(1)
x2+xy−12y2+2x+15y−3
=(x+4y)(x−3y)+2x+15y−3
a(x+4y)+b(x−3y)=2x+15y(a+b)x=2x(4a−3b)y=15y
a+b=2①4a−3b=15②
a=3,b=−1
ab=−3 であるので
(x+4y−1)(x−3y+3)
(2)
2x2−5xy−3y2+3x−2y+1
=(2x+y)(x−3y)+3x−2y+1
(2x+y)+(x−3y)=3x−2y であるので
(2x+y+1)(x−3y+1)
(3)
(a2−1)x2−2(a2+1)x+a2−1
x=a2−1(a2+1)±(a2+1)2−(a2−1)2=a2−1a2+1±4a2=a2−1a2+1±2a=a2−1(a+1)2,a2−1(a−1)2=a−1a+1,a+1a−1
これから
{(a−1)x−a−1}{(a+1)x−a+1}