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

Answer 2

x^2+mx+n=(x-3)(x+4)=x²+x-12
∴m=1, n=-12
m^2n+mn^2
=mn(m+n)
=1×(-12)×(1-12)
=132