,b = -4; a = 2,b = 6;a = -2,b = -6

或 a =12,b=1; a = -12,b=-1; a=4,b=3; a = -4,b = -3; a = 6,b = 2; a =-6,b = -2

由上述讨论可知:

m = a+b 的可能取值有

当a =1,b=12或a =12,b=1时,m = 13

当a=-1,b= -12或a = -12,b=-1时,m = -13

当a=3,b=4或a=4,b=3时,m = 7

当 a = -3,b = -4或a = -4,b = -3时,m = -7

当 a = 2,b = 6 或a = 6,b = 2时,m = 8

当a = -2,b = -6 或a = -6,b = -2时,m = -8

(注:因为ab = 12,所以a、b不可能一正一负)