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) > 0

- 4 < x < - 1 or 2 < x < 4

Interval Notation

(- 4, - 1) \cup (2, 4)

Solution steps

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

(x - 2) (x + 1) (x + 4) (x - 4) < 0

Identify the intervals

- 4 < x < - 1 or 2 < x < 4

\frac{- 12 x ^{- 2} y ^{4}}{3 x y ^{- 3}}

2 k^{2} - 5 k - 18 = 0

8 n + 2 n + 6 < - 5 + 1

w^{2} + 10 w + 21 = 0

∣ x ∣ > 2

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