問題
次の式を因数分解せよ.
解説
- (1) $(x+2)^2+5(x+2)+6$
- $x+2=t$と置くと,
$t^2+5t+6=(t+2)(t+3)$より,
$t$を戻すと$(x+4)(x+5)$.
- $x+2=t$と置くと,
- (2) $(x-y)^2-x+y-12$
- $x-y=t$と置くと,
$t^2-t-12=(t+3)(t-4)$より,
$t$を戻すと$(x-y+3)(x-y-4)$.
- $x-y=t$と置くと,
- (3) $6(x+y)^2-5(x+y)-4$
- $x+y=t$と置くと,
$6t^2-5t-4=(2t+1)(3t-4)$より,
$t$を戻すと$(2x+2y+1)(3x+3y-4)$.
- $x+y=t$と置くと,
- (4) $(x-y)^2-5(x-y)z+4z^2$
- $x-y=t$と置くと,
$t^2-5tz+4z^2=(t-z)(t-4z)$より,
$t$を戻すと$(x-y-z)(x-y-4z)$.
- $x-y=t$と置くと,
- (5) $(a+b)^2+8c(a+b)+16c^2$
- $a+b=t$と置くと,
$t^2+8ct+16c^2=(t+4c)^2$より,
$t$を戻すと$(a+b+4c)^2$.
- $a+b=t$と置くと,
- (6) $(x+y+1)(x+y-3)-12$
- $x+y=t$と置くと,
$(t+1)(t-3)-12=t^2-2t-15=(t+3)(t-5)$より,
$t$を戻すと$(x+y+3)(x+y-5)$.
- $x+y=t$と置くと,
