✔ 最佳答案
(1) c² (c+2)-c-2
(2) a³-4a-45a
(3) (a²-2ab+b²)-25
(4) (k+1)²-(k+1)-56
(5) (2x-3)²+10x-15
(6) 3(x+7)²+(x+7)(x-6)-14(x-6)²
(1)
c²(c+2) – c – 2
= c³ + 2c² - c – 2
= c(c² + 2c – 1) – 2
(2)
a³ – 4a – 45a
= a(a² - 4 – 45)
= a(a² - 49)
= a(a+7)(a-7)
(3)
a² - 2ab + b² - 25
= (a-b) ² - 25
= (a-b-5)(a-b+5)
(4)
(k+1) ² - (k+1) – 56
= (k+1+7)(k+1-8)
= (k+8)(k-7)
(5)
(2x -3) ² + 10x – 15
= 4x² - 12x + 9 + 10x – 15
= 4x² - 2x – 6
= 2(2x² - x – 3)
= 2(x+1)(2x – 3)
(6)
3(x+7)²+(x+7)(x-6)-14(x-6)²
Let y = (x+7) and z = (x-6)
3y² + yz – 14z²
= (y -2z)(3y + 7z)
= (x + 7 – 2(x-6) )*(3x + 21 + 7(x-6) )
= (x + 7 – 2x + 12) (3x + 21 + 7x – 42)
= (-x + 19)(10x – 32)
= (10x – 32)(19 –x)
2006-10-30 10:23:34 補充:
(1)c²(c+2) – c – 2= c³ + 2c² - c – 2= c(c² + 2c – 1) – 2做錯=.=改正:c²(c+2) – (c+2)= (c+2)(c² - 1)= (c+2)(c+1)(c-1)