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) \leq 0

x \leq - 3 or - 1 \leq x \leq 1 or x \geq 4

Interval Notation

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

Solution steps

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

(x - 4) (x + 3) (x + 1) (x - 1) \geq 0

Identify the intervals

x \leq - 3 or - 1 \leq x \leq 1 or x \geq 4

6 + 2 i

0.8 5^{5}

3 x^{2} - 45 = 0

\frac{1}{3} y + 7 \geq 12

8^{\frac{- 2}{3}}

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