What is (4-x)(x+3)(x^2-1)<0 ?

The solution to (4-x)(x+3)(x^2-1) 4

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(4-x)(x+3)(x^2-1)<0

(4 - x) (x + 3) (x^{2} - 1) < 0

x < - 3 or - 1 < x < 1 or x > 4

Interval Notation

(- \infty, - 3) \cup (- 1, 1) \cup (4, \infty)

Solution steps

(4 - x) (x + 3) (x^{2} - 1) < 0

Factor

(4 - x) (x + 3) (x^{2} - 1) : - (x - 4) (x + 3) (x + 1) (x - 1)

- (x - 4) (x + 3) (x + 1) (x - 1) < 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 4) (x + 3) (x + 1) (x - 1)) (- 1) > 0 \cdot (- 1)

Simplify

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

Identify the intervals

x < - 3 or - 1 < x < 1 or x > 4

3 (4 f - 5) - 2 = 6 f - 1

t^{3} t^{4} t^{2}

- 1 \leq 2 x - 5 \leq 7

16 (y - 3)^{2} + 1 = 8 (y - 3)

2401

What is (4-x)(x+3)(x^2-1)<0 ?
The solution to (4-x)(x+3)(x^2-1)<0 is x<-3\lor-1<x<1\lor x>4