問題
次の式を因数分解せよ.
解説
- (1) $x^2+(3y+1)x+(y+4)(2y-3)$
- $=$$(x+y+4)(x+2y-3)$
- (2) $x^2+3xy+2y^2-6x-11y+5$
- $=x^2+(3y-6)x+2y^2-11y+5$
$=x^2+(3y-6)x+(y-5)(2y-1)$
$=$$(x+y-5)(x+2y-1)$
- $=x^2+(3y-6)x+2y^2-11y+5$
- (3) $x^2-2xy+y^2-x+y-2$
- $=x^2-(2y+1)x+y^2+y-2$
$=x^2-(2y+1)x+(y+2)(y-1)$
$=$$(x-y-2)(x-y+1)$
- $=x^2-(2y+1)x+y^2+y-2$
- (4) $2x^2+5xy+2y^2+4x-y-6$
- $=2x^2+(5y+4)x+2y^2-y-6$
$=2x^2+(5y+4)x+(y-2)(2y+3)$
$=$$(2x+y-2)(x+2y+3)$
- $=2x^2+(5y+4)x+2y^2-y-6$
- (5) $2x^2+xy-y^2+7x-5y-4$
- $=2x^2+(y+7)x-(y^2+5y+4)$
$=2x^2+(y+7)x-(y+1)(y+4)$
$=$$(2x-y-1)(x+y+4)$
- $=2x^2+(y+7)x-(y^2+5y+4)$
- (6) $2x^2+5xy-3y^2-x+11y-6$
- $=2x^2+(5y-1)x-(3y^2-11y+6)$
$=2x^2+(5y-1)x-(y-3)(3y-2)$
$=$$(2x-y+3)(x+3y-2)$
- $=2x^2+(5y-1)x-(3y^2-11y+6)$
