Question

若多项式x^2+mx+n分解因式得(x-3)(x+4)求m^2n+mn^2的值

Answer 1

x²+mx+n=(x-3)(x+4)

x²+mx+n=x²+x-12

(m-1)x+(n+12)=0

要对任意x,等式恒成立,只有

m-1=0

n+12=0

解得m=1 n=-12

m²n+mn²

=mn(m+n)

=1×(-12)×[1+(-12)]

=132