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

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

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

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

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

Interval Notation

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

Solution steps

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

Factor

(x^{2} + 2^{2}) (x - 1)^{3} (x^{2} - 9) : (x - 1)^{3} (x + 3) (x - 3) (x^{2} + 4)

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

Identify the intervals

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

\frac{( x - 4 x ^{2} )}{( x - 1 )} > 0

x^{3} - 4 x^{2} + 3 x < 0

36 m^{2} - 180 m + 144 < 0

\frac{x}{( x - 6 )} \geq 0

\frac{4}{( x - 5 )} - \frac{2}{( x - 5 )} > \frac{( 3 x + 3 )}{( x ^{2} - 25 )}

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