3 Maths Questions about Factorization (20 marks)

1. (3x-2) ³ - 8(x+1) ³
= (3x – 2)(3x-2) ² - 8(x+1)(x+1) ²
= (3x – 2)(9x² - 12x + 4) – (8x + 8)(x² + 2x + 1)
= (27x³ - 36x² + 3x – 18x² + 14x – 8) – (8x³ + 16x² + 8x + 8x² + 16x + 8)
= 27x³ – 8x³- 36x² - 16x² - 8x² – 18x²+ 3x + 14x - 8x - 16x – 8 – 8
= 19x³ - 78x² - 7x – 16
（very unsure. @@）

2. (x+2) ² - (x+2)(y-3) - 6(y-3) ²
Put a = x+2 and b = y-3
= a² - ab – 6b²
= (a+2b)(a-3b)
= (x + 2 + 2y – 6)(x + 2 – 3y + 9)
= (x + 2y – 4)(x – 3y + 11)

3. 3(x+7) ² + (x+7)(x-6) - 14(x-6) ²
Put a = x+7 and b = x – 6
= 3a² + ab – 14b²
= (a – 2b)(3a + 7b)
= (x + 7 – 2x + 12)(3x + 21 + 7x – 42)
= (19 – x)(10x – 21)

1) (3x-2)^3 - 8(x+1)^3 USE THE EQUATION a^3-b^3=(a-b)(a^2+ab+b^2)
=(3x-2)^3- [2(X+1)]^3
=[3X-2-2(x+1)] {(3x-2)^2+(3x-2)[2(x+1)]+[2(X+1)]^2}
=(x-4) [9X^2-12X+4+6X^2+2x-4+4(x^2+2x+1)]
=(x-4) (9X^2-12X+4+6X^2+2x-4+4x^2+8x+4)
=(x-4)(19X^2-2x+4)

2)(x+2)^2 - (x+2)(y-3) - 6(y-3)^2
let a=x+2 b=y-3
a^2 -ab -6b^2
=(a-3b)(a+2b)
=[x+2- 3(y-3)] [X+2+2(Y-3)]
=(x-3y+11)(x+2y-4)

3)3(x+7)^2 + (x+7)(x-6) - 14(x-6)^2
let a=x+6 b=x-6
3a^2+ab-14b^2
=(3a+7b)(a-2b)
=[3(x+7)+7(x-6)] [x+7-2(X-6)]
=(10x-21)(19-x)

1.
(3X-2)^3 = 3X^3 + 3(3X)^2(-2) + 3(3X)(-2)^2 + (-2)^3 (binomial theorem)
= 3X^3 - 54X^2 + 36X - 8

8(x+1)^3 = 8 [ X^3 + 3(X)^2(1) + 3(X)(1)^2 + 1^3 (binomial theorem)
= 8 ( X^3 + 3X^2 + 3X + 1 )
= 8X^3 + 24^2 + 24X +8

2006-11-02 20:27:26 補充：
1. ( 3X^3 - 54X^2 36X - 8 ) - ( 8X^3 24^2 24X 8 )= - 5X^3 - 78X^2 12X -162. (x 2)^2 - (x 2)(y-3) - 6(y-3)^2= ([X 2 2(Y-3)] [X 2-3(Y-3)]= (X 2 2Y-6)(X 2-3Y 9)= (X-4 2Y)(X-3Y 11)

2006-11-02 20:29:15 補充：
2. (x 2)^2 - (x 2)(y-3) - 6(y-3)^2= ([X 2 2(Y-3)] [X 2-3(Y-3)]= (X 2 2Y-6)(X 2-3Y 9)= (X-4 2Y)(X-3Y 11)