What is (x^2-1)(x+5)(x-3)<= 0 ?

The solution to (x^2-1)(x+5)(x-3)<= 0 is -5<= x<=-1\lor 1<= x<= 3

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(x^2-1)(x+5)(x-3)<= 0

(x^{2} - 1) (x + 5) (x - 3) \leq 0

- 5 \leq x \leq - 1 or 1 \leq x \leq 3

Interval Notation

[- 5, - 1] \cup [1, 3]

Solution steps

(x^{2} - 1) (x + 5) (x - 3) \leq 0

Factor

(x^{2} - 1) (x + 5) (x - 3) : (x + 1) (x - 1) (x + 5) (x - 3)

(x + 1) (x - 1) (x + 5) (x - 3) \leq 0

Identify the intervals

- 5 \leq x \leq - 1 or 1 \leq x \leq 3

- 3 x^{2} - 0.75 x + 6 = 0

\frac{x ^{- 1} y ^{2}}{x ^{2} y ^{- 1}}

m^{2} + 16 m + 64

5 6^{2} + x^{2} = 6 5^{2}

\frac{1}{x + 4} \leq \frac{- 2}{1 - x}

What is (x^2-1)(x+5)(x-3)<= 0 ?
The solution to (x^2-1)(x+5)(x-3)<= 0 is -5<= x<=-1\lor 1<= x<= 3