What is (x^2-9)^2*(x+1)(x^2-2x-3)(x-1)<= 0 ?

The solution to (x^2-9)^2*(x+1)(x^2-2x-3)(x-1)<= 0 is x=-3\lor x=-1\lor 1<= x<= 3

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(x^2-9)^2*(x+1)(x^2-2x-3)(x-1)<= 0

(x^{2} - 9)^{2} \cdot (x + 1) (x^{2} - 2 x - 3) (x - 1) \leq 0

x = - 3 or x = - 1 or 1 \leq x \leq 3

Interval Notation

x = - 3 \cup x = - 1 \cup [1, 3]

Solution steps

(x^{2} - 9)^{2} (x + 1) (x^{2} - 2 x - 3) (x - 1) \leq 0

Factor

(x^{2} - 9)^{2} (x + 1) (x^{2} - 2 x - 3) (x - 1) : (x - 3)^{3} (x + 3)^{2} (x + 1)^{2} (x - 1)

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

Identify the intervals

x = - 3 or x = - 1 or 1 \leq x \leq 3

- x^{3} + x \leq 0

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

\frac{12 x}{x - 4} \geq \frac{6 x}{x + 6}

- 2 x^{2} + x - 1 < 0

x^{2} - 2 x < 1

What is (x^2-9)^2*(x+1)(x^2-2x-3)(x-1)<= 0 ?
The solution to (x^2-9)^2*(x+1)(x^2-2x-3)(x-1)<= 0 is x=-3\lor x=-1\lor 1<= x<= 3