Maths

1.
(a)
x² - 6 是 ax³ + ax² + bx + 12 的因式。
設 (ax³ + ax² + bx + 12) ÷ (x²- 6) = ax + c
所以 ax³ + ax²+ bx + 12 = (ax + c)(x² - 6)
ax³ + ax² + bx + 12 = ax³ + cx² - 6ax - 6c

比較兩邊常數項：
-6c = 12
c = -2

比較兩邊 x² 項：
a = c
a = -2

比較兩邊 x 項：
b = -6a
b = -6(-2)
b = 12

因此，a = -2 及 b = 12

(b)
f(x) = 0
-2x³ - 2x² + 12x + 12 = 0
x³ + x² - 6x - 6 = 0
x²(x + 1) - 6(x + 1) = 0
(x² - 6)(x + 1) = 0
x² = 6 或 x = -1
x = √6 或 x = -√6 或 x = -1


2.
(a)
f(x) = -2x³ + 5x² + x - 6

f(3/2)
= -2(3/2)³ + 5(3/2)² + (3/2) - 6
= -(27/4) + (45/4) + (3/2) - 6
= 0

故此，2x - 3 是 f(x) 的因式。
而 3 - 2x 亦是 f(x) 的因式。

(b)
用長除法：
(-2x³ + 5x² + x - 6) ÷ (-2x + 3) = x² - x - 2
所以-2x³ + 5x² + x - 6 = -(2x -3)(x² - x - 2)

f(x) = 2x - 3
-2x³ + 5x² + x - 6 = 2x - 3
-(2x - 3)(x² - x - 2) - (2x - 3) = 0
(2x - 3)[(x² - x - 2) + 1] = 0
(2x - 3)(x² - x - 1) = 0
x = 3/2 或 x = (1 ± √5)/2