What is (2-x)(x+1)(x^2-16)>= 0 ?

The solution to (2-x)(x+1)(x^2-16)>= 0 is -4<= x<=-1\lor 2<= x<= 4

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(2-x)(x+1)(x^2-16)>= 0

(2 - x) (x + 1) (x^{2} - 16) \geq 0

- 4 \leq x \leq - 1 or 2 \leq x \leq 4

Interval Notation

[- 4, - 1] \cup [2, 4]

Solution steps

(2 - x) (x + 1) (x^{2} - 16) \geq 0

Factor

(2 - x) (x + 1) (x^{2} - 16) : - (x - 2) (x + 1) (x + 4) (x - 4)

- (x - 2) (x + 1) (x + 4) (x - 4) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 2) (x + 1) (x + 4) (x - 4)) (- 1) \leq 0 \cdot (- 1)

Simplify

(x - 2) (x + 1) (x + 4) (x - 4) \leq 0

Identify the intervals

- 4 \leq x \leq - 1 or 2 \leq x \leq 4

\frac{1}{125} x^{3} + 27 y^{3}

- 3 x^{2} + 5 x - 2 = 0

\frac{2 x + 3}{x - 2} < 5 x - 2

4 x^{2} + x y - 18 y^{2}

x^{2} + 8 x - 36 = 0

What is (2-x)(x+1)(x^2-16)>= 0 ?
The solution to (2-x)(x+1)(x^2-16)>= 0 is -4<= x<=-1\lor 2<= x<= 4