兩個(gè)多項(xiàng)式的值相等,則有等式：
x³-2x²-4x-1=x(x²-mx+n)+2(mx-1/2)
x³-2x²-4x-1=x³-mx²+nx+2mx-1
x³-2x²-4x-1=x³-mx²+(2m+n)x-1
兩個(gè)多項(xiàng)式恒等,對應(yīng)的每一項(xiàng)都相等,可列方程組：
-m=-2
2m+n=-4
解方程組,得 m=2,n=-8
m-3次根號n=2-3次根號-8
=2-(-2)
=2+2
=4