ax^2±bx+c和ax^2±bx-c的因式分解

ax^2 + bx + c = (px + r)(qx + s)
ax^2 + bx - c = (px - s)(qx + r)
所以 pq = a, rs = c, ps + qr = b, pr - qs = b
因為最後兩個都等於 b，所以
ps + qr = pr - qs
==> p(r - s) = q(r + s)
所以只要符合
p = r + s 及 q = r - s
那就可以了。

例如 r = 4, s = 3, p = 7, q = 1, 即
(7x ± 4)(x ± 3) = 7x^2 ± 25x + 12
(7x - 3)(x + 4) = 7x^2 + 25x - 12
(7x + 3)(x - 4) = 7x^2 - 25x - 12

例如 r = 7, s = 4, p = 11, q = 3, 即
(11x ± 7)(3x ± 4) = 33x^2 ± 65x + 28
(11x - 4)(3x + 7) = 33x^2 + 65x - 28
(11x + 4)(3x - 7) = 33x^2 - 65x - 28

2013-02-27 16:34:32 補充：
a = pq, b = ps + qr 或 b= pr - qs, c = rs, p = r + s, q = r - s
所以只要你設定咗 r 及 s, 那 a, b, c, p, q 就出晒來了。
例如設 r = 5, s = 2, 則
q = 3, p = 7, a = 21, b = 29, c = 10, 即
21x^2 ± 29x + 10 = (7x ± 5)(3x ± 2)
21x^2 + 29x - 10 = (7x - 2)(3x + 5)
21x^2 - 29x - 10 = (7x + 2)(3x - 5)

事實上有無限多個 (infinitely many) 可能的答案.
一個一元二次的三項式, 其根為 ( -b 加或減 sqrt(b^2-4ac) )/ (2a), 所以, b^2 - 4ac 一定要是平方數, 才能使 p, q, r, s 為整數.

2013-02-25 16:19:38 補充：
再淺白一點說,
b^2 - 4ac = 1 可取 a = 2, b = 3, c = 1, 就是 2x^2 + 3x +1 = (x+1)(2x+1)
b^2 - 4ac = 4 可取 a = 3, b = 4, c = 1, 就是 3x^2 + 4x +1 = (x+1)(3x+1)
b^2 - 4ac = 9 可取 a = 4, b = 5, c = 1, 就是 4x^2 + 5x +1 = (x+1)(4x+1)
可見其中一答案可以是 a = m, b = m+1 c = 1.

2013-02-25 16:19:55 補充：
我可再舉多個: a = 1, b = m+1 c = m

2013-02-25 16:20:11 補充：
樓主, 其實我已經答了你問題.
你要的條件就是 b^2 - 4ac = k^2, where k is any integer.
即: b^2 - 4ac = 1, 4, 9, 25,... 等等的平方數.