Question

(-6x-1)(4-5x)(x+3)>0

Answer 1

$(-6x-1)(4-5x)(x+3)>0, \\ -6\cdot(-5)\cdot(x+\frac{1}{6})(x-\frac{4}{5} )(x+3)>0, \\ 30(x+\frac{1}{6})(x-\frac{4}{5} )(x+3)>0, \\ (x+\frac{1}{6})(x-\frac{4}{5} )(x+3)>0, \\ (x+\frac{1}{6})(x-\frac{4}{5} )(x+3)=0, \\ x_1=-3, x_2=- \frac{1}{6}, x= \frac{4}{5}, \\ x\in(-3;- \frac{1}{6})\cup(\frac{4}{5};+\infty)$

Answer 2

To solve the inequality $(-6x - 1)(4 - 5x)(x + 3) > 0$, you can use the concept of interval notation and the sign of each factor.

1. Find the critical points where each factor equals zero:

   • $-6x - 1 = 0$ when $x = -1/6$
   • $4 - 5x = 0$ when $x = 4/5$
   • $x + 3 = 0$ when $x = -3$

2. These critical points divide the real number line into four intervals: $(-∞, -3)$, $(-3, -1/6)$, $(-1/6, 4/5)$, and $(4/5, ∞)$.

3. Now, test a value from each interval in the inequality to determine the sign of the expression within each interval:

   • For $x < -3$, choose $x = -4$: $(-6(-4) - 1)(4 - 5(-4))(x + 3) = (23)(24)(-1) < 0$

   • For $-3 < x < -1/6$, choose $x = -1$: $(-6(-1) - 1)(4 - 5(-1))(x + 3) = (5)(9)(2) > 0$

   • For $-1/6 < x < 4/5$, choose $x = 0$: $(-6(0) - 1)(4 - 5(0))(x + 3) = (-1)(4)(3) < 0$

   • For $x > 4/5$, choose $x = 1$: $(-6(1) - 1)(4 - 5(1))(x + 3) = (-7)(-1)(4) > 0$

4. Now, consider the signs in each interval:

   • $(-∞, -3)$: The expression is negative.
   • $(-3, -1/6)$: The expression is positive.
   • $(-1/6, 4/5)$: The expression is negative.
   • $(4/5, ∞)$: The expression is positive.

5. The original inequality $(-6x - 1)(4 - 5x)(x + 3) > 0$ is satisfied when the expression is greater than zero. Therefore, the solution to the inequality is:

   $-3 < x < -\frac{1}{6}$ or $x > \frac{4}{5}$

In interval notation, this can be written as:

$(-3, -\frac{1}{6}) \cup (\frac{4}{5}, \infty)$