题目
已知多项式3x^3+mx^2+nx+42能被x^2-5x+6整除,求m+n的值.
提问时间：2020-09-04
答案
因为3x^3+mx^2+nx+42能被x^2-5x+6整除,而x^2-5x+6=(x-2)(x-3),所以x=2或x=3代入多项式后结果应该都等于0,所以有24+4m+2n+42=0,81+9m+3n+42=0,化简得,2m+n=-33,(1)3m+n=-41,(2)(2)-(1)得m=-8,所以n=-17,所以m+n=-8+(-...
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