What is -x(x+5)(x-2)^2(x+4)<= 0 ?

The solution to -x(x+5)(x-2)^2(x+4) = 0

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-x(x+5)(x-2)^2(x+4)<= 0

- x (x + 5) (x - 2)^{2} (x + 4) \leq 0

- 5 \leq x \leq - 4 or x \geq 0

Interval Notation

[- 5, - 4] \cup [0, \infty)

Solution steps

- x (x + 5) (x - 2)^{2} (x + 4) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

(- x (x + 5) (x - 2)^{2} (x + 4)) (- 1) \geq 0 \cdot (- 1)

Simplify

x (x + 5) (x - 2)^{2} (x + 4) \geq 0

Identify the intervals

- 5 \leq x \leq - 4 or x \geq 0

w^{3} - 8

\frac{3}{4} t + \frac{7}{8} t = 39

x^{2} - 2 x + 5 x - 10

p - 5 < 2

27 m^{6} + 343 n^{9}

What is -x(x+5)(x-2)^2(x+4)<= 0 ?
The solution to -x(x+5)(x-2)^2(x+4)<= 0 is -5<= x<=-4\lor x>= 0