Question

-2x(x+1)(x-4)(x2-16)<_0

Answer 1

-2x(x+1)(x-4)(x²-16)≤0    |÷(-2)
x(x+1)(x-4)(x-4)(x+4)≥0
x(x+1)(x-4)²(x+4)≥0
-∞___-___-4___+___-1___-___0__+__[4]__+__+∞
x∈[-4;-1]U[0;+∞).

Answer 2

To solve the inequality -2x(x + 1)(x - 4)(x^2 - 16) ≤ 0, we need to find the values of x that satisfy the inequality.

First, let's factor the expression completely:

-2x(x + 1)(x - 4)(x^2 - 16) = -2x(x + 1)(x - 4)(x - 4)(x + 4)

The expression is now factored into individual factors. To determine the sign of each factor, we can examine the intervals where each factor is positive or negative. We can create a sign chart to keep track of the signs:

markdownCopy code
Intervals      | -∞  | -4  | -1  |  4  | +∞
-------------------------------------------
x + 1          | -   | -   | -   | +   | +
x - 4          | -   | -   | -   | -   | +
x - 4          | -   | -   | -   | -   | +
x + 4          | -   | -   | -   | +   | +
Solution       |  -  | -4  | -1  |  4  | +∞

To determine the sign of the expression -2x(x + 1)(x - 4)(x^2 - 16), we consider the signs of the individual factors. If there is an odd number of negative factors, the expression is negative. If there is an even number of negative factors, the expression is positive. From the sign chart, we can see that there are negative factors at x = -4, x = -1, and x = 4. Since there are three negative factors, the expression is negative in the intervals (-4, -1) and (4, +∞).

Therefore, the solution to the inequality is:

-4 < x < -1 or x > 4