F. 2 Maths (最佳解答取30分)

Factorize the following expressions.


1) 9p²-q²-12p-4q

9p²-q²-12p-4q

= (3p)² - q² - 4(3p) - 4(q)

= (3p -q)(3p + q) - 4(3p + q) ........ as a²-b² = (a-b)(a+b)

= (3p + q)[(3p - q) -4]

= (3p + q)(3p - q -4)

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2) m(x+y)²-4m(x-y)²

m(x+y)² - 4m(x-y)²

= m[(x+y)² - 4(x-y)²]

= m[(x+y)² - [2(x-y)]²]

= m[(x+y) - 2(x-y)][(x+y) + 2(x-y)] ........ as a²-b² = (a-b)(a+b)

= m(x+y-2x+2y)(x+y+2x-2y)

= m(-x+3y)(3x-y)

= m(3y-x)(3x-y)

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3) 2-18(a+1)²

2 - 18(a+1)²

= 2[1-9(a+1)²]

= 2[1² - [3(a+1)]²]

= 2[1 - 3(a+1)][1 + 3(a+1)] ........ as a²-b² = (a-b)(a+b)

= 2(1 - 3a - 3)(1 + 3a + 3)

= 2(-2 - 3a)(4 + 3a)

= -2(3a + 2)(3a + 4)

2006-12-09 12:33:50 補充：
小小補充：factorizaton (因式分解) 常用到的恆等式有a²-b² = (a-b)(a+b) ..... 最常用(a-b)² = a²-2ab+b²(a+b)² = a²+2ab+b²還要注意的是：1. 答案中每個項都不可以有括號在人內。例：(x+1)(2x+3) ..正確(x+1)[2(x+1)+1) .. 不正確2. 答案中每個項中不能夠所有項都有負號（要抽 -1 出來）而且負項不可以放在首，例：-(x+1)(x-2) .. 正確-(x+1)(-x+2) .. 不正確(x+1)[-x-2) .. 不正確

1. 9p^2-q^2-12p-4q
=(3p)^2-q^2-4(3p+q)-->先將9p^2變為(3p)^和後面抽common factor4,目的是將兩邊變相似
=(3p-q)(3p+q)-4(3p+q)-->因為a^2-b^2=(a+b)(a-b),所以可將(3p)^2-q^2變成為(3p-q)(3p+q)
=(3p+q)(3p-q-4)-->抽common factor(3p+q),剩番(3p-q-4)

2. m(x+y)^2-4m(x-y)^2
=m(x^2+2xy+y^2)-4m(x^-2xy+y^2))-->先將(x+y)^2和(x-y)^2分解
=mx^2+2mxy+my^2-4mx^2+8mxy-4mxy^2)-->將m乘入括號
=-3mx^2+6mxy-3my^2
=-3m(x^2-2xy+y^2)-->抽common　factor-3m
=-3m(x-y)^2-->(x^2-2xy+y^2)最簡為(x-y)^2

3. 2-18(a+1)^2
=2[1-9(a+1)^2]-->抽common factor2
=2{1-[3(a+1)]^2}-->9是3的2次
=2(1+3a+3)(1-3a-3)-->因為a^2-b^2等於(a+b)(a-b)

2006-12-09 12:46:54 補充：
第3題的答案應為2(3a+2)(3a-2)