HELP! 幾題數學問題

1. The function is f(x) = A(x + 3)(x - 4)(x + 4)
Sub (5,12) into function, 12 = A(5 + 3)(5 - 4)(5 + 4)
A = 12/(8*1*9)
A = 1/6
So f(x) = (x + 3)(x - 4)(x + 4)/6
2. Since x - 2 < x - 1 < x + 3
Their product is greater than 0 if
(I) all of them are greater than zero OR
(II) 2 are negative and one is positive
Case (I) when the smallest is greated than zero, all are greater than zero
x - 2 > 0 => x > 2
Case (II) the largest one is greater than zero and the middle one is less than zero ( so the smallest one must be negative)
x + 3 > 0 => x > -3
x - 1 < 0 => x < 1
Together -3 < x < 1
Combining the two cases, x > 2 or -3 < x < 1
3. Let f(x) = ax^3 + bx^2 - x + 3
Remainder when divided by x - 1 is f(1) = a + b - 1 + 3 = 4
a + b = 2 ... (1)
Remainder when divided by x - 2 is f(2) = 8a + 4b - 2 + 3 = 21
8a + 4b = 20
2a + b = 5 ... (2)
(2) - (1) => a = 3
(1) => b = -1
4. If the question means (-15/x) + 1 <= 3, then
(-15/x) + 1 <= 3 and x <> 0
-15/x <= 2
if x > 0, then -15 <= 2x
-15/2 <= x which is always true when x > 0
if x < 0, then -15 >= 2x
-15/2 >= x
Therefore the solution is x > 0 or -15/2 <= x
If the question means -15/(x + 1) <= 3, then
-15/(x + 1) <= 3 and x <> -1
-5/(x + 1) <= 1
if (x + 1) > 0 ,i.e. x > -1
-5 <= x + 1
-6 <= x which is always true for x > -1
if (x + 1) < 0, i.e. x < -1
-5 >= x + 1
-6 >= x
Therefore the solution is x > -1 or x <= -6