Question

已知多项式（x^2+mx+n）（x^2-3x+4）展开后不含x^3和x^2项,(1)试求m,n的值

Answer 1

(x^3+mx+n)(x^2-3x+4)

=x^5-3x^4+4x^3+mx^3-3mx^2+4mx+nx^2-3nx+4n

=x^5-3x^4+(4+m)x^3+(n-3m)x^2+(4m-3n)x+4n

根据展开后不含x^3和x^2项,说明4+m=0,n-3m=0,m=-4 n=-12

(m+n)(m^2-mn+n^2)

(m+n)[(m-n)^2+mn]=(-16)*[(-4)^2+48]= - 1024