數學:因式分解

(1)
題目中的 2b 應為 b 之誤。

設 u = a - b
則 2(a - b)² - a + b - 15
= 2(a - b)² + (-a + b) - 15
= 2(a - b)² - (a - b) - 15
= 2u² - u - 15
= [u - 3] [2u + 5]
= [(a - b) - 3] [2(a - b) + 5]
= (a - b - 3)(2a - 2b + 5)


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(2)
設 u = 3x + 1 及 v = y - 2
則 2(3x + 1)² + (3x + 1)(y - 2) - (y - 2)²
= 2u² + uv - v²
= [u + v] [2u - v]
= [(3x + 1) + (y - 2)] [2(3x + 1) - (y - 2)]
= [3x + 1 + y - 2] [6x + 2 - y + 2]
= (3x + y - 1)(6x - y + 4)


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(3)
設 u = a + b 及 v = a - b
則 2(a + b)² + 5(b² - a²) - 12(a - b)²
= 2(a + b)² - 5(a² - b²) - 12(a - b)²
= 2(a + b)² - 5(a + b)(a - b) - 12(a -b)²
= 2u² - 5uv - 12v²
= [u - 4v] [2u + 3v]
= [(a + b) - 4(a - b)] [2(a + b) + 3(a - b)]
= [a + b - 4a + 4b] [2a + 2b + 3a - 3b]
= (5b - 3a)(5a - b)

(2) 2(3x+1)平方+(3x+1)(y-2)-(y-2)平方

2(3x+1)²+(3x+1)(y-2)-(y-2)²

設 a = (3x+1) ; b = (y-2)

2(3x+1)²+(3x+1)(y-2)-(y-2)² = 2a² + ab - b²

　　２　　１
　　１　　１

= (2a - b)(a + b)

2(3x+1)²+(3x+1)(y-2)-(y-2)² = [2(3x+1) - (y-2)][(3x+1) + (y-2)]

若要打平方，可以用" ^ "代表，例如x^2, y^5

2014-08-25 15:39:59 補充：
2. 原式=[(3x+1)+1][2(y-2)-1]

= (3x+2)(2y-5)

2014-08-25 15:45:44 補充：
3. 原式=2(a+b)^2-5(a^2-b^2)-12(a-b)^2

=2(a+b)^2-5(a+b)(a-b)-12(a-b)^2

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