中二數學兩題【求完整說明及解答】

1a)
y^2 - 11y + 24

y -8
y -3
__________
-8y-3y=-11y
∴y^2 - 11y + 24 = (y-8)(y-3)

b)
(y^2-2y)^2 - 11(y^2-2y) + 24
設x=y^2-2y
x^2 - 11x + 24
=(x-8)(x-3) [根據(a)的答案]
∴(y^2-2y)^2 - 11(y^2-2y) + 24
= (y^2 - 2y - 8) (y^2 - 2y - 3)

y -4
y +2
__________
-4y+2y=-2y

y -3
y +1
_________
-3y+y=-2y
∴ (y^2 - 2y - 8) (y^2 - 2y - 3) = (y-4)(y+2)(y-3)(y+1)
∴(y^2-2y)^2 - 11(y^2-2y) + 24= (y-4)(y+2)(y-3)(y+1)

2a)
x^2 - 4x - 21

x -7
x +3
__________
-7x+3x=-4x
∴x^2 - 4x - 21 = (x-7)(x+3)

b)
4x^2 + 11x - 3

x +3
4x -1
_________
12x-x=11x
∴4x^2 + 11x - 3 = (x+3)(4x-1)

c)
(2x^3 - 8x^2 - 42x) + (16x^2 + 44x - 12)
=2x(x^2 - 4x - 21) + 4(4x^2 + 11x - 3)
=2x(x-7)(x+3) + 4(x+3)(4x-1) ... [根據(a)及(b)的答案]
=(x+3)[2x(x-7) + 4(4x-1)] ... [抽相同因子]
=(x+3)(2x^2 - 14x + 16x - 4)
=(x+3)(2x^2 + 2x - 4)
=2(x+3)(x^2 + x - 2) ... [抽相同因子]

x +2
x -1
_______
-x+2x=x
∴(x^2 + x - 2)=(x+2)(x-1)
∴(2x^3 - 8x^2 - 42x) + (16x^2 + 44x - 12)
=2(x+3)(x^2 + x - 2)
=2(x+3)(x+2)(x-1)