与えられた数式を因数分解します。 問題6 (1) $(x+y)^2 - 5(x+y) + 6$ (2) $(a-b)^2 + 4(a-b) - 21$ (3) $(x-y)^2 + 4(x-y) + 4$ 問題7 (1) $a(x+y) - x - y$ (2) $ax - ay + b(x-y)$ (3) $xy + 3y - 2x - 6$

代数学因数分解多項式
2025/4/24

1. 問題の内容

与えられた数式を因数分解します。
問題6
(1) (x+y)25(x+y)+6(x+y)^2 - 5(x+y) + 6
(2) (ab)2+4(ab)21(a-b)^2 + 4(a-b) - 21
(3) (xy)2+4(xy)+4(x-y)^2 + 4(x-y) + 4
問題7
(1) a(x+y)xya(x+y) - x - y
(2) axay+b(xy)ax - ay + b(x-y)
(3) xy+3y2x6xy + 3y - 2x - 6

2. 解き方の手順

問題6
(1)
x+y=Ax+y = A とおくと、
A25A+6=(A2)(A3)A^2 - 5A + 6 = (A-2)(A-3)
AAx+yx+y に戻すと、
(x+y2)(x+y3)(x+y-2)(x+y-3)
(2)
ab=Aa-b = A とおくと、
A2+4A21=(A+7)(A3)A^2 + 4A - 21 = (A+7)(A-3)
AAaba-b に戻すと、
(ab+7)(ab3)(a-b+7)(a-b-3)
(3)
xy=Ax-y = A とおくと、
A2+4A+4=(A+2)2A^2 + 4A + 4 = (A+2)^2
AAxyx-y に戻すと、
(xy+2)2(x-y+2)^2
問題7
(1)
a(x+y)xy=a(x+y)(x+y)=(a1)(x+y)a(x+y) - x - y = a(x+y) - (x+y) = (a-1)(x+y)
(2)
axay+b(xy)=a(xy)+b(xy)=(a+b)(xy)ax - ay + b(x-y) = a(x-y) + b(x-y) = (a+b)(x-y)
(3)
xy+3y2x6=y(x+3)2(x+3)=(y2)(x+3)=(x+3)(y2)xy + 3y - 2x - 6 = y(x+3) - 2(x+3) = (y-2)(x+3) = (x+3)(y-2)

3. 最終的な答え

問題6
(1) (x+y2)(x+y3)(x+y-2)(x+y-3)
(2) (ab+7)(ab3)(a-b+7)(a-b-3)
(3) (xy+2)2(x-y+2)^2
問題7
(1) (a1)(x+y)(a-1)(x+y)
(2) (a+b)(xy)(a+b)(x-y)
(3) (x+3)(y2)(x+3)(y-2)