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

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

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

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

- 3 < x < 3 or x > 3

Interval Notation

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

Solution steps

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

Factor

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

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

Identify the intervals

- 3 < x < 3 or x > 3

(x + 4) (x - 6) (4 - 5) = 0

9 (x - 5) = 8 (x + 5)

7 x^{2} - 2 x + 3 = 0

\frac{- 3 - 2}{3 17}

6 x^{2} + 21 x - 12

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