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求m为何值时,多项式x2-y2+mx+5y-6能因式分解,并分解此多项式.
人气:490 ℃ 时间:2020-06-07 03:47:56
解答
∵x2-y2=(x+y)(x-y),
∴可设x2-y2+mx+5y-6=(x+y+a)(x-y+b).
∵(x+y+a)(x-y+b)=x2-xy+bx+xy-y2+by+ax-ay+ab
=x2-y2+(a+b)x+(b-a)y+ab,
∴x2-y2+mx+5y-6=x2-y2+(a+b)x+(b-a)y+ab.
a+b=m①
b−a=5②
ab=−6③

由②得b=a+5,代入③得a(a+5)=-6.
整理得:a2+5a+6=0.
解得:a1=-2,a2=-3.
∴当a=-2时,b=3,m=1;当a=-3时,b=2,m=-1.
∴当m=1时,x2-y2+x+5y-6=(x+y-2)(x-y+3);
当m=-1时,x2-y2-x+5y-6=(x+y-3)(x-y+2).
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