Question

已知多项式x^2-y^2+3x-7y+a能分成两个一次因式的乘积,求a的值

Answer 1

假设

(x+y+m)*(x-y+n)=x^2+xn-y^2+yn+xm-ym+mn

=x^2-y^2+(m+n)x+(m-n)y+mn

令 m+n=3

m-n=7

则 m=5 n=-2

a=mn=-10