一多项式[(x²+5x+6)-(ax²-bx+c)]÷3x=2x+1 余式为0,求a+b+c的值
问题描述:
一多项式[(x²+5x+6)-(ax²-bx+c)]÷3x=2x+1 余式为0,求a+b+c的值
答
看系数:[(x²+5x+6)-(ax²-bx+c)]÷3x=[(1-a)x²+(5+b)x+(6-c)]÷3x=[(1-a)/3]x+(5+b)/3+(6-c)/3x∵[(x²+5x+6)-(ax²-bx+c)]÷3x=2x+1∴[(1-a)/3]=2,1-a=6,a=-5(5+b)/3=1,5+b=3,b=-2(6-c)/3x=...